EEVblog Electronics Community Forum
General => General Technical Chat => Topic started by: rfeecs on April 29, 2022, 04:32:41 pm
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I thought it might be good to start a new thread for Veritasium's new video, the follow up to the "Big Misconception" video:
https://www.youtube.com/watch?v=oI_X2cMHNe0 (https://www.youtube.com/watch?v=oI_X2cMHNe0)
He carries out his own experiment, much like AlphaPheonix.
He gives a lot of credit to Ben Watson and other YouTubers including EEVBLOG and Electroboom, etc.
I thought a new thread might be warranted, since the thread on the previous video wandered so off topic into pseudoscience trolling that it is probably being ignored. :rant:
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One quibble is that several times he says there is an electric field inside the wires but other times he says the electric field inside the wires is zero.
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One quibble is that several times he says there is an electric field inside the wires but other times he says the electric field inside the wires is zero.
Could you elaborate on this quibble? I thought he was pretty clear when its zero and when its not zero - electrostatic versus electrodynamic conditions. :)
PS
I thought this was a very nice follow up video - especially in addressing all the specific objections people had to his description of the original thought experiment. What a lot of work to explain how an antenna works. ;)
PPS
It's still too bad though that Heaviside's work with coaxial cables didn't have time to get mentioned - that's the most immediately practical application of Poynting Theory beyond it being a mere theoretical curiosity.
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One quibble is that several times he says there is an electric field inside the wires but other times he says the electric field inside the wires is zero.
Could you elaborate on this quibble? I thought he was pretty clear when its zero and when its not zero - electrostatic versus electrodynamic conditions. :)
One point that sounded odd to me is at 8:00 he talks about a python simulation of a DC circuit. He talks about there is an electric field in the center of the wire. Maybe I misunderstood what he is trying to say.
At 11:14 he says at the instant that the switch closes, the electric field inside the conductor is no longer zero. So I guess this would be the electrodynamic condition? OK, but this condition only exists for attoseconds until the charges move and the field disappears.
We typically assume the boundary conditions for a perfect conductor include: the tangential electric field at the surface is zero, and the electric field inside the conductor is also zero. This applies both for static and dynamic conditions. But I suppose this is only an approximation and neglects the attoseconds long period when charges re-arrange themselves.
Then again, in both cases he may be assuming the wire is not a perfect conductor and has some resistance.
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What pushes the charges inside the wire if there is no electric field inside the wire and the electrons don't push themselves.
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What pushes the charges inside the wire if there is no electric field inside the wire and the electrons don't push themselves.
For DC, we could use conservation of charge (KCL) to say that if there is a current at one point in the wire you have the same current everywhere along the wire, no net electric field required.
Diffusion is another mechanism that can create current locally. In a forward biased diode, charges are moving against the electric field in some places.
Not a very satisfying answer, I know.
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What he observes during initial transient has to do with charging the energy storage device (that is the transmission line in this case).
So the small current he can see trough the lamp is the current used to charge the transmission line.
Energy will be transferred trough the wire.
He makes the same sort of mistake ignoring energy storage in his video about the faster than wind direct down wind vehicle.
I think in schools people should learn that in real life you can not get rid of energy storage same as you can not get rid of friction.
He should use a capacitor instead of a battery since that way he can see exactly how much energy was used by just measuring the voltage across the capacitor. Then he could compare that input energy with the energy delivered to the lamp.
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The biggest mistake in this video, basically amounts to denying "electrons push each other" -- well, we have mechanistic descriptions for marbles in a pipe, or water in a tube, fine. But then, surely, all that's missing is a suitable definition, to say how electrons might "push" each other, and what that means mechanistically? And, electrons have charge, so, they emit electric field, so, if it's field that pulls them along......... it's not any stretch at all to say electrons push each other along, is it?
And that's a pretty small, well -- I guess it depends how obvious this point is, versus how consistent you expect the video to be. It seems a shame to miss this one oversight, which seems pretty obvious for those in the know, and perhaps the casual viewer will even make this connection; but perhaps only the more astute will realize that this very thing is basically what the video says. So, I'm not sure if that's intentional or accidental.
In either case, that's a small aspect of the whole video, and honestly -- I'm excited that someone is taking the time, the chance -- to finally discuss things in terms of fields or transmission lines as a first principle. I feel this is something that could be taught much earlier, and mainly isn't because "we've always done it that way" which is always such a stupid reason to teach something.
And then, sure there are going to be rough edges, no one else has (apparently) taught it this way -- and so also you get the academic problem, or the Wikipedia problem, or the Stack Exchange problem, or anywhere really that you have some experts trying to teach newbies, but inevitably doing so under the strict scrutiny of other experts: the problem is, whatever you write, must be simplified to a sufficient extent that a beginner can at least start to grasp what's being discussed; but also without being so wildly oversimplified that it's wrong more than it's right, so that you end up spending all your time defending your explanation against other experts, rather than simply engaging with beginners! Psychology is a bitch like that... :palm:
And then, the other thing, is basically as I expected -- the exact setup was ill defined, and in the earlier thread I gave the plausible example of an LED and resistor, which will light perceptibly (if maybe not very well so in broad daylight) given the arrangement described. I don't know why, or that, the lightbulb was such a hangup for (some? many?) people, but this clarifies what was meant, and indeed such an example is spot on.
Tim
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Tim,
Not sure if your replay was directed to me.
Derek's main claim is that energy done not flow in wires and that is just wrong (there is no space for interpretation).
He just disregards the energy storage device (transmission line) with no mention if the first video and still no understanding of it in the second video.
Just imagine two identical large capacitors (air as dielectric or just vacuum). Say one of them is charged and the other is not then you parallel them (just get the plates in contact). Energy will be transferred from the charged capacitor to the discharged capacitor and if there are no losses total energy of the parallel capacitors will remain the same so equal to energy stored in the charged capacitor.
You ca repeat the experiment and add a lamp (any sort) in series that way some of the energy during the transfer from one capacitor to the other will be wasted as emitted photons (ideal light bulb) or mostly heat and some photos.
All energy was flowing trough conductors/wires nothing else involved and once the energy is equally split between the two capacitors there is no more light (photons and or heat).
Energy was transferred from one capacitor to the other.
In Derek's experiment energy was transferred from battery to (transmission line) thus the current trough the lamp as lamp is between the series capacitors thus it will measure the energy transfer.
You can not just ignore the energy storage device that is the transmigration line. If you do so you get to the wrong conclusions.
If you could have a transmission line that had no capacitance (possible in a theoretical analysis) then there will be no current trough the lamp until the electron wave had time to travel trough the conductor/wire.
So what he sees before electron wave gets there is the charging of the energy storage that exists with a real world transmission line. Once that is charged up after initial transition all energy delivered to the lamp will flow in wires.
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Just imagine two identical large capacitors (air as dielectric or just vacuum). Say one of them is charged and the other is not then you parallel them (just get the plates in contact). Energy will be transferred from the charged capacitor to the discharged capacitor and if there are no losses total energy of the parallel capacitors will remain the same so equal to energy stored in the charged capacitor.
Um, you can't do that "if there are no losses" bit.
https://en.wikipedia.org/wiki/Two_capacitor_paradox
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Um, you can't do that "if there are no losses" bit.
https://en.wikipedia.org/wiki/Two_capacitor_paradox
I can do whatever I want :) and I sure can imagine a circuit with no losses.
There is no paradox so not quite sure why that wiki page even exists.
The energy in each capacitor will just be half of the energy stored in the charged capacitor at the beginning of the test that means the same energy when you add the energy in both capacitors.
"Thus the final energy Wf is equal to half of the initial energy Wi. Where did the other half of the initial energy go?"
The above quote from Wikipedia is just silly and written by someone that did not understood what he was writing.
The final energy is the same as the initial energy and can be tested even with real components (there will be a small deficit due to restive losses) but it will still be very close to the same energy.
Just get two identical capacitors with low self discharge charge one of them then parallel that with the discharged one and measure the voltage to get the final energy.
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Um, you can't do that "if there are no losses" bit.
https://en.wikipedia.org/wiki/Two_capacitor_paradox
I can do whatever I want :) and I sure can imagine a circuit with no losses.
There is no paradox so not quite sure why that wiki page even exists.
The energy in each capacitor will just be half of the energy stored in the charged capacitor at the beginning of the test that means the same energy when you add the energy in both capacitors.
"Thus the final energy Wf is equal to half of the initial energy Wi. Where did the other half of the initial energy go?"
The above quote from Wikipedia is just silly and written by someone that did not understood what he was writing.
The final energy is the same as the initial energy and can be tested even with real components (there will be a small deficit due to restive losses) but it will still be very close to the same energy.
Just get two identical capacitors with low self discharge charge one of them then parallel that with the discharged one and measure the voltage to get the final energy.
(Assuming that I am not missing something, and the joke isn't on me)
I don't like playing with high-value charged capacitors, but the math is strongly on my side.
Capacitance (F) is Volts (V) per Coulomb of charge (Q).
C = V / Q
Energy stored in a capacitor is
P = Q * V / 2
This can be proven with Calculus if you want. NOTE Voltage is not Energy - a high voltage spark from static does not have more energy than an arc welder even though the voltage much, much higher!
So a 1F capacitor charged to 1V holds 0.5 Joule of Energy.
An experimenter charges a 1F capacitor to 1V, and then connects a second 1F capacitor to the first one in a loss-less way.
They now have two 1F capacitors, each with half a Coulomb each, and because V = C * Q they each will measure at 0.5V.
The energy in each capacitor is now 0.5 * 0.5 / 2 = 0.125 J, and for both capacitors that gives 0.25J in total.
Half of the energy has disappeared (well, in reality it will be lost into the environment somehow, maybe with a loud POP or a sizzle followed with swearing).
The only way out of this is if you can magic up additional 0.42 Coulomb of charge out of nowhere to make each 1F capacitor measure at 0.71 V - because V = 0.71 and Q = 0.71 then P = V * Q = 0.5J. But that can't happen because charge is a physical quantity and is conserved..
EDIT: sed s/power/energy/g
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Your reasoning is wrong from the start:
1. C = q/V not V/q
2. P is power. It is not the amount of energy stored in the capacitor. The stored energy is E = 1/2CV^2.
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Your reasoning is wrong from the start:
1. C = q/V not V/q
2. P is power. It is not the amount of energy stored in the capacitor. The stored energy is E = 1/2CV^2.
3. The final voltage across each of the capacitors will be 1/sqrt(2) V.
Yes, shouldn't post after dinner. A half pint of cider wipes me out.
(1) Yeah, my bad. Was standing on my head at the time of writing.
(2) Both are the same thing - see http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html (http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html) if you want to double-check that
U = 1/2*Q^2/C = 1/2 * Q * V = 1/2 * C * V^2
(3) So each 1U capacitor will have V = 0.71, and as C = q/V and so q = 0.71? Where does the extra 0.21 Coulombs of charge come from for each capacitor?
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Just went out to test.... definitely halves the voltage, not x 0.71
https://www.youtube.com/watch?v=nKw2pUudlKA (https://www.youtube.com/watch?v=nKw2pUudlKA)
Not done under rigorous conditions - can't be bothered heating up the soldering iron. The caps are just bits and bobs from the junk draw with the same value and rating, so aren't perfectly matched, and so on.
EDIT: Updated video to have two meters.
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Indeed, just what does move the electrons in the wire if the electric field in that wire is zero? Now, one end of the wire is connected to the + battery terminal and the other end to the - battery terminal. He clearly shows an electric field from those battery terminals that surrounds the battery so a free electron or other charged particle would travel in that field from one battery terminal to the other (assume a vacuum).
But somehow there is NO E field in that small path from one battery terminal to the other where the wire happens to be? How does that work? Oh, he says a super conductor so there is no resistance. But wouldn't a vacuum also have zero resistance? If he can assume a super conductor, then I can assume a perfect vacuum, can't I? And charged particles are KNOWN to travel astronomical distances in the vacuum of space, for the most part without slowing down. So that perfect vacuum, with no resistance has a field around the battery, but that single, narrow path where the superconducting wire happens to be does not. Really?
When an electron leaves the - battery terminal and enters the wire, that end of the wire has acquired an excess negative charge, super conductor or not. And since nothing can travel faster then c, then it will take at least the length of the conductor divided by c for the new distribution of the electrons in the wire to reach the other end. And by comparison to the end with the extra electron, that other end will have a + charge. For a short period of time we have a wire with one end more + and the other end more -. And there just MUST be an electrical field in that wire, super conductor or not.
A STATIC, EXTERNAL E field may not be able to penetrate that super conductor, but when the extra charge in inside the super conductor then a field MUST exist until the speed limit of the universe (c) allows that imbalance to work itself out. I am not an expert on super conductors, so I do not know it this takes place at c or at some other velocity, but it can mot be completely instantaneous.
He shows some nice graphics that he says represent these fields. He even shows one graphic that seems to show an electric field inside the wire but apologizes for the arrows being on the outside. WHAT THE HECK????? Is he contradicting himself?
My next point is about when he shows the step nature of the current. He shows a small, but definite step in the Voltage or current when the switch is first closed and then the rise to the full values when the electrons have had time to travel around the wire path. He explains this by the capacitance between the wire close to the battery and the wire close to the light or resistor. He then explains how that capacitance charges as the current in the wire moves from the battery to the far ends of the wire where it has a U turn. This is probably correct. It looks a lot like the model of a transmission line (twin lead, coaxial cable, etc.). He says that this produces a lit lamp when the field has had time to propagate from the battery to the lamp at the speed of light. BUT, the distance involved, which is one meter or less, is far too short to allow that speed to be measured with any accuracy. Especially with the instruments he is using. So neither he nor we really know if this is at c or at some lower velocity. But forget that. There is another problem here and it is in exactly what he says to explain this action: the capacitance between the wires.
If I were to connect a couple of capacitors from the wires near the battery/switch and the same wires near the lamp, and made them large value capacitors, like 1F or bigger, not a single one of us would be at all surprised if that lamp lit up at full brilliance for a period of seconds after the switch was closed. As the capacitors charge to the battery Voltage we expect a momentary current to flow. It is flowing through the capacitors in the same manner that virtually every capacitor in every electric and electronic circuit in the world or universe conducts a changing Voltage or current. So YES, I must admit that there is an E field between the plates of the capacitors, but the current is also flowing IN THE WIRES that connect these capacitors to our circuit. And the electrical energy is flowing in those wires. Just try this experiment with even a microscopic opening in those wires. That current flow will be greatly reduced in proportion to the capacitance of that gap in the wires to the gap in the large value capacitors.
I say that there is no essential difference between the capacitance in my capacitors and the capacitance between the wires. The difference is in the amount of current that each size of capacitance allows to flow.
And if you replace the simple wire with shielded coaxial cable, then there is NO field outside of that shield. The conductors near the battery/switch will no longer be able to extend their field to the conductors near the lamp. The small step he sees on his scope will no longer be there and the speed of propagation IN the coaxial cable will be entirely what determines when the lamp will light. If the switch is a single pole one at only one of the battery terminals, then the entire loop of wire will begin at either + or - battery Voltage. And when the switch is closed, then the impulse from that battery terminal will travel down the coax at the speed of propagation in that particular type of coaxial cable (check out the cable manufacturer's specs. as they can and do differ) and the light will come on only when that time period has passed EVEN IF THE LAMP IS ONLY MILLIMETERS AWAY FROM THE BATTERY/SWITCH.
He will probably argue that this distorts the E field by confining it between the outer and inner conductors of the coax and makes it longer. I guess that is so. But his small step argues against the full current/power being transmitted by the E field. That E field does not suddenly get stronger when 1/c seconds have passed. In fact, it is exactly the same as it was JUST prior to that time period passing. The E field can only transmit a small fraction of the current/power. The small size of his initial step shows that. For the full power to be transmitted the current/power must travel THROUGH THE WIRE.
And this is EXACTLY what his experiment is actually showing.
A further experiment could be done with two types of wire: standard wire and coaxial cable. If you were to establish the initial lighting of the lamp with the standard wire and the lamp comes to full brightness after 1/c, then the E field would be completely established. Then if you simultaneously switch both ends of the wires to substitute the coaxial cable for the standard wire, although the E field is still collapsing, the lamp would immediately go out and not light up again until the current from the battery has had another 1/c time period to travel through the coaxial cable. If the initial E field was transmitting the current/power, then the bulb would fade out as the field collapses. But no, it immediately goes off because the wire and the current IN the wire has been disconnected. That collapsing E field will not, can not keep it on.
What pushes the charges inside the wire if there is no electric field inside the wire and the electrons don't push themselves.
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Remember Ohm's Law, in the physics description: J = s E,
where J (vector) is the current density (A/m2)
E (vector) is the electric field or voltage gradient (V/m)
and s (usually a lower-case sigma) is the conductivity ( \$\Omega\$m)-1.
In an isotropic conductor (like a copper wire), s is a scalar, but in an anisotropic conductor (like crystalline graphite) it can be a tensor.
PS: Just as with a normal resistor in circuit, you can treat this as a voltage gradient (field) caused by a current flow, or a current flow caused by a voltage gradient (field) along the length of the conductor.
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I can do whatever I want
We've noticed.
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Just went out to test.... definitely halves the voltage, not x 0.71
Not done under rigorous conditions - can't be bothered heating up the soldering iron. The caps are just bits and bobs from the junk draw with the same value and rating, so aren't perfectly matched, and so on.
Your test is correct. The interpretation of the test is incorrect.
Yes you will have exactly half the voltage when you connect a charged capacitor to an identical discharged capacitor.
You need to consider this is a real experiment and you have ESR (a resistor in series with an ideal capacitor).
Thus when you connect the two you form a resistor divider and so when energy is transferred half of it is lost as heat on the ESR of the two capacitors.
Maybe repeat the test with two different capacitors then you will see that energy you end up with will no longer be half but more or less depending on witch capacitor is charged.
If you do not have a different value capacitor maybe you have a third capacitor of same time. Just parallel two capacitors charge those then connect a third capacitor in parallel (the one that is discharged) then you will see significantly more than half of the initial energy is still stored in the 3 capacitors and you can also do the experiment the other way around and see that significantly less than half of the energy will remain.
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Your test is correct. The interpretation of the test is incorrect.
Yes you will have exactly half the voltage when you connect a charged capacitor to an identical discharged capacitor.
You need to consider this is a real experiment and you have ESR (a resistor in series with an ideal capacitor).
Thus when you connect the two you form a resistor divider and so when energy is transferred half of it is lost as heat on the ESR of the two capacitors.
Maybe repeat the test with two different capacitors then you will see that energy you end up with will no longer be half but more or less depending on witch capacitor is charged.
If you do not have a different value capacitor maybe you have a third capacitor of same time. Just parallel two capacitors charge those then connect a third capacitor in parallel (the one that is discharged) then you will see significantly more than half of the initial energy is still stored in the 3 capacitors and you can also do the experiment the other way around and see that significantly less than half of the energy will remain.
Your reaction to the paradox betrays your lack of understanding --
The above is a non sequitur. In the limiting case, two capacitors at equal voltage, connected together, lose no energy. It's not about the absolute energy, it's about where it goes. The zero-full case is as illustrative as any other, and using other ratios or initial conditions just adds more busy work to the problem without making any change to the central fact.
Indeed, one needs only a single capacitor, and some reflection upon what it means to discharge into a short circuit. In this case, we shouldn't expect charge to be conserved (it's shorted out), but you would apparently expect energy to be conserved, while it is not.
The complete answer is this: when inductance is included, then the capacitor(s) resonate with the inductance. If the circuit is later broken precisely at a peak or valley (when I_L = 0), the initial condition is restored and terminal voltage is +/- initial. In this case, energy is conserved, but so is the voltage, and |charge| has not changed. (Or +/- whatever the difference is, for the multiple capacitor case.) There is no state in which the circuit can be stopped, that gives a voltage other than this, that does not result in energy loss somehow (into ESR, switch loss, or it's stored in the inductor).
The case in reality of course always includes resistance, even if it's radiation loss, or poorly understood loss mechanisms of superconductors (example, polished niobium resonators at 2K and 500MHz have a very large Q factor -- indeed somewhat better than quartz crystals, but still far from infinite). In this case, as t --> infty, the difference, the AC / transient component, will always die out, leaving the intermediate case that satisfies charge conservation and not energy conservation (because the energy always finds a way out).
Tim
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Your reaction to the paradox betrays your lack of understanding --
The above is a non sequitur. In the limiting case, two capacitors at equal voltage, connected together, lose no energy. It's not about the absolute energy, it's about where it goes. The zero-full case is as illustrative as any other, and using other ratios or initial conditions just adds more busy work to the problem without making any change to the central fact.
Indeed, one needs only a single capacitor, and some reflection upon what it means to discharge into a short circuit. In this case, we shouldn't expect charge to be conserved (it's shorted out), but you would apparently expect energy to be conserved, while it is not.
The complete answer is this: when inductance is included, then the capacitor(s) resonate with the inductance. If the circuit is later broken precisely at a peak or valley (when I_L = 0), the initial condition is restored and terminal voltage is +/- initial. In this case, energy is conserved, but so is the voltage, and |charge| has not changed. (Or +/- whatever the difference is, for the multiple capacitor case.) There is no state in which the circuit can be stopped, that gives a voltage other than this, that does not result in energy loss somehow (into ESR, switch loss, or it's stored in the inductor).
The case in reality of course always includes resistance, even if it's radiation loss, or poorly understood loss mechanisms of superconductors (example, polished niobium resonators at 2K and 500MHz have a very large Q factor -- indeed somewhat better than quartz crystals, but still far from infinite). In this case, as t --> infty, the difference, the AC / transient component, will always die out, leaving the intermediate case that satisfies charge conservation and not energy conservation (because the energy always finds a way out).
Tim
Let me know what part you do not agree with.
a) Energy is transferred from the charged capacitor to the discharged capacitor.
b) Since they are real components they have ESR so during the transfer process exactly half of the energy is lost for two identical capacitors.
c) When capacitors are not equal more or less than half of the energy is lost depending on with of the capacitors is the the charged one at the beginning of the test.
You do not need to include the small amount of inductance (witch is also an energy storage device) to do a correct energy balance.
If you could eliminate the ESR (like using superconductors) then energy at the end of the experiment will be the same with the initial energy.
Assuming you agree with the above do you agree with the fact that energy transfer is done trough wires ?
If you do not agree please be specific.
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I'm happy that he did acknowledge the issue with units in answer C), which should have been 1 m/c.
Was this just unit laziness or an intentional bit of misdirection? Can't say. I'm inclined to believe laziness given the graphic which show the distances in kelvin mega (KM) rather than kilo meters (km).
A more universal answer for C) would have been d/c. Where d is the distance between the parallel wires.
By using 1/c, one might at first think it is the reciprocal of the speed of light, except for the meter term left unresolved.
If the distance were say 18.5 m, then answer C) would have been a more obvious choice if it were 18.5 m/c.
Still, I thought it was a good follow up video and appreciated his shout outs to many of the other response videos, each of which I'd seen and appreciated.
This is how we advance understanding of the art. Pose a thought and let others have a think and take pot shots at it. If the idea survives it has some merit, if not, well maybe one of the responses will be correct or trigger a new line of thought, and so on. :)
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If you consider this trilemma to be exhaustive, I have a bridge to sell you?
Those are not equivalent to what you claimed above... the confusion seems to be over whether the capacitors have half energy each, or half energy total. Perhaps that post was poorly worded/typoed, I don't know.
Charge conservation dictates the voltage go down by half (again, for the equal case; don't bring other ratios into it, it's beside the point), therefore the energy in each is 1/4 and the total energy is 1/2 initial. This energy is dissipated in the resistance, or is stored in the inductance half the time. In no case will the capacitors have sqrt(2)/2 voltage (half initial energy) in them, which would be a very peculiar ratio indeed to see in a discrete circuit of such simple values.
Tim
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If you consider this trilemma to be exhaustive, I have a bridge to sell you?
Those are not equivalent to what you claimed above... the confusion seems to be over whether the capacitors have half energy each, or half energy total. Perhaps that post was poorly worded/typoed, I don't know.
Charge conservation dictates the voltage go down by half (again, for the equal case; don't bring other ratios into it, it's beside the point), therefore the energy in each is 1/4 and the total energy is 1/2 initial. This energy is dissipated in the resistance, or is stored in the inductance half the time. In no case will the capacitors have sqrt(2)/2 voltage (half initial energy) in them, which would be a very peculiar ratio indeed to see in a discrete circuit of such simple values.
Tim
They will have half of energy each if you do not have ESR and capacitors have the same capacity.
If capacity is different then energy stored in each will be proportional (still no ESR).
Only in the special case where you have identical capacitors with identical ESR that you end up with half the initial energy as the other half was lost as heat due to ESR.
That is why I mentioned that energy storage should be a more important subject in school and be treated the same way as friction or restive losses.
Both mistakes done by Derek. This one about electricity and the one about faster than wind direct down wind vehicle are related to not understanding what energy storage is and conservation of energy.
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Then you will agree the final voltages will be sqrt(2)/2, not 0.5, or 1 (initial)?
Please provide waveforms showing this.
Tim
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Then you will agree the final voltages will be sqrt(2)/2, not 0.5, or 1 (initial)?
Please provide waveforms showing this.
Tim
Are you confusing voltage with energy ?
For an ideal pair of identical capacitors (no ESR) one charged and one discharged, when paralleled the energy will remain the same thus voltage will drop to sqrt(2)/2
For a real pair of identical capacitors (ESR different from zero), when you parallel them the total energy will be 0.5 as the other half was lost as heat in the ESR the voltage will also be half in this particular example but is not to be confused with energy that also just happened to be half due to the symmetry of the example.
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They will have half of energy each if you do not have ESR and capacitors have the same capacity.
No you won't. You need to take a long bath, relax and reflect on this. In particular how the resistance (in Ohms) between the two caps makes no difference in the outcome, only the time it takes to get there.
If capacity is different then energy stored in each will be proportional (still no ESR).
Only in the special case where you have identical capacitors with identical ESR that you end up with half the initial energy as the other half was lost as heat due to ESR.
With idea components (2 ideal capacitors and two ideal wires), there is only one place the excess energy can go - into the EM field, causing oscillation between all the energy being in one cap, and then being in the other, until the radiated EM waves removes half the energy from the system
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Personally, I think the inital question Veritasium posed was misleading and that is what caused all the controversy. He shouldn't have asked how long before the light bulb lit up because for a practical engineer that is when you have adequate load current across the bulb.
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No you won't. You need to take a long bath, relax and reflect on this. In particular how the resistance (in Ohms) between the two caps makes no difference in the outcome, only the time it takes to get there.
Take a piece of paper and draw a diagram showing two ideal capacitors with a series resistance. Then make the calculations and see what happens.
If you have 1Ws stored in the one capacitor then you parallel that with another identical capacitor that has no energy stored the energy will be split in half so each capacitor will have 0.5Ws total in both will be 1Ws as there is no loss.
Real capacitors have ESR thus half of the energy will be lost as heat thus in that case at the end each capacitor will have just 0.25Ws total 0.5Ws for both and the other 0.5Ws will have ended as heat.
With idea components (2 ideal capacitors and two ideal wires), there is only one place the excess energy can go - into the EM field, causing oscillation between all the energy being in one cap, and then being in the other, until the radiated EM waves removes half the energy from the system
What excess energy ? In ideal case where you parallel two capacitors that have no ESR there will be no lost energy. The energy will just be redistributed in the two capacitors.
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I suggest exactly the same, but you focus on charge (the actual conserved quantity), not energy.
Draw +s and -s on the plates it that helps.
Half the charge in each cap gives half the voltage, giving each capacitor a quarter of the energy of the original capacitor's charge.
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I suggest exactly the same, but you focus on charge (the actual conserved quantity), not energy.
Draw +s and -s on the plates it that helps.
Half the charge in each cap gives half the voltage, giving each capacitor a quarter of the energy of the original capacitor's charge.
This is where you are wrong. The important thing is energy as energy is what needs to be conserved.
If you claim that in an ideal capacitor case the energy is not conserved you need to explain where it was lost.
In real case half of the energy (half in case of two identical capacitors one fully charged and one fully discharged so just special symmetrical case) ends up as heat due to ESR.
Like I mentioned already do a test with different capacitors then try to explain the results. You will not be able to do so if you do not understand what happens.
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One point that sounded odd to me is at 8:00 he talks about a python simulation of a DC circuit. He talks about there is an electric field in the center of the wire. Maybe I misunderstood what he is trying to say.
The python simulation is here: http://tinyurl.com/SurfaceCharge (http://tinyurl.com/SurfaceCharge)
I had a look at it and I now realize that it is simulating a resistive wire, hence the electric field in the center of the wire is not zero.
I thought he was assuming a perfectly conducting wire, as he stated in the first video. Now I see in this video he is assuming a real wire with some resistance.
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I fully understand your position, but it is the wrong one. Energy is not conserved in an electric circuit.
Do the math as you approach ESR = 0. For all values of ESR except 0 half the energy is lost. At zero everything is vaporized by an infinitely short current pulse of infinite amps.
This debate is offering you an 'Aha!' moment. It is up to you if you take it.
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I fully understand your position, but it is the wrong one. Energy is not conserved in an electric circuit.
Do the math as you approach ESR = 0. For all values of ESR except 0 half the energy is lost.
This debate is offering you an 'Aha!' moment. It is up to you if you take it.
Energy is always conserved.
In the real case (as you seen from your own test) half of the energy you started with remains stored in the two capacitors and the other half ends up as heat.
So all energy is accounted for.
It is irrelevant what the ESR is as long as it is different from zero as energy will be lost and the smaller the ESR value the faster the charging takes place so high current for a short period or smaller current for a longer period if ESR value is higher.
If you disagree that energy (half in this symmetrical example) is lost in the ESR as heat then please let me know where that energy ended up ?
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I fully agree half the energy is lot from the circuit into the environment it one way or another. The caps are at half the voltage.
The disagreement I have is with those who say that the caps are at 0.71 of the initial voltage, and all the energy is still in the circuit. They are missing an insight.
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I fully agree half the energy is lot from the circuit into the environment it one way or another. The caps are at half the voltage.
The disagreement I have is with those who say that the caps are at 0.71 of the initial voltage, and all the energy is still in the circuit. They are missing an insight.
For an ideal capacitor setup that 0.71 of initial voltage will be correct as there will be no losses. And even in real world you can build capacitors with superconducting plates and they will also not have any loss.
But whatever you take that loss in to consideration real world or not the fact still is that a transmission line has capacitance.
The claim Derek makes as energy not traveling trough wires is wrong. That small current and energy transfer to the lamp is just because the transmission line capacitance is charging.
If you ignore that energy storage you get to the wrong conclusion that energy is not traveling trough wires and that is wrong.
Same problem with energy storage applies to the direct down wind faster than wind vehicle and there the concussion Derek makes as as outrageous basilica claiming (not directly) that system is an overunity device (more than a perpetuum mobile).
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It's still too bad though that Heaviside's work with coaxial cables didn't have time to get mentioned - that's the most immediately practical application of Poynting Theory beyond it being a mere theoretical curiosity.
How did he use the mere theoretical curiosity, and to prove what?
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I fully agree half the energy is lot from the circuit into the environment it one way or another. The caps are at half the voltage.
The disagreement I have is with those who say that the caps are at 0.71 of the initial voltage, and all the energy is still in the circuit. They are missing an insight.
For an ideal capacitor setup that 0.71 of initial voltage will be correct as there will be no losses.
Nope, if there are no losses it will oscillate forever. How could it not? there is nothing to dampen it down and the system starts in an unstable state.
There is no steady state where the paired capacitors have the same total energy and total charge as the original charged plus uncharged capacitors.
And even in real world you can build capacitors with superconducting plates and they will also not have any loss.
Are you implying that a superconducting system would not radiate EM waves? Even with 'ideal' circuit the energy is lost:
https://www.researchgate.net/publication/243492397_The_two-capacitor_problem_with_radiation (https://www.researchgate.net/publication/243492397_The_two-capacitor_problem_with_radiation)
By treating radiative effects in the simplest approximation, we show that the paradox is really nothing more than an inappropriately applied lumped-parameter model. In particular, we show that in the zero-resistance circuit, radiation fully accounts for all of the energy lost.
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He should leave this alone. Lol.
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Are you implying that a superconducting system would not radiate EM waves? Even with 'ideal' circuit the energy is lost:
By treating radiative effects in the simplest approximation, we show that the paradox is really nothing more than an inappropriately applied lumped-parameter model. In particular, we show that in the zero-resistance circuit, radiation fully accounts for all of the energy lost.
Quite some years ago experiments with super conductors where performed where a current flow was induced in a super conductor ring and that current was there with no losses.
Of course each time you make a measurement you will influence the current conditions but you account for those.
Instructors are also energy storage devices (maybe harder to understand but still energy storage devices).
After you charge a capacitor there is no loss other than the small leakage current inside the capacitor.
Same after you charge an inductor and there is a constant current flowing trough it all the losses are restive and same as with capacitor there are some small leakage losses.
When you disconnect the supply from a capacitor the stored energy will be provided to the load.
Same as when you reduce the current the field will collapse providing that stored energy to the circuit.
Any capacitor will have some small amounts of distributed inductance and the same is true for an inductor that will have small amounts of distributed capacitance.
Both the inductor and capacitor are energy storage devices.
If Derek did not closed the circuit at the ends when closing the switch he will still have seen that small amount of transient current that is due to energy storage being charged but nothing after that.
Energy is transferred trough the wires both when circuit is open at the ends and when it is closed.
When it is open only the amount of energy needed to charge the transmission line will flow (trough wires) and when circuit is closed energy will continue to flow trough wires.
His main claim is that energy is not delivered to lamp (load) trough wires and that is just not the case.
Is also easy to demonstrate that the energy transfer in real world is not 100% efficient thus you will need to see heat generate in the medium that transfers the energy. A thermal camera can easily show that the wire is where that loss happens as it will heat up and not the medium around the wire.
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Interestingly, superconductors don't seem to be entirely perfect at DC even, but certainly aren't at AC.
A simple demonstration is thus:
When a chunk of YCBO is cooled below Tc, why doesn't it become suddenly a perfect mirror?
Indeed it remains black, an extremely lossy surface at optical frequencies!
It stands to reason that, somewhere between DC and light (literally!), there is a point of increasing losses.
As it turns out, this point falls particularly in the deep IR to THz range, corresponding to the binding energy of Cooper pairs in the material (so, on par with thermal energy, and thus falling in the thermal to cryo IR range of the spectrum). For frequencies below this cutoff, some degree of superconducting behavior is expected, and above this, none.
(Indeed, the population of Cooper pairs is somewhat limited, and they can be momentarily broken by a bright flash, at least in films where the penetration depth of light is sufficient to do so. Thus, superconductivity can be optically switched.)
It additionally happens that, for type 2 superconductors (like YCBO), the AC losses extend all the way down (as a limiting case) to DC, in a sense: for less than some critical field strength, it remains superconducting, but above it, some flux is permitted through the material (violating the Meissner effect, at least in bulk; presumably, local domains remain free of internal field, and this occurs at defect sites?). An effect known as flux pinning. It's hysteresis loss, in very much the same way that magnetic materials exhibit hysteresis loss, a predominantly AC effect but which extends down to DC in the limit.
Type 1 superconductors are generally quite good quality at modest AC frequencies, but still have nonzero losses. As I mentioned earlier, Nb resonators have a Q factor in the 10^7 range -- quite high, but still far from infinite.
So, even given superconductors, there is no lossless condition.
But in any case, this is another distraction -- if the claimed energy conservation exists, then we must be able to measure it for short time scales, before the dissipative time constant has elapsed. This is true whether a truly ideal situation exists (the TC is simply infinity, so the measurement can be made at any time) or with decidedly nonideal real capacitors (which might have a short TC, less even than the LC resonance period, so we shall select parts to ensure this is not the case, and a measurement can be made before significant dissipation has occurred).
Then, you can contrive a circuit where such a waveform will be present, at least transiently, no? Illustrating the claimed sqrt(2)/2 voltage ratio, that is?
Even if the system is decidedly nonideal, the claim remains strong and testable: for equal capacitors and 0/nominal charge, the measured (DC or cycle-averaged) voltage must be somewhat greater than the predicted 0.5, and no more than the asserted sqrt(2)/2. Even if we measure a value within this range, we have proven that something other than the charge-conservation prediction has occurred.
This is one of the best cases science has to offer -- a clear and concise claim, with an easily measurable and easily discriminated result. I'm a bit excited to see the results, honestly. :-+
Tim
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Superconducting magnets in MRI machines run in "persistent" mode. There is a small section of superconducting wire that is allowed to heat up above the transition temperature, and an external current supply connected there ramps up a current flowing in the rest of the coil, since the (low, but finite) resistance of the "hot" section allows the current to flow around the remainder. When the desired current is reached, the small section is allowed to cool back to below the transition point, and the current "chases its tail" around the shorted coil. It decays very, very slowly, presumably due to imperfections in the weld where the coil winding was completed.
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I feel that we are getting away from the main topic.
Claim made by Derek is that energy is not flowing trough wire.
His proof seems to be that some current flows trough the lamp/load before the electron wave had time to go around the long loop.
He seems to ignore that the transmission line has capacitance and that in order to charge that current will flow.
It is irrelevant if capacitor/transmission line is real so it has ESR or ideal. The difference is just the extra losses while charging.
I think the power supply he used was around 20V so on that 1.1kOhm resistor (ignoring the small resistance of the transmission line) 20V/1100Ohm = 0.01818A * 20V = 0.36W vs the power delivered during transmission line charging 4V/1100Ohm = 0.0036A * 4V = 0.0145W
That is about 25x less energy delivered compared to the point when electron wave that travels trough the wire gets to the Load. And there is a fairly sharp transition not a gradual one. All of this results are against his claim of energy being delivered outside the wire.
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@electrodacus I 100% agree with above. He asked Question A and when he got it worng answered Question B. Without the wires and a closed circuit you can't deliver adequate power (current) to light up that bulb. He back tracked in the second video to show a low powered LED to get out of jail. The animations and everything showed an incandescent bulb and it was obvious viewer will assume delivery of reasonable rated load current or at least 80% plus thereof. If he had asked the question how long before there would be a field effect on the conductor 1m apart I would have answered the question same way. With respect to the original question, Veritasium is still worng and in my mind he has lost so much respect for not being able to admit it. Now that I learned from eevblog live stream this morning he consulted Dave and others who are in the YouTube egnineering space I am also dissapointed he wasn't told whats what. The last nail in the coffin for him is that even in the small scale model experiment he showed, he didn't include a light but rather a resistor because he know as well as we do that the light bulb won't light up in the time he claimed.
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@electrodacus I 100% agree with above. He asked question A and when he got it worng answered Question B. Without the wires and a closed circuit you can't deliver adequate power (current) to light up that bulb. He back tracked in the second video to show a low powered LED to get out of jail. The animations and everything showed a incandescent bulb and it was obvious viewer will assume delivery of rated load current or at least 80% plus thereof. If he had asked the question how long before there would be a field effect on the conductor 1m apart I would have whol heartdly answered the question same way. With respect to the original question, Veritasium is still worng and my my mind he has lost so much respect for not being able to admit it.
I do not think he will not admit if he understood the problem.
I think he just does not understand what energy is.
I had a fairly long email discussion with him about the faster than wind direct down wind vehicle and I was not able to convince him that he was wrong and it was basically the same problem of understanding what energy is.
The LED light was to show that there will be some visible light at that power level. But he may have intentionally selected a fairly thick pipe in order for the transmission line capacitance to be high.
He could have saved a lot of cost and just use some thin enamel wire but then the line capacitance will have been so low that he will not been able to even distinguish that signal from the noise.
He can still get the same with the ends of the transmission lines open but then it will have been just a pulse of a few nano seconds and nothing after that as energy is not flowing outside the wire (the main point he tries to make).
With the open ends his circuit will be a a loop made of a battery, a switch, a capacitor, load/lamp, and another capacitor closing the circuit.
The load energy transfer is then a byproduct of the two capacitors being charged and energy available to load/lamp is limited to a fraction of a second only as long as it is needed to charge the capacitors.
With the ends closed the exact same thing happens just that energy is properly and indefinitely delivered to load/lamp after the electron wave gets there.
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I thought a new thread might be warranted, since the thread on the previous video wandered so off topic into pseudoscience trolling that it is probably being ignored. :rant:
I mostly stopped reading the orignal thread.
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Now that I learned from eevblog live stream this morning he consulted Dave and others who are in the YouTube egnineering space I am also dissapointed he wasn't told whats what.
I told him that no matter what he did he wouldn't please everyone or be able cover every aspect of this in the way they want. And he was very close to actually ditching all his work and not doing a follow up video because of this.
This video was shot and was supposed to have been released before xmas, and I suspect it might have been my talk with him a few days before xmas that made him delay it for all this time, as I was the last one he talked to I think, and he was quite on the edge about publishing a response.
He has an almost hour long video of us discussing it.
I think he did a very good job of showing what he wanted to show, which was the traditional physics Maxwell/Poynting approach. Whether or not that's others cup of tea, then well, everyone has a camera and can upload to youtube....
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In case someone wants to understand how supraconductors work, you have 2 sets of charges acting in parallel:
- electrons which flow with some resistance and are modeled with a resistor
- Cooper pairs, which can be accelerated at will without any loss, and are modeled with a perfect inductance.
It's true up to THz (and up to critical field), and the colder the material the more electrons turn into Cooper pairs.
So perfect conductor at DC, extremely good at RF (Q up to 10^10) not so good beyond.
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Derek's claim exact quote:
"Energy doesn't flow in wires"
Proceeds to show a DC circuit where the energy source (a battery) is connected with wires to the load (light bulb).
If "energy doesn't flow in wires" why he even needs the wires ?
He even shows in his second video examples of inductive chargers ???. These inductive chargers have nothing to do with his main example where wires are used to connect the source to the load.
Even during that initial transient the coupling is capacitive not inductive.
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Derek's claim exact quote:
"Energy doesn't flow in wires"
Proceeds to show a DC circuit where the energy source (a battery) is connected with wires to the load (light bulb).
If "energy doesn't flow in wires" why he even needs the wires ?
He even shows in his second video examples of inductive chargers ???. These inductive chargers have nothing to do with his main example where wires are used to connect the source to the load.
Even during that initial transient the coupling is capacitive not inductive.
So then how does the energy pass across the empty space between the switch and the lamp in the initial transient few nanoseconds?
How does it get from the bottom two wires to the top two wires?
"Capacitive coupling" doesn't equate to the energy only flows in wires.
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While the new video is a good effort and gets into a lot more details than the initial one, there still are points that are definitely not addressed.
Some of them come from - as was mentioned in the other thread early - a chicken-and-egg problem.
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So then how does the energy pass across the empty space between the switch and the lamp in the initial transient few nanoseconds?
How does it get from the bottom two wires to the top two wires?
"Capacitive coupling" doesn't equate to the energy only flows in wires.
Replace the battery (source) with a charged capacitor so just imagine a simple two plate capacitor that has excess of electrons on one plate and a deficit of electrons on the other.
Then have two other much smaller capacity capacitors that are empty (no stored energy) in series with a light bulb or resistor (same thing).
That group of two capacitors and a resistor all in series can be equivalent with a single capacitor so when you connect (using wires) that to the charged capacitor is like modifying the existing charged capacitor by increasing his capacity thus charges will be redistributed.
If you can imagine all that you will notice that there was no flow outside wires (this includes the capacitor electrodes).
You can also imagine just a charged capacitor on witch you increase the plate area. While you do so the energy stored will remain the same for ideal case while in real case part of it is lost due to plates having a non zero resistance so when electrons move to rearrange you will have some IR losses.
All energy travels trough the plates in this case so trough conductors/wires.
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While the new video is a good effort and gets into a lot more details than the initial one, there still are points that are definitely not addressed.
Some of them come from - as was mentioned in the other thread early - a chicken-and-egg problem.
There is no chicken-and-egg problem as there you ask witch one was first but here you start with either the chicken or the egg so that problem of what was first is non existent.
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So then how does the energy pass across the empty space between the switch and the lamp in the initial transient few nanoseconds?
How does it get from the bottom two wires to the top two wires?
"Capacitive coupling" doesn't equate to the energy only flows in wires.
... All energy travels trough the plates in this case so trough conductors/wires.
Apparently you mean current, not energy.
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Apparently you mean current, not energy.
If there is no current there is no energy.
There can be current and no energy flow if current flow is constant and it travels trough a superconductor loop.
I think that the charged capacitor where you change the size of the plates is the most simplified example I can think of that represents the initial few moments after the switch is closed.
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Apparently you mean current, not energy.
If there is no current there is no energy.
So there is no energy in electromagnetic waves?
There's no energy in a laser beam?
No energy in sunlight?
Then how does solar energy work? You seem to be in that business.
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While the new video is a good effort and gets into a lot more details than the initial one, there still are points that are definitely not addressed.
Some of them come from - as was mentioned in the other thread early - a chicken-and-egg problem.
Something like do moving charges cause fields or do fields cause charges to move?
Perhaps he addressed that in saying electrons don't push each other.
He comes down on the side of fields cause charges to move.
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I don't think he adequately debunks the idea that charges gain potential energy (V) and then that potential energy is converted to kinetic energy. The P=IV model.
He doesn't mention potential energy, just kinetic energy. He argues that the electrons drift too slowly to carry the energy.
But the (retarded) potential moves at the speed of light.
I imagine he didn't want to mention potential because it introduces another concept that might be complicated to explain.
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So there is no energy in electromagnetic waves?
There's no energy in a laser beam?
No energy in sunlight?
Then how does solar energy work? You seem to be in that business.
The resistor (lamp) is not powered by lasers or sunlight. Such a lamp powered by laser or sunlight will be a a mirror.
A capacitor and an inductor are energy storage devices.
Storing energy is not the same with using energy to do work (convert one type of energy in to another).
If the ends of that loop in Derek's video where open then whose long parallel wires will only be a capacitor so when switch will have been closed that capacitor will have charged the exact same way it happened with the closed loop tested.
The difference will have been that after those few ns of charging this capacitor there will be no current flow thus no voltage drop on that 1.1kOhm resistor and so no power available for the resistor/lamp thus no energy.
The capacitor will have remained charged so if you look at the balance of energy you will see this
Energy delivered from battery = IR loss on the internal battery ESR + IR loss in wires + IR loss in the 1.1kOhm resistor + energy stored in the capacitor made by the parallel wires.
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Energy delivered from battery = IR loss on the internal battery ESR + IR loss in wires + IR loss in the 1.1kOhm resistor + energy stored in the capacitor made by the parallel wires.
You are using a lumped model that is not adequate.
This circuit is three dimensional. There is a one meter gap between the switch and the load. You haven't explained how energy gets across that gap.
You can look at the simulation in the video. When the switch is closed a spherical wave propagates out in all directions. It should be very clear that some of that energy is going out away from the bulb. Some of the energy is lost. Your model doesn't account for the energy radiated away.
Consequently your equation is wrong.
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While the new video is a good effort and gets into a lot more details than the initial one, there still are points that are definitely not addressed.
Some of them come from - as was mentioned in the other thread early - a chicken-and-egg problem.
Something like do moving charges cause fields or do fields cause charges to move?
That's one of them, yes.
Perhaps he addressed that in saying electrons don't push each other.
He comes down on the side of fields cause charges to move.
So, if he said so, that must be settled? ;D
To extend the matter a bit, what about gravity?
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You are using a lumped model that is not adequate.
This circuit is three dimensional. There is a one meter gap between the switch and the load. You haven't explained how energy gets across that gap.
You can look at the simulation in the video. When the switch is closed a spherical wave propagates out in all directions. It should be very clear that some of that energy is going out away from the bulb. Some of the energy is lost. Your model doesn't account for the energy radiated away.
Consequently your equation is wrong.
There is no energy radiated away. This is a copy paste of a replay I just made to someone else but is applies to your question also.
Not sure how much you understand a battery so is best to replace that with a charged capacitor as it is simpler to understand than a battery.
----------------[RESISTOR]--------------------
-------------------{-CAP+}--s/ ------------------
The open loop above is just a charged capacitor "CAP" not connected to anything if the switch is open (ignoring the super small switch capacitance).
As soon as you close the switch "s/" you are paralleling the charged "CAP" with the two series capacitors formed by the lines on each side and those caps are in series with the resistor but that is not very relevant (it is just like having a wire there).
So what you have when the switch is closed is a closed loop made up of 3 capacitors in series. You can consider those two discharged capacitors in series as a single capacitor and then simplification will be a charged capacitor in parallel with a discharged capacitor.
_________I I__________
I I
I I
I I
I_________I I__________I
There is no current flowing trough the capacitor dielectric and yes an electric field will be formed there as the capacitor charges but that is due to the electrons moving from the charged capacitor. There will not be a field at the discharged capacitor before the electrons from the charged capacitor get there.
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There is no energy radiated away.
I disagree. |O
I wonder how you think radios work. By energy storage I suppose.
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I disagree. |O
I wonder how you think radios work. By energy storage I suppose.
Have you properly read my replay ?
A constant electric field will not be detectable by a radio.
A radio works by charging and discharging energy. But the current discussion is much simpler than that.
The circuit simplifies to charging an empty capacitor with a charged capacitor. If you disagree with the fact that this all that is two parallel capacitors (simplified) then let me know where do you think the omission is.
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I disagree. |O
I wonder how you think radios work. By energy storage I suppose.
Have you properly read my replay ?
A constant electric field will not be detectable by a radio.
A radio works by charging and discharging energy. But the current discussion is much simpler than that.
The circuit simplifies to charging an empty capacitor with a charged capacitor. If you disagree with the fact that this all that is two parallel capacitors (simplified) then let me know where do you think the omission is.
I have already done that. I have said there is radiation. You say there isn't. Enough said.
The simulation shows what happens.
If I wanted to model it, I would start where the battery/switch is connected to two wires. These wires form an antenna, or an odd looking transmission line. They can be thought of as a skinny bi-cone antenna. The infinite bi-cone looks like a transmission line and has constant impedance. It has spherical symmetry and the wave propagates out spherically. In this case, the wires are not conical but straight, but the propagation is approximately spherical and the impedance will change along the line but levels off to a slow increase in impedance.
When the switch closes, there is a transient voltage change, as in a Heaviside step function. This transient is what starts the energy propagating out in all directions, roughly spherically.
After a period of time, the wave front hits the top pair of wires. Now we have another antenna / transmission line. The electric field across the load will cause a current and voltage wave to propagate along this line in a similar manner as the source antenna. Clearly the signal is much smaller because the field has spread out spherically.
The two antennae are clearly coupled and form another set of transmission lines, the twin line that has been often mentioned. So to properly model this I would consider these to be coupled lines. The odd mode impedance of the twin line is well known. The even mode impedance will be formed by the bi-cone type lines.
Yes, this model is ridiculously more complicated that your capacitor model. But it can model the fact that the lower pair of wires initially have a higher current than the upper pair of wires and give the correct current for both wires.
Your model cannot do this.
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I disagree. |O
I wonder how you think radios work. By energy storage I suppose.
Have you properly read my replay ?
A constant electric field will not be detectable by a radio.
A radio works by charging and discharging energy. But the current discussion is much simpler than that.
The circuit simplifies to charging an empty capacitor with a charged capacitor. If you disagree with the fact that this all that is two parallel capacitors (simplified) then let me know where do you think the omission is.
Is what you are saying "If we pick a simplified model, it simplifies to this simple model, that doesn't explain the details of the physical results, but the model agrees with the model, so the model is good".
I'm wait for the "I'll add in ESR resistors and parasitic inductance to take my first-order approximation to a second order approximation, so we get virtual coils and resistors along with the virtual capacitors I've added because my model demands it, not because there are actual coils and capacitors in the experimental apparatus."
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I don't think he adequately debunks the idea that charges gain potential energy (V) and then that potential energy is converted to kinetic energy. The P=IV model.
He doesn't mention potential energy, just kinetic energy. He argues that the electrons drift too slowly to carry the energy.
But the (retarded) potential moves at the speed of light.
I imagine he didn't want to mention potential because it introduces another concept that might be complicated to explain.
Electron's potential energy is converted into heat in a resistor.
And he didn't debunk it because it is true.
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I don't think he adequately debunks the idea that charges gain potential energy (V) and then that potential energy is converted to kinetic energy. The P=IV model.
He doesn't mention potential energy, just kinetic energy. He argues that the electrons drift too slowly to carry the energy.
But the (retarded) potential moves at the speed of light.
I imagine he didn't want to mention potential because it introduces another concept that might be complicated to explain.
Electron's potential energy is converted into heat in a resistor.
And he didn't debunk it because it is true.
Yes, of course it's true. But his "misconception number 1" is "electrons carry energy from battery to bulb". He tries to prove this without mentioning potential energy which seems to be part of the most common explanation.
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I have already done that. I have said there is radiation. You say there isn't. Enough said.
The simulation shows what happens.
If I wanted to model it, I would start where the battery/switch is connected to two wires. These wires form an antenna, or an odd looking transmission line. They can be thought of as a skinny bi-cone antenna. The infinite bi-cone looks like a transmission line and has constant impedance. It has spherical symmetry and the wave propagates out spherically. In this case, the wires are not conical but straight, but the propagation is approximately spherical and the impedance will change along the line but levels off to a slow increase in impedance.
When the switch closes, there is a transient voltage change, as in a Heaviside step function. This transient is what starts the energy propagating out in all directions, roughly spherically.
After a period of time, the wave front hits the top pair of wires. Now we have another antenna / transmission line. The electric field across the load will cause a current and voltage wave to propagate along this line in a similar manner as the source antenna. Clearly the signal is much smaller because the field has spread out spherically.
The two antennae are clearly coupled and form another set of transmission lines, the twin line that has been often mentioned. So to properly model this I would consider these to be coupled lines. The odd mode impedance of the twin line is well known. The even mode impedance will be formed by the bi-cone type lines.
Yes, this model is ridiculously more complicated that your capacitor model. But it can model the fact that the lower pair of wires initially have a lower current than the upper pair of wires and give the correct current for both wires.
Your model cannot do this.
We are discussing about a field and you mention radiation. The infrared radiation for example from a lamp or resistor or from battery is not what transfers the energy from battery to lamp.
Do you agree that battery (or charged capacitor) is the only energy source in that example ?
If you do then just move the switch far from the battery and lamp/load then see what is the time required for the lamp to see any current.
You can not claim that energy comes from the switch if you agreed that battery/charged capacitor is the only source.
When the switch closes it just connects one of the plate of the charged capacitor to a discharged capacitor (transmission line).
There is no electric field before closing the switch and first electrons will move in to the wire or out of the wire (depending witch plate you connected) and just because of that you will start to have an electric field.
Here is the super simplified example
________________ common plate
+++++++
_______/ ________
-----------
This simplified example of Derek's experiment.
Those are real capacitors so the plates have both resistance and inductance. Capacitor on the left is the charged one while the one on the right is discharged.
What do you think happens when you close the switch ?
-
What do you think happens when you close the switch ?
:horse:
-
Here is the super simplified example
________________ common plate
+++++++
_______/ ________
-----------
This simplified example of Derek's experiment.
Those are real capacitors so the plates have both resistance and inductance. Capacitor on the left is the charged one while the one on the right is discharged.
What do you think happens when you close the switch ?
You say "There is no electric field before closing the switch". I don't think this is true. Here is the same diagram with two negative charges electrons sitting between the plates of your capacitor(s). What way do they move? and why do they move, and what provides the energy for them to move?
________________ common plate
+ + + + + + +
- -
_______/ ________
-----------
Also the positive charges are way too close together. Their mutual repulsion will cause them to spread out over the plate pretty much uniformly.
-
What do you think happens when you close the switch ?
:horse:
Here is the exact example if you do not like the simplification
________________________________________________
+++++++++++++++++++++ ___________________
1KOhm___________________
_____________________/ _________________________
-----------------------------------
Those are real capacitors so the plates have both resistance and inductance. Capacitor on the left is the charged the equivalent of the battery and the two capacitors on the right are representing the transmission line and they are in series with 1KOhm resistor. The ends are not closed as is not relevant for the first few ns.
What do you think happens when you close the switch ?
-
Here is the exact example if you do not like the simplification
Can you please upgrade to at least crayons and paper? Because this is what the rest of us are looking at:
-
You say "There is no electric field before closing the switch". I don't think this is true. Here is the same diagram with two negative charges electrons sitting between the plates of your capacitor(s). What way do they move? and why do they move, and what provides the energy for them to move?
________________ common plate
+ + + + + + +
- -
_______/ ________
-----------
Also the positive charges are way too close together. Their mutual repulsion will cause them to spread out over the plate pretty much uniformly.
Also the unconnected wire will have potentials at either end, as one end is closer to a large static charge.
That is incorrect and I noticed the same wrong explanation in Derek's video. The electrons and holes will be equal and they will extend just up to the switch even on the common plate.
Only after the switch is closed electrons and holes will move symmetrically on top and bottom plate.
Keep in mind the drawing is nowhere near to scale. There may be just 0.1mm between plates and plates may be a few meters long for the charged capacitor and then for the discharged capacitor that represents the transmission line the distance between plates may be order of magnitude higher but even if the same 0.1mm there charge will end at the switch both on top and bottom plates.
-
Here is the exact example if you do not like the simplification
Can you please upgrade to at least crayons and paper? Because this is what the rest of us are looking at:
Sorry I did not expect the forum will scale with monitor resolution (you likely watch this on your phone).
I made is much shorter hopefully it will fit.
-
You say "There is no electric field before closing the switch". I don't think this is true. Here is the same diagram with two negative charges electrons sitting between the plates of your capacitor(s). What way do they move? and why do they move, and what provides the energy for them to move?
________________ common plate
+ + + + + + +
- -
_______/ ________
-----------
Also the positive charges are way too close together. Their mutual repulsion will cause them to spread out over the plate pretty much uniformly.
Also the unconnected wire will have potentials at either end, as one end is closer to a large static charge.
That is incorrect and I noticed the same wrong explanation in Derek's video. The electrons and holes will be equal and they will extend just up to the switch even on the common plate.
Only after the switch is closed electrons and holes will move symmetrically on top and bottom plate.
Keep in mind the drawing is nowhere near to scale. There may be just 0.1mm between plates and plates may be a few meters long for the charged capacitor and then for the discharged capacitor that represents the transmission line the distance between plates may be order of magnitude higher but even if the same 0.1mm there charge will end at the switch both on top and bottom plates.
Lol - if that is the case, when you close the switch, what makes the electrons move across the switch? It seems you are saying that they are quite happy being all bunched up in a huddle on the left...
You never did say what happens to the two charges floating in the middle of the capacitor (assuming they are free to move if they have any force on them).
-
Lol - if that is the case, when you close the switch, what makes the electrons move across the switch? It seems you are saying that they are quite happy being all bunched up in a huddle on the left...
You never did say what happens to the two charges floating in the middle of the capacitor (assuming they are free to move if they have any force on them).
Yes that is exactly what I'm saying and I'm fairly sure that I'm correct. They will move across the switch when switch is closed and symmetrically on the top plate.
After the right side is charged they will be also uniformly distributed across the entire new capacitor that has now higher capacity.
This is assuming that distance between plates is the same on the left side and right side else if say on the right side the distance between plates is larger (lower capacity) there will be proportionally less electrons on that side.
-
Lol - if that is the case, when you close the switch, what makes the electrons move across the switch? It seems you are saying that they are quite happy being all bunched up in a huddle on the left...
You never did say what happens to the two charges floating in the middle of the capacitor (assuming they are free to move if they have any force on them).
Yes that is exactly what I'm saying and I'm fairly sure that I'm correct. They will move across the switch when switch is closed and symmetrically on the top plate.
After the right side is charged they will be also uniformly distributed across the entire new capacitor that has now higher capacity.
This is assuming that distance between plates is the same on the left side and right side else if say on the right side the distance between plates is larger (lower capacity) there will be proportionally less electrons on that side.
No, the charges floating in the middle between the plates - the extra '-' signs... I'll go back to your original diagram, so I'm not changing two things at once.
________________ common plate
+++++++
- - << These guys which way do the move and why.
_______/ ________
-----------
If they are free to move, which way to they go and why?
-
No, the charges floating in the middle between the plates - the extra '-' signs... I'll go back to your original diagram, so I'm not changing two things at once.
________________ common plate
+++++++
- - << These guys which way do the move and why.
_______/ ________
-----------
If they are free to move, which way to they go and why?
Not sure why you had to draw those ones as they are already on the diagram. And yes they are incorrectly drawn under the bottom plate (limitations of this CAD tool :) ).
The ++++ and ------ symbols in my diagram represents holes and electrons in the plates/wires as that is what is of interest in this discussion.
They will move when switch is closed and they will move symmetrically on the top and bottom plate's.
To maybe help you understand that there is symmetry between the charges on the top and bottom plate imagine there is also a switch on the top plate exactly above the one on the bottom and now thing what will happen if you start with both switches open and only close the top one.
Do you think there will be any current trough that closed top switch ? If not that means charges remain arranged the same as they where before you closed that switch.
-
No, the charges floating in the middle between the plates - the extra '-' signs... I'll go back to your original diagram, so I'm not changing two things at once.
________________ common plate
+++++++
- - << These guys which way do the move and why.
_______/ ________
-----------
If they are free to move, which way to they go and why?
Not sure why you had to draw those ones as they are already on the diagram. And yes they are incorrectly drawn under the bottom plate (limitations of this CAD tool :) ).
The ++++ and ------ symbols in my diagram represents holes and electrons in the plates/wires as that is what is of interest in this discussion.
They will move when switch is closed and they will move symmetrically on the top and bottom plate's.
To maybe help you understand that there is symmetry between the charges on the top and bottom plate imagine there is also a switch on the top plate exactly above the one on the bottom and now thing what will happen if you start with both switches open and only close the top one.
Do you think there will be any current trough that closed top switch ? If not that means charges remain arranged the same as they where before you closed that switch.
They will not move symmetrically, because there is no symmetry to begin with.
Electrostatics is relatively easy so I wrote a quick solver for distribution of 50 + charges and 50 - charges on 10m and 5m wires one 1m apart. (so a 10:1 aspect ratio).
You can make what you want of this, but the net horizontal force on each charge (other than those at the ends of the 'wires', where they are constrained) is zero.
-
They will not move symmetrically, because there is no symmetry to begin with.
Electrostatics is relatively easy so I wrote a quick solver for distribution of 50 + charges and 50 - charges on 10m and 5m wires one 1m apart. (so a 10:1 aspect ratio).
You can make what you want of this, but the net horizontal force on each charge (other than those at the ends of the 'wires', where they are constrained) is zero.
You do not call that symmetry ?
Do you understand that my drawings where not done at scale?
Make the same again with 5m and 10m plates but 1mm apart (is even much lower than 1mm for any typical capacitor).
Or if you want to be even more realistic make the first 5m that overlap 0.1mm apart and then the remaining 5m 0.5m from the center as that will be the transmission line.
Still I feel that you try to look at other aspects only not to talk about the main one. No electron movement no current.
When you close the bottom switch is when the electrons will start to move and you will have energy transfer from the source (charged capacitor) to the load in this case a discharged capacitor that is the equivalent of the transmission line (so yes it also have some inductance and resistance).
Edit: looking closer at your charge density distribution it is still no where close to accurate for those dimensions.
Edit 2: and just to be clear there are free electrons in any conductor so even a discharged capacitor will have free electrons in both plates the difference is that a discharged capacitor will have the exact same density of free electrons in both plates.
In a charged capacitor what you will represent is just the excess on one plate and deficit on the other plate and they will be symmetrical in all aspects.
The electrons and holes want to be as close as possible to each other.
-
Yes, one capacitor can charge another, and there is energy that can be taken from the system, one way or another (e.g. heating of components). Does it have any utility here? No, not that I can see.
In fact, I've lost track of exactly what you were trying to prove... I only joined in again as you were commenting that there was no electric field before the switch was closed, when there clearly one is present.
Can't be bothered. Enjoy whatever.
-
Yes, one capacitor can charge another, and there is energy that can be taken from the system, one way or another (e.g. heating of components). Does it have any utility here? No, not that I can see.
In fact, I've lost track of exactly what you were trying to prove... I only joined in again as you were commenting that there was no electric field before the switch was closed, when there clearly one is present.
Can't be bothered. Enjoy whatever.
There is electric field in the charged capacitor (the one representing the battery)
There is no electric field in the discharged capacitor.
Edit: A good visualization of the capacitor
https://www.youtube.com/watch?v=f_MZNsEqyQw (https://www.youtube.com/watch?v=f_MZNsEqyQw)
-
So I set this up with three meters, two caps and a switch on the bench.
You have the switch open. You short out all the caps so everything is at 0V.
+------------+
| |
---- 0V ---- 0V
---- ----
| |
Gnd +----/ ------+
You charge up the capacitor on the left hand side, Gnd to the switch side, +10V to the wire common between the two capacitors.
You now have 10v across the left hand side cap, 0V across the right hand side cap, and 10V over the switch terminals .
+------------+
| |
---- 10V ---- 0V
---- ----
| |
GND +----/ ------+
10V
Q1) How did that 10V measured over the switch get there :-//? My suggestion: as no charge has moved across the capacitor, no current is moving so it can't be magnetic. So unless I want to add a new fundamental field it must be electric field across the capacitor. This has caused charges to move around to balance out the electric field, until there is zero voltage gradient over the capacitor, resulting in 10V over the opened switch.
I've measured this all on the bench - there is 10V across the switch.
You close the switch a small spark is heard and the 10V of potential difference is now gone.
+------------+
| |
---- 5V ---- 5V
---- ----
| |
GND +------------+
0V
You now have 5V across both caps - and half the energy goes out of the system as heat.
What part am I missing?
-
Hontas Farmer is back still saying the Derek is both right and wrong acording to QFT/QED
https://www.youtube.com/watch?v=VsxXX5cGamY (https://www.youtube.com/watch?v=VsxXX5cGamY)
-
Dereks's video at 21:10 Re. Rick Hartley about fields is 100% correct for high speed PCB design. But that does NOT apply at DC, not at all, not even one tiny bit.
-
I posted this comment on Derek's video, slighly expanded:
Riddle me this, a new thought experiment: Let's go extreme and say you have a 100mm diameter copper conductor at pure steady state DC delivering a small amount of power to a pure resitive load, say 1W, but go as low as you want. No transients, no skin effect, no nothing, just pure steady state DC into a resistive load.
Is there NO energy WITHIN this comicly large wire? None? It's all on the OUTSIDE of the wire in the fields at DC? Really? REALLY?
The classicial field theory math might work at DC, but I just can't get over the feeling that it doesn't pass the sniff test at DC. I don't get The Vibe I get with AC and transients. Quantum Electrodynamics and probability theory in the electron fields within the wire better passes the sniff test at DC.
Can someone please convince me that there is no energy flow within this 100mm diameter wire at all, and that all the energy flows outside the wire at DC.
-
Hontas Farmer is back still saying the Derek is both right and wrong acording to QFT/QED
https://www.youtube.com/watch?v=VsxXX5cGamY (https://www.youtube.com/watch?v=VsxXX5cGamY)
Double-posting, Dave?
https://www.eevblog.com/forum/chat/veritasium-(yt)-the-big-misconception-about-electricity/msg4150432/#msg4150432 (https://www.eevblog.com/forum/chat/veritasium-(yt)-the-big-misconception-about-electricity/msg4150432/#msg4150432)
Anyway, I commented there and won't repeat it here.
The classicial field theory math might work at DC, but I just can't get over the feeling that it doesn't pass the sniff test at DC. I don't get The Vibe I get with AC and transients. Quantum Electrodynamics and probability theory in the electron fields within the wire better passes the sniff test at DC.
Think for a bit. If you are convinced that energy doesn't flow in the wires for AC (or RF for that matter), why would it for DC?. Doesn't it take more energy to push electrons to and fro than to let them go indefinitely in the same direction?
-
Double-posting, Dave?
https://www.eevblog.com/forum/chat/veritasium-(yt)-the-big-misconception-about-electricity/msg4150432/#msg4150432 (https://www.eevblog.com/forum/chat/veritasium-(yt)-the-big-misconception-about-electricity/msg4150432/#msg4150432)
Yes, because there are now two threads, and many people are not reading the other thread any more.
-
I posted this comment on Derek's video, slighly expanded:
Riddle me this, a new thought experiment: Let's go extreme and say you have a 100mm diameter copper conductor at pure steady state DC delivering a small amount of power to a pure resitive load, say 1W, but go as low as you want. No transients, no skin effect, no nothing, just pure steady state DC into a resistive load.
Is there NO energy WITHIN this comicly large wire? None? It's all on the OUTSIDE of the wire in the fields at DC? Really? REALLY?
The classicial field theory math might work at DC, but I just can't get over the feeling that it doesn't pass the sniff test at DC. I don't get The Vibe I get with AC and transients. Quantum Electrodynamics and probability theory in the electron fields within the wire better passes the sniff test at DC.
Can someone please convince me that there is no energy flow within this 100mm diameter wire at all, and that all the energy flows outside the wire at DC.
Easy:
1. The given problem is a dynamics problem. Therefore we're only concerned with AC behavior.
2. He does discuss field within a conductor (though not how it gets there -- skin effect bridges the "AC" and "DC" regimes, as it turns out). And we can solve for the energy density of that field. It is terribly small in comparison, but nonzero.
The fact that most of the system's energy is stored in the magnetic field around the wires, hints further that "energy flows outside the wires", but this is an interpretation, and one can take either point; they're equivalent, once everything's quiescent, i.e. you can solve from one given the other (and material properties, boundary conditions, all that).
That is, the E-field inside the wire, given its resistance, tells the current density, and the wire diameter give the total current, and the wire placement is given, so the magnetic field can be solved.
Tim
-
""The given problem is a dynamics problem. Therefore we're only concerned with AC behavior.""
Does the video make this clear. I missed it.
He does a poor job of separating the transient from the steady state conditions
-
Double-posting, Dave?
https://www.eevblog.com/forum/chat/veritasium-(yt)-the-big-misconception-about-electricity/msg4150432/#msg4150432 (https://www.eevblog.com/forum/chat/veritasium-(yt)-the-big-misconception-about-electricity/msg4150432/#msg4150432)
Yes, because there are now two threads, and many people are not reading the other thread any more.
So let me double post this from the other thread, because I have yet to find a satisfying answer:
Ok, let me bring up this argument I put forward a few dozen pages ago (I will simplify it even more):
I have a mass of 1 kg in position P at 0 meters over sea level.
I take this mass 1000 meters away to drop it from a cliff into a hole deep 10 meters.
The potential energy of the mass is converted into kinetic energy and then this is uses to generate heat. Let's say I 'generated' 1 joule of energy.
Has this energy traveled along the 1000 meters path?
What if I changed my mind and headed in a different direction and after 1000 meters I dropped the mass into a hole twice as deep?
Would the 2 joule energy have traveled instead?
Has energy ever traveled along the path?
How much?
1 joule? 2 joule? 100 joule? m c^2 joule?
-
Q1) How did that 10V measured over the switch get there :-//? My suggestion: as no charge has moved across the capacitor, no current is moving so it can't be magnetic. So unless I want to add a new fundamental field it must be electric field across the capacitor. This has caused charges to move around to balance out the electric field, until there is zero voltage gradient over the capacitor, resulting in 10V over the opened switch.
You closed the switch just not the one you are concentrating at. When you connected the multimeter you basically closed a switch that is in series with 1MOhm resistor so electrons will flow from the charged capacitor to the discharged one trough your multimeter.
I've measured this all on the bench - there is 10V across the switch.
You close the switch a small spark is heard and the 10V of potential difference is now gone.
+------------+
| |
---- 5V ---- 5V
---- ----
| |
GND +------------+
0V
You now have 5V across both caps - and half the energy goes out of the system as heat.
What part am I missing?
You are missing the ESR. The electron wave can move at the speed of light and how many of them can flow per unit of time depends on resistance in this case mostly the ESR and some of your wire resistance.
The reason you do not have 7.07V at the end of the experiment so same energy is because half of the energy was dissipated as heat in the ESR + wires.
If you could measure the temperature accurate enough you will see that capacitors and wires heated up by the exact amount representing half the original energy.
-
Hontas Farmer is back still saying the Derek is both right and wrong according to QFT/QED
She says QED is in fact simpler than classical field theory?
What if you need to actually calculate something for a macro system like this? Given the dimensions of the wires (diameter, spacing), the input waveform (step function), the load impedance, what is the current at the load and the source (vs time)?
Clearly you can calculate this with classical field theory. You can calculate the impedances from the dimensions. You can do an accurate simulation.
She says for this problem, using classical theory is even harder. You have to do all sorts of computer calculations. But for QED its just probability. Easy.
The fact is with classical theory you can actually get a numerical answer that you can use. Can you do that with QED? Or does the complexity immediately get out of hand for even a much simpler system, like a single molecule? Here's a video discussing the difficulty of that:
https://www.youtube.com/watch?v=55c9wkNmfn0 (https://www.youtube.com/watch?v=55c9wkNmfn0)
There are lots of models. Each has limitations. Use the right tool for the job.
-
So let me double post this from the other thread, because I have yet to find a satisfying answer:
Ok, let me bring up this argument I put forward a few dozen pages ago (I will simplify it even more):
I have a mass of 1 kg in position P at 0 meters over sea level.
I take this mass 1000 meters away to drop it from a cliff into a hole deep 10 meters.
The potential energy of the mass is converted into kinetic energy and then this is uses to generate heat. Let's say I 'generated' 1 joule of energy.
Has this energy traveled along the 1000 meters path?
What if I changed my mind and headed in a different direction and after 1000 meters I dropped the mass into a hole twice as deep?
Would the 2 joule energy have traveled instead?
Has energy ever traveled along the path?
How much?
1 joule? 2 joule? 100 joule? m c^2 joule?
Potential energy is just, what it is. It's a property of the mass's altitude (in this case, or velocity too if you want to count that), not some internal state that comes along for the ride. It's not like it's got a battery in it. :) However, if you include yourself in the system, you're a battery (of sorts), that's transporting energy.
Or if you have a definition to apply it to, like: "transporting energy" sounds like integral path length times potential energy along that path, in which case, you'd be transporting potential energy with respect to that 10m drop any time you're above its floor, and negative potential while below.
Tim
-
I posted this comment on Derek's video, slighly expanded:
Riddle me this, a new thought experiment: Let's go extreme and say you have a 100mm diameter copper conductor at pure steady state DC delivering a small amount of power to a pure resitive load, say 1W, but go as low as you want. No transients, no skin effect, no nothing, just pure steady state DC into a resistive load.
Is there NO energy WITHIN this comicly large wire? None? It's all on the OUTSIDE of the wire in the fields at DC? Really? REALLY?
The classicial field theory math might work at DC, but I just can't get over the feeling that it doesn't pass the sniff test at DC. I don't get The Vibe I get with AC and transients. Quantum Electrodynamics and probability theory in the electron fields within the wire better passes the sniff test at DC.
Can someone please convince me that there is no energy flow within this 100mm diameter wire at all, and that all the energy flows outside the wire at DC.
In both cases DC and AC/transients all energy transfer from source to load is trough wires not outside the wires.
His test setup can be reduced to one charged capacitor (as the source instead of the battery) and one discharged capacitor (simulating the transmission line with the ends open).
Due to the fact that capacitors have an ESR half of the energy will be lost during the energy transfer from one capacitor to another and that loss will be as heat in the wires and capacitor plates that are also wires.
There is no field in the empty capacitor and the field will start to form as electrons are transferred and an imbalance of electrons is formed between the two plates.
-
Can we lock the other thread or something? This discussion is super confusing to read when we've got entire lines of thought being double-posted in both threads. It's super confusing to follow who is responding to what. :(
"Let's see - is this the BIG thread talking about how Veritasium is right but actually wrong or the LITTLE thread talking about how Veritasium is wrong but actually right?" :o
-
I have a mass of 1 kg in position P at 0 meters over sea level.
I take this mass 1000 meters away to drop it from a cliff into a hole deep 10 meters.
The potential energy of the mass is converted into kinetic energy and then this is uses to generate heat. Let's say I 'generated' 1 joule of energy.
Has this energy traveled along the 1000 meters path?
If we take the Poynting style conservation of energy argument, we know that energy is created in the volume of space around where you 'generated' it. Energy was 'dissipated' in the volume of space around where you dropped it. So we can say energy flowed through the space between. But we can't say exactly the path that the energy took.
Edit:
However, if we decide that the potential energy is located where the mass of the rock is located (like we do by saying that the fields have energy), then energy flowed along with the rock. How much potential energy the rock contains is a problem without knowing the baseline zero potential energy of the system.
I hate analogies. Water analogy, rock analogy, whatever. :rant:
-
I have a mass of 1 kg in position P at 0 meters over sea level.
I take this mass 1000 meters away to drop it from a cliff into a hole deep 10 meters.
The potential energy of the mass is converted into kinetic energy and then this is uses to generate heat. Let's say I 'generated' 1 joule of energy.
Has this energy traveled along the 1000 meters path?
If we take the Poynting style conservation of energy argument, we know that energy is created in the volume of space around where you 'generated' it. Energy was 'dissipated' in the volume of space around where you dropped it. So we can say energy flowed through the space between. But we can't say exactly the path that the energy took.
Edit:
However, if we decide that the potential energy is located where the mass of the rock is located (like we do by saying that the fields have energy), then energy flowed along with the rock. How much potential energy the rock contains is a problem without knowing the baseline zero potential energy of the system.
I hate analogies. Water analogy, rock analogy, whatever. :rant:
Yes this sort of analogies are not very useful.
Still I will try to give an answer
Before you let that mass drop there is that 1J of stored potential energy.
When you let that drop the potential energy is converted gradually in to kinetic energy (not all potential energy will end up as kinetic energy since part will be lost as heat due to friction with air).
Say 0.9J ended up as kinetic energy and if you have a perfectly non elastic collision with ground (not possible in real world) then all that kinetic energy will end up as heat (it just reminded me of a funny you tube video where someone cooked some meat using a robot hand to slap it).
If collision is elastic then not all kinetic energy will end up as heat at initial contact just a part will end up as heat the other part will be stored in compressing the mass (assuming the mass is elastic only) and then that stored energy will be converted back to kinetic energy as the mass will be accelerated back up.
In perfect elastic collision and no friction with air the mass will get back up to 1000m and since there is no loss it will be a perpetuum mobile (not possible in real life).
Yes the energy has traveled along the path. You can see with a thermal camera how jut the air that got in contact with the mass heated up and there is no heat below the mass where the mas did not get to and also no heat on the ground until the mass gets there to deliver energy same as electron needs to get there to deliver energy and electron travels trough wires thus energy is delivered trough wire.
-
If "energy doesn't flow in wires" why he even needs the wires ?
Well, as I understand it, a medium is needed to hold charges, and if there are no charges, there's no field?
The question then is more about fields making charges move rather than charges moving creating fields. Or something.
-
I have a mass of 1 kg in position P at 0 meters over sea level.
I take this mass 1000 meters away to drop it from a cliff into a hole deep 10 meters.
The potential energy of the mass is converted into kinetic energy and then this is uses to generate heat. Let's say I 'generated' 1 joule of energy.
Has this energy traveled along the 1000 meters path?
If we take the Poynting style conservation of energy argument, we know that energy is created in the volume of space around where you 'generated' it. Energy was 'dissipated' in the volume of space around where you dropped it. So we can say energy flowed through the space between. But we can't say exactly the path that the energy took.
Edit:
However, if we decide that the potential energy is located where the mass of the rock is located (like we do by saying that the fields have energy), then energy flowed along with the rock. How much potential energy the rock contains is a problem without knowing the baseline zero potential energy of the system.
I hate analogies. Water analogy, rock analogy, whatever. :rant:
It was not meant to be an analogy.
I am considering the mechanical system only.
In more detail: one perfectly flat frictionless path goes from point A (the source) to point B (the load). We can put the weight on the path and with an infinitesimal push we make it travel D meters to point B.
Now, at point B we have a cusps that leads to lets' say 20 different holes of different depths, from 1 to 20 meters. Chance determines what the final path will be. But once the weight falls into one of the holes, all of its gravitational potential energy from height 0m to the depth of the hole gets converted into heat (to simplify things).
At the bottom of each hole there is a path (horizontal and perfectly frictionless) that leads back to the source.
At the source the weight is lifted by a machine to sea level and the cycle repeats.
Not an analogy, I repeat. It is a mechanical system.
Does the energy travel through the one forward path at sea level?
If so: HOW MUCH ENERGY does travel along the path?
Remember, I do not know which hole the weight will fall into until the weight falls into it.
If we cannot get an agreement on this mechanical system, how can we get an agreement on the electromagnetic system where the depth of the hole is determined by the charges themselves (by creating a surface distribution that obeys the constitutive relation in the wires and resistor)?
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If so: HOW MUCH ENERGY does travel along the path?
If we assume that potential energy is located in the rock, and arbitrarily say sea level is zero energy, then zero energy travels along the surface path, and negative energy travels back along the underground path.
Between the two paths (passing through a plane perpendicular to the paths), there is a net energy flow from A to B. The amount of energy is the energy converted to heat. So it depends on which hole it goes down.
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The fact that most of the system's energy is stored in the magnetic field around the wires, hints further that "energy flows outside the wires", but this is an interpretation, and one can take either point; they're equivalent, once everything's quiescent, i.e. you can solve from one given the other (and material properties, boundary conditions, all that).
One thing that people are not considering is how the hydraulic analogy for electricity--although it helps people to have an initial grasp of what is going on in the wires--is ingrained in the collective minds of engineers and hobbyists preventing them from really understanding the phenomenon at hand.
Derek briefly approaches that in the misconception section of his video.
One of the problems with analogies is that you have to have a deep understanding of model's (in this case hydraulic) system. Derek seems to have done his homework. He points out that unlike a hydraulic system (where the water from the pump to the load has high pressure and low velocity and the water in the return pipe has lower pressure and higher velocity), with an electric circuit, there's no difference in current density or drift speed for the electrons going in and coming out of the battery.
What is making the electrons drift is not some kind of pressure. It is just a portion of the electric field generated by the battery that does not contribute to the transfer of energy from the battery to the load. So electrons in the wire are not being pushed by each other like in a fluid. They're just parading in response to an external cause (the electric field) exactly like in the load.
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As I have posted elsewhere, inside an ohmic conductor (tautologically defined as one that obeys Ohm's Law), the physics equation (analogous to I = V/R) relates the current density vector J in A/m2 to the E-field or voltage gradient E in V/m by the conductivity s in (\$\Omega\$m)-1 .
For an isotropic conductor, s is a scalar, but for an anisotropic conductor (such as crystalline graphite), s is a tensor. (In textbooks, this is usually written as a lower-case sigma.)
J = s E
Just as with a resistor from Mouser, this can be considered as the voltage (gradient) produced by the current (density), or vice-versa.
Of course, if there be skin effect, E is a function of position within the conductor.
Non-ohmic conductors (PN diodes, vacuum diodes, etc.) have their own defining equations.
[corrected for typo about skin effect]
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One of the problems with analogies is that you have to have a deep understanding of model's (in this case hydraulic) system. Derek seems to have done his homework. He points out that unlike a hydraulic system (where the water from the pump to the load has high pressure and low velocity and the water in the return pipe has lower pressure and higher velocity),
Are you sure about that? :)
Evidently the return line is smaller diameter; which is fine, that works too. This perhaps highlights possible confusion over current density vs. total flow as well?
The most direct analogy for magnetic or electric fields in a hydraulic system, I think, would be the expansion or deflection of the pipes themselves? But this isn't so much a general effect as dependent on geometry, mounting etc. Much as mechanical deflection of electrical cables, it's a higher order effect you can all but ignore until very high levels. Neh.
But really, it comes down to wave mechanics, which should be no surprise. So the more unusual aspects, of inertial flow, acoustic waves, etc., will be similarly lost on those who simply aren't familiar with them. (But, at that level, also even worse, because Navier-Stokes is hella nonlinear. Maxwell's is largely linear in practice; that we should be so lucky as EEs!)
What is making the electrons drift is not some kind of pressure. It is just a portion of the electric field generated by the battery that does not contribute to the transfer of energy from the battery to the load. So electrons in the wire are not being pushed by each other like in a fluid. They're just parading in response to an external cause (the electric field) exactly like in the load.
But like I said at the top of this thread -- electrons can be considered to push each other, given a suitable definition of "push". It's hardly a stretch of physics to say electrons repel, and no one needs QED to describe that (really? really?..). This remains true in the charge-balanced condition of a conductor, it's just over a much shorter range (except for the slight remaining charge imbalance, evident at the conductors' surface, and the field in space between them), and the absence therefore gives us a complementary "pull" as well.
Tim
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I posted this comment on Derek's video, slighly expanded:
Riddle me this, a new thought experiment: Let's go extreme and say you have a 100mm diameter copper conductor at pure steady state DC delivering a small amount of power to a pure resitive load, say 1W, but go as low as you want. No transients, no skin effect, no nothing, just pure steady state DC into a resistive load.
Is there NO energy WITHIN this comicly large wire? None? It's all on the OUTSIDE of the wire in the fields at DC? Really? REALLY?
The classicial field theory math might work at DC, but I just can't get over the feeling that it doesn't pass the sniff test at DC. I don't get The Vibe I get with AC and transients. Quantum Electrodynamics and probability theory in the electron fields within the wire better passes the sniff test at DC.
Can someone please convince me that there is no energy flow within this 100mm diameter wire at all, and that all the energy flows outside the wire at DC.
There is no experimental evidence of it. Same for AC, or RF. So no, no one can convince you.
I have a mass of 1 kg in position P at 0 meters over sea level.
I take this mass 1000 meters away to drop it from a cliff into a hole deep 10 meters.
The potential energy of the mass is converted into kinetic energy and then this is uses to generate heat. Let's say I 'generated' 1 joule of energy.
Has this energy traveled along the 1000 meters path?
If we take the Poynting style conservation of energy argument, we know that energy is created in the volume of space around where you 'generated' it. Energy was 'dissipated' in the volume of space around where you dropped it. So we can say energy flowed through the space between. But we can't say exactly the path that the energy took.
Edit:
However, if we decide that the potential energy is located where the mass of the rock is located (like we do by saying that the fields have energy), then energy flowed along with the rock. How much potential energy the rock contains is a problem without knowing the baseline zero potential energy of the system.
I hate analogies. Water analogy, rock analogy, whatever. :rant:
It was not meant to be an analogy.
I am considering the mechanical system only.
In more detail: one perfectly flat frictionless path goes from point A (the source) to point B (the load). We can put the weight on the path and with an infinitesimal push we make it travel D meters to point B.
Now, at point B we have a cusps that leads to lets' say 20 different holes of different depths, from 1 to 20 meters. Chance determines what the final path will be. But once the weight falls into one of the holes, all of its gravitational potential energy from height 0m to the depth of the hole gets converted into heat (to simplify things).
At the bottom of each hole there is a path (horizontal and perfectly frictionless) that leads back to the source.
At the source the weight is lifted by a machine to sea level and the cycle repeats.
Not an analogy, I repeat. It is a mechanical system.
Does the energy travel through the one forward path at sea level?
If so: HOW MUCH ENERGY does travel along the path?
Remember, I do not know which hole the weight will fall into until the weight falls into it.
If we cannot get an agreement on this mechanical system, how can we get an agreement on the electromagnetic system where the depth of the hole is determined by the charges themselves (by creating a surface distribution that obeys the constitutive relation in the wires and resistor)?
The relativistic answer (convention) is that a momentum density p of matter corresponds to an energy flux of pc².
If you want general relativity then the answer is, well, complicated. https://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html
Classical physics is simple and says you have a potential energy of -GMm/r, you're not moving 0 energy, you're just converting: potential->kinetic->heat.
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Well, as I understand it, a medium is needed to hold charges, and if there are no charges, there's no field?
The question then is more about fields making charges move rather than charges moving creating fields. Or something.
In this example (Derek's setup) the charges are needed to create the field as the only electric field is inside the battery
There is no moving object in this experiment.
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Can we lock the other thread or something? This discussion is super confusing to read when we've got entire lines of thought being double-posted in both threads. It's super confusing to follow who is responding to what. :(
"Let's see - is this the BIG thread talking about how Veritasium is right but actually wrong or the LITTLE thread talking about how Veritasium is wrong but actually right?" :o
The other thread IMO was essentially hijacked by aetherist with all these alternative theories, it's why I stopped reading it.
No reason to lock the other thread, they'll just come here.
This thread seems to be a clean slate based on the new video as the OP mentioned.
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Why is it that, in the two capacitor paradox problem, the two capacitors are connected in parallel with each other and the switch connected in series as shown in the circuit below?
(https://upload.wikimedia.org/wikipedia/commons/8/86/Two_capacitor_paradox.svg)
Source: Wikipedia en.wikipedia.org/wiki/Two_capacitor_paradox
Why not have the capacitors connected in series with each other (back to back, of course) and the switch connected in parallel? After the switch is closed, the net voltage across the two capacitors will be zero since both capacitors would be equally charged and zero current would then be flowing.
I don't think it would make any difference to the initial and final charge/energy calculations.
Also, are the capacitors regarded as ideal voltage sources or ideal current sources? The fact that the capacitors lose/gain voltage and charge when the switch is closed suggests that they are (ideal) current sources and not (ideal) voltage sources.
\$I = C \frac{dV}{dt}\$ suggests that for a constant current the terminal voltage of a capacitor must either increase or decrease depending on the direction of current flow (assuming that capacitance is constant and NOT infinite).
In order to prevent the terminal voltage of the capacitor from decreasing the capacitance would need to be decreasing to maintain a constant voltage because voltage is energy per unit charge \$V = \frac{E}{Q}\$
If the charge on the capacitor is decreasing then \$\frac{dQ}{dt} < 0\$
Also, if the terminal voltage is decreasing then \$\frac{d{^2}E}{dQ{^2}} < 0\$
The output power is zero before the switch is closed, then peaks before both capacitors are charged and then returns to zero.
If the capacitors had infinite capacitance and were already charged to some voltage then the voltages of the capacitors would be constant across their terminals independent of flowing current. This is the definition of an ideal voltage source.
On the subject of mechanical analogies, there is a post in a related discussion thread:
https://www.eevblog.com/forum/projects/where-does-12-come-from-in-capacitor-energy-calculation/msg1671035/#msg1671035 (https://www.eevblog.com/forum/projects/where-does-12-come-from-in-capacitor-energy-calculation/msg1671035/#msg1671035)
Using this mechanical analogy, it's as though the two capacitors fuse together when the switch is closed as opposed to colliding elastically where momentum and energy would be conserved.
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Why is it that, in the two capacitor paradox problem, the two capacitors are connected in parallel with each other and the switch connected in series as shown in the circuit below?
(https://upload.wikimedia.org/wikipedia/commons/8/86/Two_capacitor_paradox.svg)
Source: Wikipedia en.wikipedia.org/wiki/Two_capacitor_paradox
Not sure why that Wiki page even exists as there is no paradox. If you understand what capacitors are and how they work you can know exactly what happens and why.
If that is made with superconductors so no resistance in wires or capacitor plates then energy will be conserved.
If those two capacitors are identical (same capacity) the voltage after switch is closed will be 0.707 * Vi
If the circuit is not made with superconductors meaning wires, capacitor plates or both have any resistance the voltage after switch is closed will be 0.5 * Vi
The reason final energy is half of the initial is because the other half was dissipated as heat due circuit resistance.
Due to resistance you have Vi / 2 in the moment switch is closed even before any energy starts to be transferred as you basically have a 1/2 resistor divider at that middle point on a symmetrical circuit.
The value of the resistance is irrelevant and will only influence the speed at witch the energy is transferred from charged capacitor to the discharged one still same half of the energy will be wasted as heat.
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If those two capacitors are identical (same capacity) the voltage after switch is closed will be 0.707 * Vi
That's not how physics works. You're implying that energy can be created out of nothing to buffer an energy deficit.
Using the water analogy, if you have two buckets of the same capacity to represent the two capacitors, then one bucket is filled and the other is empty. This is the initial condition. If you pour 70.7% of the water from the filled bucket into the empty bucket, then there will be 29.3% of the water remaing in the initially-filled bucket. How can there also be 70.7% water remaining in the initially-filled bucket when only 29.3% of the water remains in that bucket?
That is absurd and contradictory. That's the definition of a paradox.
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That's not how physics works. You're implying that energy can be created out of nothing to buffer an energy deficit.
Using the water analogy, if you have two buckets of the same capacity to represent the two capacitors, then one bucket is filled and the other is empty. This is the initial condition. If you pour 70.7% of the water from the filled bucket into the empty bucket, then there will be 29.3% of the water remaing in the initially-filled bucket. How can there also be 70.7% water remaining in the initially-filled bucket when only 29.3% of the water remains in that bucket?
That is absurd and contradictory. That's the definition of a paradox.
What extra energy are you seeing ?
There is no extra energy. Half of the energy stored in the charged capacitor is transferred to the identical capacity discharged capacitor and that will result in 0.707 * Vi.
Voltage is not energy.
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Well, as I understand it, a medium is needed to hold charges, and if there are no charges, there's no field?
The question then is more about fields making charges move rather than charges moving creating fields. Or something.
In this example (Derek's setup) the charges are needed to create the field as the only electric field is inside the battery
There is no moving object in this experiment.
I'm not sure you got what I meant, nor that I got what you meant. And nobody talked about moving objects here.
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I'm not sure you got what I meant, nor that I got what you meant. And nobody talked about moving objects here.
Maybe you understand magnetic field better. If you add a permanent magnet somewhere in Derek's setup. Will that magnetic field do any work if neither the magnet nor the wires are moving relative to each other?
Similarly in a setup like the one Derek made where none of the components move relative to each other from where will you have an electric field.
When electrons will start to move you will get both an electric field and also a magnetic field.
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That's not how physics works. You're implying that energy can be created out of nothing to buffer an energy deficit.
Using the water analogy, if you have two buckets of the same capacity to represent the two capacitors, then one bucket is filled and the other is empty. This is the initial condition. If you pour 70.7% of the water from the filled bucket into the empty bucket, then there will be 29.3% of the water remaing in the initially-filled bucket. How can there also be 70.7% water remaining in the initially-filled bucket when only 29.3% of the water remains in that bucket?
That is absurd and contradictory. That's the definition of a paradox.
What extra energy are you seeing ?
There is no extra energy. Half of the energy stored in the charged capacitor is transferred to the identical capacity discharged capacitor and that will result in 0.707 * Vi.
Voltage is not energy.
Really? This? Again? Charge is the conserved quantity. In a capacitor the voltage is proportional to the charge (due to the very definition of Capacitance) so in this case voltage is conserved too. Electrical energy does not have to be conserved in an electric circuit - it is often converted to some other form of non-electrical energy (heat. light, motion, radio waves and sometimes smoke).
You will never, never, never, never, never, never, never, ever get a stable state with 0.707 * Vi.
If you don't believe me take it up with this random internet guy from Princeton:
https://physics.princeton.edu/~mcdonald/examples/twocaps.pdf
"If there were no damping (dissipative) mechanism, the circuit would then oscillate forever"
"[if there is a damping (dissipative) mechanism] eventually a static charge distribution results, with charge Qi/2 and voltage Vi/2, on each capacitor."
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Really? This? Again? Charge is the conserved quantity. In a capacitor the voltage is proportional to the charge (due to the very definition of Capacitance) so in this case voltage is conserved too. Electrical energy does not have to be conserved in an electric circuit - it is often converted to some other form of non-electrical energy (heat. light, motion, radio waves and sometimes smoke).
You will never, never, never, never, never, never, never, ever get a stable state with 0.707 * Vi.
If you don't believe me take it up with this random internet guy from Princeton:
https://physics.princeton.edu/~mcdonald/examples/twocaps.pdf
"If there were no damping (dissipative) mechanism, the circuit would then oscillate forever"
"[if there is a damping (dissipative) mechanism] eventually a static charge distribution results, with charge Qi/2 and voltage Vi/2, on each capacitor."
Seems like understanding what energy is is a big problem in general. This leads to a lot of confusion.
Here I will provide you with a full example for the identical parallel capacitor problem.
(https://upload.wikimedia.org/wikipedia/commons/8/86/Two_capacitor_paradox.svg)
Vi = 3V
C = 1F
So energy contained in the charged capacitor = 0.5 * 1F * 3V2 = 4.5Ws (4.5J if you prefer).
In ideal case so no resistance in the circuit half of this energy will be transferred to the empty capacitor that is otherwise identical.
Now you can say you have a 2F equivalent capacitor that contains 4.5Ws (energy can not go anywhere else as there is no resistance in this ideal circuit).
or if you open the switch after current flow has stopped you have two 1F capacitors each containing 2.25Ws
Voltage on the now parallel capacitors in ideal case will be 3V * 0.707 = 2.121V
You can test that by calculating the energy stored in now the 2F equivalent capacitor
0.5 * 2F * 2.1212 = 4.5Ws
So energy is conserved because it is an ideal circuit with no losses.
Now for the real circuit will need to simplify things to keep them simple as a real capacitor will be exactly the same as a transmission line so capacitance inductance and resistance. We can ignore the inductance as that also is an energy storage device so there will be no losses in this circuit because of that or not significant anyway.
Say the capacitors have 1Ohm equivalent resistance and this includes the terminals and wires to keep things super simple.
I do not have a drawing but you can just imagine 1Ohm resistor in series with each capacitor.
When you close the switch the voltage at the switch will already be half of the charged capacitor voltage so 1.5V in this example that is thanks to the resistor voltage divider.
Should be simple to understand why half of the energy will be dissipated on this resistance in the circuit no matter what the resistor value is.
It is half only because the two capacitors are identical else it will be more or less than half the energy that will be lost during transfer.
With half the energy remaining in the capacitors after current has stopped flowing you have just 1.5V
0.5 * 2F * 1.5V2 = 2.25Ws
The other 2.25Ws where dissipated on the resistors as energy was flowing trough the conductors and they have a resistance.
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You can test that by calculating the energy stored in now the 2F equivalent capacitor
0.5 * 2F * 2.1212 = 4.5Ws
Yes, I do understand why in all cases half the energy is lost. The 'ideal' case is not realizable.
0.707 Vi is never a solution, even with ideal components. You can take a different path through the math of the system and end up with a different answer, which is 0.50 Vi. With ideal components the system is either in a constant state of change (oscillating) or cannot be solved to a consistent answer.
Edit: Sorry about the random edits
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Is the "ideal" system in a steady state (i.e. without oscillation?)
If so can you please tell me how much charge is in each of those ideal 1F capacitors, (or the ideal 2F combined one, if you like that better)? I am pretty sure the formula is Q = CV.
As we have no source of charges in this system (aside from the initial charged cap) if this is more than the initial charge, where these additional charges have appeared from? No charges are able to cross either capacitor, and we can't just magic up +s and -s out of nowhere.
And yes, I do understand why in all cases half the energy is lost. The 'ideal' case is not realizable, and your solution of 0.707 Vi is inconsistent, because the Lumped Element Model is just an approximation of reality.
Yes there is no oscillation.
You can get super close to ideal if you use an efficient DC-DC converter to transfer the energy from the charged capacitor to the discharged capacitor.
I think you did the experiment with the parallel capacitors (unless I remember wrong).
Get maybe two super capacitors as they contain more energy and it will be easier to find a DC-DC converter to work with those.
There may be some energy harvesting ultra low power DC-DC converters. The parallel capacitors is like using a linear regulator.
Please understand that energy conservation can not be broken and you do not care about the charge but about the energy. You have the impression that charge is linear but you need more energy to push the second electron in compared to the first one.
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You can test that by calculating the energy stored in now the 2F equivalent capacitor
0.5 * 2F * 2.1212 = 4.5Ws
Yes, I do understand why in all cases half the energy is lost. The 'ideal' case is not realizable.
0.707 Vi is never a solution, even with ideal components. You can take a different path through the math of the system and end up with a different answer, which is 0.50 Vi. With ideal components the system is either in a constant state of change (oscillating) or cannot be solved to a consistent answer.
Edit: Sorry about the random edits
You used the wrong math and likely also why you think the system will oscillate.
Just test with a DC-DC converter and you will see very close to ideal is possible. If the DC-DC converter was 100% efficient (so ideal) then you get 0.707 Vi but you do not need a DC-DC converter if there is no resistance to get the same result.
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Please understand that energy conservation can not be broken.
Total energy is conserved, but Electrical Energy isn't. If you drive a lightbulb, you convert electrical energy into heat and light. If you drive a speaker you get noise and heat. If you have a solar cell you can convert light into Electrical Energy. If you have a turbine you can convert gravitational potential energy of water into Volts and Amps.
You have the impression that charge is linear but you need more energy to push the second electron in compared to the first one.
That is exactly the point. The half of the charge that comes out of the charged capacitor has 3/4th of the energy in the capacitor.
What you are saying is we take half the energy (which is the 'last' 29% of the charge that took it from 0.707 Vi to Vi ) and put it in the other capacitor and they are now balanced! It isn't - because there is 71% of the charge in one capacitor and 29% of the charge in the second, causing a potential differences and potential differences cause more charge to flow. Once the switch is pushed the only stable state is with 50% of the charge in each capacitor, and by very nature of capacitance 0.5 Vi.
(And of course if you have an ultra efficiency DC-DC convertor to make the transfer then you might get close to 71%, but we don't. We have two ideal caps, some ideal wire and an ideal switch).
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hamster_nz
Your last repay is so ridiculously wrong that is not worth my time to replay.
Just test to move the energy with a DC-DC converter that is at least 80% efficient.
If you start as in my example with 3V by the time the charged capacitor drops to 2.121V you got out half the energy and even with a 80% efficient DC-DC converter you will have 2.25Ws * 0.8 = 1.8Ws so around 1.9V on the second capacitor.
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EEVblog wrote:
Dereks's video at 21:10 Re. Rick Hartley about fields is 100% correct for high speed PCB design. But that does NOT apply at DC, not at all, not even one tiny bit.
Try making a magnetic sensor using for instance a Hall-effect sensor on a PCB and run a PCB trace close to the sensor running a DC current. Without a proper DC return path the resulting magnetic field will affect the sensor.
So even at DC E and M fields may affect circuits and precaution have to be taken.
CJ
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Yes there is no oscillation.
Oooh, a new testable claim!
So you will have waveforms for us very shortly, no? The circuit is quite simple; you can use whatever RLC network between the two bulk caps as you see fit. It must be a linear circuit; it would seem unfair to the initial claim to allow such. After all, the initial claim is just two capacitors jammed together somehow, can't get more linear than that.
So this should be trivial to set up and demonstrate. You have the necessary equipment, no?
I asked for an experiment a few days ago, I assume you've just been busy and haven't gotten around to it. Surely it will take hardly any setup to do.
Looking forward to your results!
You can get super close to ideal if you use an efficient DC-DC converter to transfer the energy from the charged capacitor to the discharged capacitor.
Oh; tut tut tut -- it can of course be done with a converter, but a converter is a nonlinear element! Indeed an ideal converter has a negative (and hyperbolic at that) resistance input characteristic (for nonzero power flow, fixed load resistance/power), a great many things are possible with that, which are not possible in a linear system.
I trust you didn't misspeak earlier, that you mean a nonlinear element is in fact necessary to perform this experiment, right?
Tim
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Oh; tut tut tut -- it can of course be done with a converter, but a converter is a nonlinear element! Indeed an ideal converter has a negative (and hyperbolic at that) resistance input characteristic (for nonzero power flow, fixed load resistance/power), a great many things are possible with that, which are not possible in a linear system.
I trust you didn't misspeak earlier, that you mean a nonlinear element is in fact necessary to perform this experiment, right?
Tim
What do you mean by "it can of course be done with a DC-DC converter" ?
Do you agree that you can transfer much more of the energy from the charged capacitor to the discharged capacitor and thus final voltage is higher than half and very close to ideal (0.707 * Vi) ?
I will attempt to make an analogy so you can visualize what happens. As any analogy it will have some limitations.
Imagine two identical barrels one full with water and the other empty and you want to split the water between the two barrels.
You have two choices:
1) connect a permeable hose between the two barrels but the hose will leak exactly half of the water (resistor) , if superconductor then no leak.
2) connect a flexible/stretchy but also permeable hose (inductor) and a valve (transistor) that you input some small amount of energy to control (energy from the water not external).
If you leave the valve open (allowing water to flow) then there is no leakage for the volume of water that stretches the hose but when the house is stressed at max based on water pressure (voltage) it will work exactly like a resistor so half the water is lost trough leakage.
Then before the hose is fully stretchered out you close the valve and allow the water in the stretched pipe to be pushed in the empty barrel and that way less water is lost trough leakage (ending as heat).
With choice 2 the DC-DC converter you just use an intermediary storage to transfer the energy from one place to the other in a much more efficient way.
The DC-DC converter will get close enough to ideal / superconductor case where there is no resistance.
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But that's an active circuit. You didn't say anything about control before. Did you forget to mention control before? Then who's controlling the superconductor?
Tim
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But that's an active circuit. You didn't say anything about control before. Did you forget to mention control before? Then who's controlling the superconductor?
Tim
You understand that a DC-DC converter has no stored energy. So he will be supplied from the existing energy in that stored capacitor.
Even if this active circuit uses some of the energy that we want to transfer it is way more efficient than just paralleling the two capacitors directly.
I do not get your confusion.
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Shall I quote your numerous claims that capacitors can just be "paralleled directly"?
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Shall I quote your numerous claims that capacitors can just be "paralleled directly"?
Of course they can be paralleled directly.
What I was saying (main point of the entire discussion) is that energy is transferred trough the wire and not outside the wires.
Since normal wires have resistance you end up with half the initial energy as the other half was lost as heat in the wire thus you end up with just half the Vi in both capacitors.
If you had an ideal setup with no resistance so all conductors are superconductors then you have no loss and end up with 0.707 * Vi in both capacitors.
Then some of you claimed that is not true that with ideal circuit you end up with that high voltage as you confuse voltage with energy.
Since is hard and expensive to setup an experiment with superconductors I offered a much easier solution and that is a DC-DC converter.
Using a 80% efficient DC-DC converter CC-CV working in CC mode will result in almost 90% of the energy still being present in the two capacitors after the energy is split equally between them so close enough to 0.707 * Vi to prove my point.
If you agree that moving the energy from charged capacitor to discharged capacitor will result in a higher final voltage in both capacitors than using wires then my point is made. If you do not then all I can say is you need to do the experiment.
By adding a DC-DC converter to the circuit you do not bring any external energy. You just more efficiently move the existing energy from one place to another.
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OK so the DC-DC is just an alternative implementation?
I want to do it without a DC-DC. Can you show me an experimental setup (using superconductors if necessary) to prove the effect?
Tim
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OK so the DC-DC is just an alternative implementation?
I want to do it without a DC-DC. Can you show me an experimental setup (using superconductors if necessary) to prove the effect?
Tim
Yes DC-DC is an alternative way to transport energy from the charged capacitor to the discharged one.
The proof is already there with the normal capacitors. After you parallel the capacitors you end up with only half the energy the other half escaped as heat due to circuit resistance.
There is no electric field in the discharged capacitor and there will only be a field as soon as an electron gets to one of the plates and simultaneously an electron will leave the opposite plate.
So energy transfer is done trough wires.
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Do you have a setup I can test?
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Do you have a setup I can test?
What is exactly that you want to test ?
Is that moving energy from one charged capacitor to an identical discharged capacitor using a DC-DC converter will result in final voltage being significantly more than 1/2 and close to 0.707 ?
If you think that is not true then you are basically saying that capacitor energy equation is wrong.
E = 0.5 * C * V2
I already made the example with Vi = 3V and C = 1F
Initial energy in charged capacitor is
0.5 * 1F * 3V2 = 4.5Ws
If energy is conserved after the switch was closed and you basically have a 2F capacitor (two 1F in parallel).
0.5 * 2F * (3V * 0.707)2 = 4.5Ws
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(https://upload.wikimedia.org/wikipedia/commons/8/86/Two_capacitor_paradox.svg)
Two capacitors cant exist like this. This circuit needs 2 switches, not 1 switch.
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(https://upload.wikimedia.org/wikipedia/commons/8/86/Two_capacitor_paradox.svg)
Two capacitors cant exist like this. This circuit needs 2 switches, not 1 switch.
:) what?
Why will you need a second switch ? redundancy :)
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If you assume the bottom 'wire' between the two capacitors has zero impedance, then it's essentially just one capacitor. Isn't it?
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If you assume the bottom 'wire' between the two capacitors has zero impedance, then it's essentially just one capacitor. Isn't it?
One capacitor is charged at Vi so if you consider that the switch is a capacitor (with it is in real world) the capacitance is so small and negligible that all voltage will be on the (switch/capacitor) and the charge transferred to the discharged capacitor is extremely negligible.
So you can see a real switch as an ideal switch in parallel with a very small value capacitor.
When you close the switch you are basically shorting the small capacity capacitor (switch capacity).
For all intents and purposes you can consider the empty capacitor on the right as containing no charge.
Also if you where to add another switch on the bottom you are just adding another small capacitor in series and the difference between one switch and two switches is likely not even measurable with normal lab equipment.
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(https://upload.wikimedia.org/wikipedia/commons/8/86/Two_capacitor_paradox.svg)
Two capacitors cant exist like this. This circuit needs 2 switches, not 1 switch.
:) what?
Why will you need a second switch ? redundancy :)
If u had 2 open switches, with a charged capacitor on leftside & non-charged capacitor on righthandside, then if u closed one switch i think that the 2 capacitors would not end up with the same (say half each) charge. They would have i think a very mixed arrangement of 4 different charges.
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What is exactly that you want to test ?
Is that moving energy from one charged capacitor to an identical discharged capacitor using a DC-DC converter will result in final voltage being significantly more than 1/2 and close to 0.707 ?
As you said earlier, and as I said earlier: without a DC-DC converter.
Can you do it?
If energy is conserved
Is a very, very big 'if'.
So can you do it? With just wires, superconducting or otherwise? As you said to do it earlier? I'm really curious to see what you propose. I haven't seen a 'no', no indication that it be impossible.
Granted, it's been a challenge getting anything of substance out of you at all... I'm beginning to think you don't actually know what you're talking about. ;)
Tim
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(https://upload.wikimedia.org/wikipedia/commons/8/86/Two_capacitor_paradox.svg)
Two capacitors cant exist like this. This circuit needs 2 switches, not 1 switch.
:) what?
Why will you need a second switch ? redundancy :)
If u had 2 open switches, with a charged capacitor on leftside & non-charged capacitor on righthandside, then if u closed one switch i think that the 2 capacitors would not end up with the same (say half each) charge. They would have i think a very mixed arrangement of 4 different charges.
If you consider the switch ideal (no capacitance) then if you close just one of two switches absolutely nothing will happen.
Real switches do have some capacitance depending on how they are constructed but that will be so small so many orders of magnitude that they can be ignored.
If you where to go to extreme and say that switch capacitance is the same as the two main capacitors in the test say 1F as in the example then when you charge the capacitor on the left to 3V you are also charging the two series capacitors the one from the switch and that one drawn in diagram as discharged thus you will see 1.5V on the right side capacitor.
To get rid of even the smallest switch capacitance that you can ignore anyway. You can just short the empty capacitor before closing the switch to get rid of the infinitesimal amount of charge it got there when you charged the one on the left.
Adding another switch will not help with anything as it will be just another small capacitor in series.
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As you said earlier, and as I said earlier: without a DC-DC converter.
Can you do it?
Do you have the equipment to construct two capacitors form super conductive materials to do such a test ?
Also the equations show clearly what the result will be.
If energy is conserved
Is a very, very big 'if'.
So can you do it? With just wires, superconducting or otherwise? As you said to do it earlier? I'm really curious to see what you propose. I haven't seen a 'no', no indication that it be impossible.
Granted, it's been a challenge getting anything of substance out of you at all... I'm beginning to think you don't actually know what you're talking about. ;)
Tim
Energy is always conserved it is the law. If you do not agree with that then it is you that will need to prove it.
In the case of identical capacitors with resistance half of the energy is in the capacitors after the switch is closed and half is dissipated as heat from the conductors (that includes wires and capacitor plates).
You are lacking fundamental understanding thus you do not understand my replays.
It is sad to see influential science communicators and university professors not understanding the subject that they should learn others.
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I do. I performed the test.
I observed an oscillating waveform, with the average voltage being 0.5 Vi, and the peak sine amplitude being 0.5 Vi. Put another way, the capacitors slosh alternately between 0 and 1 Vi. At no point are they simultaneously 0.71 Vi; and the current in the wire connecting them (inductor, really), when one is at 0.71 Vi, is nonzero, so opening the switch at that instant would result in energy loss.
Can you explain my findings?
Tim
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I do. I performed the test.
I observed an oscillating waveform, with the average voltage being 0.5 Vi, and the peak sine amplitude being 0.5 Vi. Put another way, the capacitors slosh alternately between 0 and 1 Vi. At no point are they simultaneously 0.71 Vi; and the current in the wire connecting them (inductor, really), when one is at 0.71 Vi, is nonzero, so opening the switch at that instant would result in energy loss.
Can you explain my findings?
Tim
Let me guess. You used some simulation tool that was not setup properly or the simulation tool was wrongly designed.
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Discussing the case where you have switches in both the top and bottom wires...
If you consider the switch ideal (no capacitance) then if you close just one of two switches absolutely nothing will happen.
Well, not absolutely nothing. If only one switch is closed, there will always be 10V measured over the other switch, so the act of closing one of the switches changes the voltages on the right hand side.
With both switches open the voltages on the right hand side are not at all clearly defined, as the two circuits are not connected.
(I'm waiting for some "the quantum wave function collapsed because you need to a meter to make the measurements" handwaving... from the person who adds DC-DC converters)
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Discussing the case where you have switches in both the top and bottom wires...
If you consider the switch ideal (no capacitance) then if you close just one of two switches absolutely nothing will happen.
Well, not absolutely nothing. If only one switch is closed, there will always be 10V measured over the other switch, so the act of closing one of the switches changes the voltages on the right hand side.
With both switches open the voltages on the right hand side are not at all clearly defined, as the two circuits are not connected.
(I'm waiting for some "the quantum wave function collapsed because you need to a meter to make the measurements" handwaving... from the person who adds DC-DC converters)
We are talking about energy transfer.
I think I mentioned before that your multi-meter is basically 1Mohm resistor so it is like closing the switch and measuring the voltage drop on that 1MOhm resistor.
When you have two switches if you had a high resolution oscilloscope you will see a spike when you connect the oscilloscope probe as the probe impedance closes that circuit and there is a circuit because the other open switch (a real one has capacitance) thus you rearrange some of the charges.
You are concentrating on insignificant details that you do not even understand instead of the main subject of how the energy is transferred (trough wire my claim or outside of the wire Derek's claim that you seems to be defending without any basis).
Also your do not provide any feedback on where my math is wrong. Math that shows the amount of energy in a capacitor.
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Also your do not provide any feedback on where my math is wrong. Math that shows the amount of energy in a capacitor.
I have done that many times.... but one more time...
Can we agree that the total charge on the top half of the system is:
Qtotal = Cc1 Vc1 + Cc2 Vc2
So here is the initial state of the system:
Qtotal = 1F * 3V + 1F * 0V = 3 Coulomb.
I hope both agree that no charge can cross either capacitor regardless of the position of the switch. So the total charges on the top and bottom halves must remain the same before and after the switch is closed. If not, please let me know what mechanism is taking electrons out of the top half and moving them to the bottom half, as I am sure we both agree that no positively charged atoms are moving around either.
The switch is closed, and allowing for the incorrect assumption that everything settles down to a stable state when using ideal components:
Qtotal = Cc1 Vc1 + Cc2 Vc2
Qtotal = 1F * 1.5V + 1F * 1.5 = 3 Coulomb.
This is the same as you would have if you consider the two 1F capacitors now a single 2F capacitor:
Qtotal = 2F * 1.5V = 3 Coulomb.
We have the same amount of charge, but are left with only half the voltage over each capacitor. The incorrect assumption that the system becomes stable has let us down, as half the energy is missing, but ideal capacitors and wires have no resistive element to dissipate it.
A simpler example that shows how flawed your position is
What about a single charged ideal capacitor (no ESR), ideal wires (zero ohms) and a single ideal switch (zero ohms when closed, perfect isolation when open), that shorts the capacitor. If, as you say, energy must be conserved, what happens if the capacitor has been charged to 3V and the switch is closed?
What happens with "close to ideal wires", that have zero ohms resistance, but a small amount of inductance?
The inductance is important, as it gives a solution where the total energy in the system is conserved.
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Also your do not provide any feedback on where my math is wrong. Math that shows the amount of energy in a capacitor.
I have done that many times.... but one more time...
Can we agree that the total charge on the top half of the system is:
Qtotal = Cc1 Vc1 + Cc2 Vc2
So here is the initial state of the system:
Qtotal = 1F * 3V + 1F * 0V = 3 Coulomb.
I hope both agree that no charge can cross either capacitor regardless of the position of the switch. So the total charges on the top and bottom halves must remain the same before and after the switch is closed. If not, please let me know what mechanism is taking electrons out of the top half and moving them to the bottom half, as I am sure we both agree that no positively charged atoms are moving around either.
The switch is closed, and allowing for the incorrect assumption that everything settles down to a stable state when using ideal components:
Qtotal = Cc1 Vc1 + Cc2 Vc2
Qtotal = 1F * 1.5V + 1F * 1.5 = 3 Coulomb.
This is the same as you would have if you consider the two 1F capacitors now a single 2F capacitor:
Qtotal = 2F * 1.5V = 3 Coulomb.
We have the same amount of charge, but are left with only half the voltage over each capacitor. The incorrect assumption that the system becomes stable has let us down.
A simpler example that shows how flawed your position is
What about a single charged ideal capacitor (no ESR), ideal wires (zero ohms) and a single ideal switch (zero ohms when closed, perfect isolation when open), that shorts the capacitor. If, as you say, energy must be conserved, what happens if the capacitor has been charged to 3V and the switch is closed?
What happens with "close to ideal wires", that have zero ohms resistance, but a small amount of inductance?
The inductance is important, as it gives a solution where the total energy in the system is conserved.
As I mentioned you are confusing charge with energy.
The important law is called Conservation of energy and not conservation of charge.
Please do a energy balance not a charge balance and you will understand what I'm saying.
When you do the transfer from one capacitor to the other you will be losing half of the energy as heat.
Once you understand the difference between charge and energy it should be easier to also understand what I'm saying.
3 Coulomb is like saying 0.833mAh so you say nothing about energy.
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If one ignores the inductance in the circuit, and just treats it as having a finite resistance R in series with the two equal capacitors:
The final distribution of charge (1/2 on each capacitor) is independent of R, but the time required to equilibrate decreases as R decreases.
Since the final distribution is independent of R, it will still be true in the limit as R --> 0.
Also, the total energy after equilibrium in the two-capacitor system will also be 1/2 the original total energy in only one capacitor, independent of R and therefore in the limit as R --> 0.
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A simpler example that shows how flawed your position is
What about a single charged ideal capacitor (no ESR), ideal wires (zero ohms) and a single ideal switch (zero ohms when closed, perfect isolation when open), that shorts the capacitor. If, as you say, energy must be conserved, what happens if the capacitor has been charged to 3V and the switch is closed?
What happens with "close to ideal wires", that have zero ohms resistance, but a small amount of inductance?
The inductance is important, as it gives a solution where the total energy in the system is conserved.
As I mentioned you are confusing charge with energy.
The important law is called Conservation of energy and not conservation of charge.
Please do a energy balance not a charge balance and you will understand what I'm saying.
When you do the transfer from one capacitor to the other you will be losing half of the energy as heat.
Once you understand the difference between charge and energy it should be easier to also understand what I'm saying.
3 Coulomb is like saying 0.833mAh so you say nothing about energy.
But voltage is also clearly defined, through the capacitance formula of Q = C V, and you can calculate energy directly from charge and capacitance U = 1/2 Q^2 / C
And the answer in the "one cap, one switch" situation?
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If one ignores the inductance in the circuit, and just treats it as having a finite resistance R in series with the two capacitors:
The final distribution of charge (1/2 on each capacitor) is independent of R, but the time required to equilibrate decreases as R decreases.
Since the final distribution is independent of R, it will still be true in the limit as R --> 0.
Yes 1/2 of the charge but 1/4 of energy. Half of the total energy was dissipate as heat on the series resistance no matter the value of the series resistance.
Please understand that discussion is about transfer of energy.
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If one ignores the inductance in the circuit, and just treats it as having a finite resistance R in series with the two capacitors:
The final distribution of charge (1/2 on each capacitor) is independent of R, but the time required to equilibrate decreases as R decreases.
Since the final distribution is independent of R, it will still be true in the limit as R --> 0.
Yes 1/2 of the charge but 1/4 of energy. Half of the total energy was dissipate as heat on the series resistance no matter the value of the series resistance.
Please understand that discussion is about transfer of energy.
1/4 of the energy in each capacitor, for a total energy in the two-capacitor system half the original energy.
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And the answer in the "one cap, one switch" situation?
Do you mean a charged cap shorted by a switch ?
If so then all energy contained in the capacitor ends up as heat.
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1/4 of the energy in each capacitor, for a total energy in the two-capacitor system half the original energy.
Yes exactly.
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One of the problems with analogies is that you have to have a deep understanding of model's (in this case hydraulic) system. Derek seems to have done his homework. He points out that unlike a hydraulic system (where the water from the pump to the load has high pressure and low velocity and the water in the return pipe has lower pressure and higher velocity), with an electric circuit, there's no difference in current density or drift speed for the electrons going in and coming out of the battery.
Well in the case you described the "load" (a reducer/nozzle) is analogous to a transformer, and there's no difference in mass flow going in and coming out of the pump. The analogy is tight, here :-+
What is making the electrons drift is not some kind of pressure. It is just a portion of the electric field generated by the battery that does not contribute to the transfer of energy from the battery to the load. So electrons in the wire are not being pushed by each other like in a fluid. They're just parading in response to an external cause (the electric field) exactly like in the load.
What is making the electrons drift is a kind of pressure, called electric potential. They are being pushed by each other, but unlike in fluids, it happens at long range and can easily go through some solid matter.
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1/4 of the energy in each capacitor, for a total energy in the two-capacitor system half the original energy.
Yes exactly.
I have a suggestion. Forget the two capacitor question. Veritasium's experiment uses a battery. Maybe start a new thread about two capacitors? It's off topic.
We don't have to understand batteries beyond they are ideal voltage sources. They maintain a constant voltage across their terminals. Not complicated. Go from there.
Here's a nice video to get you started, 1940's style Veritasium:
https://www.youtube.com/watch?v=JHSPRcRgmOw (https://www.youtube.com/watch?v=JHSPRcRgmOw)
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1/4 of the energy in each capacitor, for a total energy in the two-capacitor system half the original energy.
Yes exactly.
I have a suggestion. Forget the two capacitor question. Veritasium's experiment uses a battery. Maybe start a new thread about two capacitors? It's off topic.
We don't have to understand batteries beyond they are ideal voltage sources. They maintain a constant voltage across their terminals. Not complicated. Go from there.
You are wrong about the fact that it is of topic.
There is no difference in this context between a battery and a capacitor as they are both the source of energy in the experiment.
Understanding what energy is and how capacitors work (energy storage devices) is critical to understand the problem.
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If those two capacitors are identical (same capacity) the voltage after switch is closed will be 0.707 * Vi
One other thing I want to point out is that if the final voltage on both capacitors is 70.7% of the initial voltage then you will NOT be conserving charge, but only conserving energy instead. Initially, the charge is Q = CV before the switch is closed. After the switch is closed, if the final voltage is 0.707*Vi on both capacitors then the final charge will be C*V*0.707 + C*V*0.707 = 1.414*Q where Q is the initial charge. Where does the extra 41.4% charge come from if it wasn't there initially?
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You are wrong about the fact that it is of topic.
There is no difference in this context between a battery and a capacitor as they are both the source of energy in the experiment.
Understanding what energy is and how capacitors work (energy storage devices) is critical to understand the problem.
There is a difference between a battery and a capacitor.
I understand how capacitors work.
Now, energy is another issue. :-// But I can always look up the textbook definition.
What is your definition of energy?
What did you think of the video?
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The unasked question so far is:
Those pipes are hollow. What if you put a thick wire inside those pipes, electrically connected only at both ends of the pipe.
You balance the cross-sectional area such that the wire and pipe have the same mm^2. After a constant current power supply is connected and the system reaches a steady state. How much current would flows through the inner wire, and how much through the outer pipe?
Three different schools of thought:
1) The pipe is a (oddly shaped) Faraday cage. There will be minimal potential difference across the inner wire, so minimal current will flow in it. The bulk of the current flows in the pipe. If connected in parallel outside of the pipe then more current would flow.
2) The current flow is based on the cross-sectional area of the conductor and its bulk resistance, so half the power will be in the inner wire, and half in the outer pipe. The same current would flow the wire was connected in parallel outside the pipe.
3) Well, it's not that simple, as the field from the pipe will still be interacting with the wire and the other way around. You would need to know about a lot of other things about the materials and geometries before you could answer this.
I'm firmly in camp 2 for steady state. But during the transient when current source is connected I'm quickly jumping into "camp 3".
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Well, not absolutely nothing. If only one switch is closed, there will always be 10V measured over the other switch, so the act of closing one of the switches changes the voltages on the right hand side.
Don't let him distract you with added switches, or odd ratios of capacitor: again, it adds nothing to the problem, and only deflects from the central point.
Tim
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If those two capacitors are identical (same capacity) the voltage after switch is closed will be 0.707 * Vi
One other thing I want to point out is that if the final voltage on both capacitors is 70.7% of the initial voltage then you will NOT be conserving charge, but only conserving energy instead. Initially, the charge is Q = CV before the switch is closed. After the switch is closed, if the final voltage is 0.707*Vi on both capacitors then the final charge will be C*V*0.707 + C*V*0.707 = 1.414*Q where Q is the initial charge. Where does the extra 41.4% charge come from if it wasn't there initially?
The important part is conservation of energy not charge.
You start with 4.5Ws of stored energy in the charged capacitor 0.5 * 1F * 3V2. That is what you need to justify not charge unless you are confusing charge with energy and that seems to be the case.
Initial charge 1F * 3V = 3 Coulomb = 0.833mAh
4.5Ws / 3600s = 1.25mWh
3V * 0.833mAh = 2.5mWh if the voltage will have been constant but voltage drops linearly so is half that 1.25mWh as calculated with the formula for Energy in a capacitor.
At the end of the experiment you can have for ideal case
2.121V on each capacitor
So 1.25mWh initial energy divided between the two capacitors so 0.625mWh in each
If you care about charge on each capacitor (not relevant) 2.121 Coulomb / 3600 = 0.589mAh
(2.121V /2) * 0.589mAh = 0.625mWh
As you see both energy and charge checks out in my calculation as I did the correct assumption that energy is conserved not charge.
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I do. I performed the test.
I observed an oscillating waveform, with the average voltage being 0.5 Vi, and the peak sine amplitude being 0.5 Vi. Put another way, the capacitors slosh alternately between 0 and 1 Vi. At no point are they simultaneously 0.71 Vi; and the current in the wire connecting them (inductor, really), when one is at 0.71 Vi, is nonzero, so opening the switch at that instant would result in energy loss.
Can you explain my findings?
Tim
Let me guess. You used some simulation tool that was not setup properly or the simulation tool was wrongly designed.
I mean, you're welcome to scrutinize whatever models I use. But my oscilloscope speaks for itself:
(https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/?action=dlattach;attach=1477765;image)
From the left, the voltage steps down corresponding to the switch turning on. Subsequently, the capacitor voltages (Ch1, Ch3) oscillate as predicted. I'm not seeing any trend towards 0.71 Vi here.
Tim
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The unasked question so far is:
Those pipes are hollow. What if you put a thick wire inside those pipes, electrically connected only at both ends of the pipe.
You balance the cross-sectional area such that the wire and pipe have the same mm^2. After a constant current power supply is connected and the system reaches a steady state. How much current would flows through the inner wire, and how much through the outer pipe?
As with everything it seems, there is a video for that:
https://www.youtube.com/watch?v=5SiKCgQ9mi8 (https://www.youtube.com/watch?v=5SiKCgQ9mi8)
He says the currents starts out on the outside of the conductor, then moves inside with time.
A constant current source would be high impedance. Not sure that's the best choice for a high speed signal.
So who's going to test it?
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I mean, you're welcome to scrutinize whatever models I use. But my oscilloscope speaks for itself:
(https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/?action=dlattach;attach=1477765;image)
From the left, the voltage steps down corresponding to the switch turning on. Subsequently, the capacitor voltages (Ch1, Ch3) oscillate as predicted. I'm not seeing any trend towards 0.71 Vi here.
Tim
What do you have on channel 1 and 3 and also what are you triggering on channel 4?
Also what is the exact circuit ?
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Channels 1 and 3 are the voltages at the two capacitors in the circuit:
(https://upload.wikimedia.org/wikipedia/commons/8/86/Two_capacitor_paradox.svg)
The scope is grounded to one side of the switch, hence the voltage step at turn-on. This also indicates the initial voltage. Ch4 is a sync signal triggering the switch.
Digital averaging and bandwidth limiting (100MHz) have also been used to improve signal quality.
The cursor at 4.2µs shows the half wave time.
Tim
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Channels 1 and 3 are the voltages at the two capacitors in the circuit:
(https://upload.wikimedia.org/wikipedia/commons/8/86/Two_capacitor_paradox.svg)
The scope is grounded to one side of the switch, hence the voltage step at turn-on. This also indicates the initial voltage. Ch4 is a sync signal triggering the switch.
Digital averaging and bandwidth limiting (100MHz) have also been used to improve signal quality.
The cursor at 4.2µs shows the half wave time.
Tim
A photo of the setup will be super useful.
I do not get how your setup is made and there are to many unknowns.
What capacitors are you using value and type. How are the capacitors connected (wires if so how long).
How isolated is the system ? Like any power supply connected (even if just one of the wires from the power supply is connected).
What sort of switch do you use ? Is that maybe a mosfet (what model) and how is that controlled ? Is is a photo diode optocoupler ? It may be supper important and likely the reason for osculations if your circuit is not properly isolated.
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Capacitors: about 20nF each.
Wire: I could measure it, but actually, you have enough information now to calculate this.
The system is isolated on battery power. Anyway, it's only powered before the switch turns on.
Will my setup produce the fabled 0.71 Vi you have predicted?
Tim
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Capacitors: about 20nF each.
Wire: I could measure it, but actually, you have enough information now to calculate this.
The system is isolated on battery power. Anyway, it's only powered before the switch turns on.
Will my setup produce the fabled 0.71 Vi you have predicted?
Tim
20nF ??? so you have build some LC oscillator.
I was for good reason using 1F as an example but at least hundreds or thousands of uF will be useful for such a test.
Someone else here had a video with a test using some larger Electrolytic capacitors. I was expecting you use some similar setup and maybe you had some strange isolation problem due to non isolated switch driver. But matched inductance and capacitance also make sense and that is most likely you case as 20nF is basically nothing.
If you use large capacitors Large electrolytic or super-capacitors then you can use a DC-DC converter with constant current control and get very close to 0.7 Vi
The probing was also strange as I have no idea why you will connect the oscilloscope ground to the switch. But is fairly clear that you have build an LC resonator so not at all what is shown in that diagram two parallel capacitors and a switch.
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20nF ??? so you have build some LC oscillator.
I was for good reason using 1F as an example but at least hundreds or thousands of uF will be useful for such a test.
Sure, but I don't have superconducting capacitors that large. Are you saying this effect disappears for small values, then? By what mechanism? That's quite a peculiar effect if so!
Someone else here had a video with a test using some larger Electrolytic capacitors. I was expecting you use some similar setup and maybe you had some strange isolation problem due to non isolated switch driver. But matched inductance and capacitance also make sense and that is most likely you case as 20nF is basically nothing.
If you use large capacitors Large electrolytic or super-capacitors then you can use a DC-DC converter with constant current control and get very close to 0.7 Vi
The probing was also strange as I have no idea why you will connect the oscilloscope ground to the switch. But is fairly clear that you have build an LC resonator so not at all what is shown in that diagram two parallel capacitors and a switch.
But I was told a DC-DC converter is not required, this effect will be observed for superconducting wires?
I have wired the diagram exactly as shown, I'm not sure what is missing?
Tim
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Sure, but I don't have superconducting capacitors that large. Are you saying this effect disappears for small values, then? By what mechanism? That's quite a peculiar effect if so!
Are you saying those 20nF capacitors are made of superconductor material ? Even the wires ?
Do you also have the means to cool them down to whatever temperature is needed to become superconductors ? You will not be able to use a mosfet switch in that case anyway.
But I was told a DC-DC converter is not required, this effect will be observed for superconducting wires?
I have wired the diagram exactly as shown, I'm not sure what is missing?
Tim
The effect can be observed for superconducting wires (this need to include the switch and the capacitor plates). Unless you have access to some university lab with this sort of equipment I doubt this is an option for you.
You did not wired as shown in diagram as you added significant inductance connecting the super small 20nF capacitors with wires. There is no inductor drawn in the schematic for a good reason.
And yes you can not get rid of inductance or capacitance but when the diagram shows capacitors only you understand that inductance in that circuit will need to be negligibly small and that is not possible when you connect 20nF capacitors with wires.
So use some few mF electrolytic capacitors and low power and efficient DC-DC converter as it is way easier and less expensive to setup than superconductors.
I also want to insist on the fact that I already proved multiple times with the correct equations that what I say is correct.
Main equation is the one for Energy stored in a capacitor = 0.5 * C * V2
initial condition 3V charged capacitor identical 1F capacitors
0.5 * 1 * 32 = 4.5Ws
end result after switch is closed and steady state is reached ideal case or very close with efficient DC-DC
0.5 * 2 * 2.1212 = 4.5Ws
end result with resistance in series
0.5 * 2 * 1.52 = 2.25Ws with the other half of the energy lost as heat due to circuit resistance.
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So use some few mF electrolytic capacitors and low power and efficient DC-DC converter as it is way easier and less expensive to setup than superconductors.
It is also 'cheating'. I too could get any answer I want if I am free to add to the system. If I said "let me put an inductor in there, a diode, a switch and a trained imp that can push the switch really quickly" would you not agree that that is not the same problem any more?
You are now worrying about wires (even superconducting ones) having inductance. Why worry? A moving charge induces a magnetic field. That is how the underlying physical reality works - a wire with zero self-inductance cannot exist! (well, is infinitely short... so no longer a wire)
Do you agree that charges moving between the capacitors will induce changes in the magnetic field? If so, there is the source of the L for the LC resonator - it's not a flawed component, it is built into nature.
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Are you saying those 20nF capacitors are made of superconductor material ? Even the wires ?
Do you also have the means to cool them down to whatever temperature is needed to become superconductors ? You will not be able to use a mosfet switch in that case anyway.
FETs work fine at superconducting temperatures, just need the right ones. ;)
The effect can be observed for superconducting wires (this need to include the switch and the capacitor plates). Unless you have access to some university lab with this sort of equipment I doubt this is an option for you.
You did not wired as shown in diagram as you added significant inductance connecting the super small 20nF capacitors with wires. There is no inductor drawn in the schematic for a good reason.
But you specifically requested superconductors; there is no resistance in the circuit. And it is, well, wired -- I'm not sure where you suggest I go and get noninductive wires from, I mean, that would be preposterous! ...Wouldn't it?
So I've done the nearest thing to the diagram I can do, and assumed the wire links mean wires, that, well, sometimes have inductance to them. I can't exactly help it, okay!
And yes you can not get rid of inductance or capacitance but when the diagram shows capacitors only you understand that inductance in that circuit will need to be negligibly small and that is not possible when you connect 20nF capacitors with wires.
Negligibly small in relation to......what? :o
It's not like there's an RC time constant in there. You said no resistance, I went to quite extreme lengths to eliminate it! But then what can be left?!
So use some few mF electrolytic capacitors and low power and efficient DC-DC converter as it is way easier and less expensive to setup than superconductors.
I also want to insist on the fact that I already proved multiple times with the correct equations that what I say is correct.
Main equation is the one for Energy stored in a capacitor = 0.5 * C * V2
Yes yes yes I get the refrain, if I wanted perfect energy balance I could do that. But you don't understand, this isn't an application -- this is a curiosity. Surely you have a curiosity about this conundrum as well? -- Else, why stick in this thread so long?
Your claim that a sqrt(2) should pop up, is quite a curious one, and as a scientist, you should be as interested to disprove it, as I am to prove it! Yes, to disprove ones' own ideas, such a strange notion to some people, but it is the scientific method; there is none easier to fool than the self, as a famous scientist once said.
If you're curious, I reconstructed the experiment with much shorter, non-superconducting wires, and obtained this waveform:
(https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/?action=dlattach;attach=1477816;image)
It again does not show a sqrt(2) ratio, but shows the perfect 0.5 as predicted by charge balance. Energy is of course conserved because the excess is dissipated in the resistance, and no worries about whether the energy flowed around or through the wires, it got where it needed to go all the same.
Tim
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And yes you can not get rid of inductance or capacitance but when the diagram shows capacitors only you understand that inductance in that circuit will need to be negligibly small and that is not possible when you connect 20nF capacitors with wires.
I think this is the core of the problem. You cannot make a circuit with current flowing where both inductance and resistance are negligible. It is not physical to ignore both at the same time, and when you try to you get incorrect answers.
There are two things that can happen: The circuit will have significant resistance and the circuit will reach a steady state at half the initial voltage with a time constant RC. Or the circuit will not have significant resistance and the circuit will oscillate at it's resonance frequency given by w= 1/sqrt(LC). Reducing L will increase the resonant frequency but the overall behavior will stay the same. The first case does not conserve electrical energy because it has resistance. The second does conserve electrical energy but never reaches a steady state, instead it (ideally) oscillates forever. Both conserve charge of course, globally and locally (i.e., the net charge on each metal island remains the same).
I also want to insist on the fact that I already proved multiple times with the correct equations that what I say is correct.
Main equation is the one for Energy stored in a capacitor = 0.5 * C * V2
Of course the correct answer has to conserve both charge and (total) energy, not just one or the other.
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Is late and I do not want to answer all the silly questions you (all) have.
You have no understanding on the basics so my answers are done considering you at least understand some basic parts.
Like the fact that a 20nF capacitor has a very small capacity and so even a relatively short wire will have an inductance large enough that you form a resonant circuit.
If you use a few mF then even with fairly long wires the inductance is super negligible and it is not a problem.
The mosfet can work at low temperatures but it will not be a superconductor so even if you have superconductor wires and capacitors adding the mosfet that will have a resistance will cancel everything.
I mentioned no resistance in the circuit and superconductors just to inform you that is not just theoretical but also possible in real life to setup such a test setup but for probably very few physics labs in the world not for any of you to attempt. Not to mention none of you even have the knowledge necessary to attempt this.
The DC-DC converter is fairly easy to do but likely not for last few people that replayed to me.
The correct answer is just energy conservation not charge conservation as I already showed.
If any of you have an engineering degree then you should be ashamed for not knowing what energy is.
Do not think I will have any time in the next few day's. I may check once a day and replay if there is a serious question for someone that understand the subject.
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Like the fact that a 20nF capacitor has a very small capacity and so even a relatively short wire will have an inductance large enough that you form a resonant circuit.
Any indicance and capacitance will (at least in an ideal situation) form a resonant circuit.
If you use a few mF then even with fairly long wires the inductance is super negligible and it is not a problem.
Only to the extent that the parasitic L and R of a real world 1 mF capacitor are likely to be larger. But an ideal 1 mF capacitor plus a 10 nH inductor will resonate at about 50 kHz. A more realistic 100 nH inductance will lower the frequency to 15 kHz.
The DC-DC converter is fairly easy to do but likely not for last few people that replayed to me.
Everybody here agrees that the DC DC converter will do exactly as you have described. Nobody is trying because there isn't any disagreement on this topic so no point arguing. Our contention is that simply shorting a capacitor out will do something different and Tim did experiments to demonstrate that the circuit either oscillates or decays to V/2.
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Removed as it didn't really answer Daves question.
CJ
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Like the fact that a 20nF capacitor has a very small capacity and so even a relatively short wire will have an inductance large enough that you form a resonant circuit.
Any indicance and capacitance will (at least in an ideal situation) form a resonant circuit.
If you use a few mF then even with fairly long wires the inductance is super negligible and it is not a problem.
Only to the extent that the parasitic L and R of a real world 1 mF capacitor are likely to be larger. But an ideal 1 mF capacitor plus a 10 nH inductor will resonate at about 50 kHz. A more realistic 100 nH inductance will lower the frequency to 15 kHz.
The DC-DC converter is fairly easy to do but likely not for last few people that replayed to me.
Everybody here agrees that the DC DC converter will do exactly as you have described. Nobody is trying because there isn't any disagreement on this topic so no point arguing. Our contention is that simply shorting a capacitor out will do something different and Tim did experiments to demonstrate that the circuit either oscillates or decays to V/2.
All are good points but you likely did not read all the earlier replays.
People did not understood that doing a test with superconductor materials is not trivial unless they work at some large university lab and even then it will be expensive. And did not got the idea that all the elements in the circuit will need to have zero resistance in order for all of the energy to remain in the two capacitors at the end of the test.
You are the first to mention that the DC-DC converter will do what I described. The others before where either not convinced as they where confusing charge with energy and thinking that everything is already conserved just with two normal capacitors.
And some others suggested that adding instructors and active components will somehow add energy from exterior.
The circuit will always decay to V/2 in a real setup (not ideal without resistance) as it was demonstrated in a video posted earlier.
But he concluded that energy was conserved as he confused charge with energy.
Exactly half of the energy was lost as heat in the circuit resistance and that will always be the case as long as resistance is different from zero no matter the resistance value.
The main point is that a transmission line is a capacitor or the other way around a capacitor is a transmission line and what Derek observed in his experiment as current trough that 1.1Kohm load resistor is just due to transmission line capacitance.
There will be a current trough the 1.1kOhm resistor because is in series with the supply the switch and two smaller value capacitors formed by the transmission line.
So his claim that energy travels outside the wires is false in all instances not just DC but even at initial transient when switch is closed or in AC
The electric field will only be present after charges are present.
Also the fact that half of the energy is lost as heat (heat in the conductors) should be clue enough that energy was transferred trough wires.
Electrons are those that transfer the energy from source to load and not as Derek claims electric field.
I'm also a bit rude and I apologies for that. I feel frustrated trying to explain this and I still have PTSD from a few months ago when I tried unsuccessfully to explain how the faster than wind direct down wind vehicle works and there the exact same type of mistake was made and energy storage was not considered.
The short answer to that problem is that energy is stored in pressure differential (air is a compressible fluid) while vehicle was well below wind speed and that stored energy is what allowed the vehicle to temporarily exceed wind speed. All experiments end before that stored energy is used up and so they concluded that vehicle will continue to accelerate basically forever.
Seeing people with phd in physics not understanding energy conservation was hard.
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All are good points but you likely did not read all the earlier replays.
People did not understood that doing a test with superconductor materials is not trivial unless they work at some large university lab and even then it will be expensive. And did not got the idea that all the elements in the circuit will need to have zero resistance in order for all of the energy to remain in the two capacitors at the end of the test.
I have just one more question. You've seen the waveform; I ask you: after one half-cycle, where is the energy? How much of it? After one full cycle, where is the energy? How much of it? And this repeats on, ad infinitum; it is, as you say, an oscillator.
Where does the energy remain at the end of the test?
Tim
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All are good points but you likely did not read all the earlier replays.
People did not understood that doing a test with superconductor materials is not trivial unless they work at some large university lab and even then it will be expensive. And did not got the idea that all the elements in the circuit will need to have zero resistance in order for all of the energy to remain in the two capacitors at the end of the test.
I have just one more question. You've seen the waveform; I ask you: after one half-cycle, where is the energy? How much of it? After one full cycle, where is the energy? How much of it? And this repeats on, ad infinitum; it is, as you say, an oscillator.
Where does the energy remain at the end of the test?
Tim
Your was a normal circuit with resistance so it will not oscillate forever as energy will all be dissipated as heat after just a few oscillation cycles.
You seemed to have an almost matched resonant circuit meaning the capacitor and inductor energy storage capacity was almost identical thus the large fluctuation.
If you had order of magnitude larger capacitance than inductance then you will not even notice any oscillation unless you had super high vertical resolution oscilloscope.
If you had the means to build a superconductor experiment then you will have build the capacitors as in one of my earlier examples as below
____________________________
____________/ ______________
this sort of physical setup will have extremely low inductance relative to capacitance (very small distance between plates and a good dielectric material between) then you will not see any oscillation unless you had super high vertical resolution equipment as oscillation amplitude will be some infinitesimal small value around the 0.707 Vi so maybe 4'th or 5'digit type fluctuation that will remain forever as there is no resistance for the energy to be dissipated as in non superconductor circuits.
An inductor opposes the change in current flow so the opposite of a capacitor that opposes the change in voltage.
The capacitor creates a electron imbalance between the plates that in turn result in an electric field inside the capacitor.
With the inductor an magnetic field is generated in the space surrounding the inductor and that field is where energy is stored (stored not dissipated).
So at constant current you will have a constant magnetic field (same as you have on a permanent magnet) but if the current reduces the magnetic field will reduce but the energy stored in that magnetic field is converted back to current flow.
The best analogy I can came up with will be a flexible stretchy hose where the elastic force represents the magnetic field.
If you flow water trough a long flexible hose (hose is already filed with water but not under any pressure) then when you open a valve connected say to a barrel filled with water (barrel the analogy for capacitor) the hose if super stretchy (high inductance) will oppose the flow of water as it will start to stretch before it allows water to flow on the other end and so at constant water flow (constant current flow) the hose is stretch is maintained at some level but if you increase the flow the hose will stretch even more storing even more energy (it is not lost) and then that energy can be recovered as the flow drops as the hose will put pressure on the water trying to maintain the earlier flow.
Do not try to push this analogy further as there are differences beyond this. It is just an analogy for the inductor as an energy storage device.
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So you propose a superconducting resonator? Yeah, those resonate too. :)
Examples that come to mind are the cavities used in particle accelerators, with Q factors up to 10^8 (it's not quite infinite because there's always some loss at AC), and superconducting qubits which, being small enough and cold enough that quantum mechanics is quite relevant, have ground states that are effectively resonators in perpetual motion (and for which, bulk measures like Q aren't so meaningful).
The bulk metal forms of these resonators might not be called high inductance, but the fact that they resonate at 100s or 1000s of MHz makes that irrelevant.
Low inductance is not no inductance!
The permeability of free space has units of per-length. Anything that has nonzero physical size, must necessarily have inductance! Even free space itself, or else waves wouldn't propagate (that, or some wierd causality shit that would be even more bonkers if true..).
It seems your gap in knowledge comes down to magnetic aspects:
- Length corresponds to inductance (notice I hinted earlier that the waveform and capacitance were sufficient to solve for the wire length -- evidently around 71m. Hm, it's quite a bit less than that actually, I think; I was lazy and just coiled it up on a spool, magnifying the effect.)
- Energy is stored in the magnetic field, proportional to current flow.
- Energy conservation is true, AND charge conservation is true. Both must be true jointly. However, it happens to be a hell of a lot easier to lose energy to dissipation or radiation into the surroundings, than charge into the surroundings!
- We can assess the behavior of a series RLC circuit (which this is, necessarily: see points above) based on the ratio of Zo = sqrt(L/C) to R. When Zo > R, some oscillation will be evident; when Zo = R, critically damped; Zo < R, overdamped (RC dominant).
- This is a continuum relationship and no distinction appears for R --> 0.
- As a special case, for R = 0, any combination of L and C will resonate; the damping factor is 0 regardless!
So I maintain that my waveform was obtained from a superconducting apparatus until proven otherwise. ;D
I mean, how would you know? Given the above information, can you solve for the resistance (if any) in my circuit?
And there's nothing wrong with the waveforms; half the time, the energy difference (the "missing" 0.5 Ei) is stored in the inductance as current flow. The other half it's in one or the other capacitor, hence the voltages alternate between 0 and Vi. Energy is always conserved! And charge is always conserved too, which is why this process averages 0.5 Vi during the wave, and as the AC transient decays (when R > 0), the energy difference is dissipated as heat. The fact that the capacitors end with 0.25 Ei each, 0.5 Ei total, is also no coincidence; perhaps less satisfying than having no dissipation, but the dissipation itself is a necessity (for such simple circuits; else, we must go to great lengths if we wish to avoid it -- such as DC-DC converters!) and so this is the result, no sqrt(2) to be found.
As for the sqrt(2), there is a separate chain of logic which should sound immediately. Such special ratios are EXTRAORDINARILY rare from simple systems. Impossible even, for suitable definition of "simple systems". Such ratios are more likely to be found in, say, properties of signals -- take the peak to RMS ratio of a sine for example, or its integral which picks up a factor of pi -- but not from such simple, finite, geometric relationships like two capacitors rubbed together. This is ultimately a deep truth about numbers themselves, you can't get an irrational (like sqrt(2)) from a rational (like 1/2) without going to some lengths first (sqrt(2) is an algebraic number).
Or, if we could easily construct such ratios -- it would certainly make transformer design easier. We could easily and accurately match 50 to 75 ohms, for example: a 1.5:1 impedance ratio. But we cannot: a 1.22474... turns ratio is needed. We can only get arbitrarily close. (The continued fraction representation of this ratio goes [1; 4, 2, 4, 2, ...]; large numbers in the continued fraction are desirable as they represent points of especially good (but still not perfect!) fit, but repeating sequences like this don't give any especially good stopping points.)
Tim
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So you propose a superconducting resonator? Yeah, those resonate too. :)
Examples that come to mind are the cavities used in particle accelerators, with Q factors up to 10^8 (it's not quite infinite because there's always some loss at AC), and superconducting qubits which, being small enough and cold enough that quantum mechanics is quite relevant, have ground states that are effectively resonators in perpetual motion (and for which, bulk measures like Q aren't so meaningful).
The bulk metal forms of these resonators might not be called high inductance, but the fact that they resonate at 100s or 1000s of MHz makes that irrelevant.
Low inductance is not no inductance!
The permeability of free space has units of per-length. Anything that has nonzero physical size, must necessarily have inductance! Even free space itself, or else waves wouldn't propagate (that, or some wierd causality shit that would be even more bonkers if true..).
It seems your gap in knowledge comes down to magnetic aspects:
- Length corresponds to inductance (notice I hinted earlier that the waveform and capacitance were sufficient to solve for the wire length -- evidently around 71m. Hm, it's quite a bit less than that actually, I think; I was lazy and just coiled it up on a spool, magnifying the effect.)
- Energy is stored in the magnetic field, proportional to current flow.
- Energy conservation is true, AND charge conservation is true. Both must be true jointly. However, it happens to be a hell of a lot easier to lose energy to dissipation or radiation into the surroundings, than charge into the surroundings!
- We can assess the behavior of a series RLC circuit (which this is, necessarily: see points above) based on the ratio of Zo = sqrt(L/C) to R. When Zo > R, some oscillation will be evident; when Zo = R, critically damped; Zo < R, overdamped (RC dominant).
- This is a continuum relationship and no distinction appears for R --> 0.
- As a special case, for R = 0, any combination of L and C will resonate; the damping factor is 0 regardless!
So I maintain that my waveform was obtained from a superconducting apparatus until proven otherwise. ;D
I mean, how would you know? Given the above information, can you solve for the resistance (if any) in my circuit?
And there's nothing wrong with the waveforms; half the time, the energy difference (the "missing" 0.5 Ei) is stored in the inductance as current flow. The other half it's in one or the other capacitor, hence the voltages alternate between 0 and Vi. Energy is always conserved! And charge is always conserved too, which is why this process averages 0.5 Vi during the wave, and as the AC transient decays (when R > 0), the energy difference is dissipated as heat. The fact that the capacitors end with 0.25 Ei each, 0.5 Ei total, is also no coincidence; perhaps less satisfying than having no dissipation, but the dissipation itself is a necessity (for such simple circuits; else, we must go to great lengths if we wish to avoid it -- such as DC-DC converters!) and so this is the result, no sqrt(2) to be found.
As for the sqrt(2), there is a separate chain of logic which should sound immediately. Such special ratios are EXTRAORDINARILY rare from simple systems. Impossible even, for suitable definition of "simple systems". Such ratios are more likely to be found in, say, properties of signals -- take the peak to RMS ratio of a sine for example, or its integral which picks up a factor of pi -- but not from such simple, finite, geometric relationships like two capacitors rubbed together. This is ultimately a deep truth about numbers themselves, you can't get an irrational (like sqrt(2)) from a rational (like 1/2) without going to some lengths first (sqrt(2) is an algebraic number).
Or, if we could easily construct such ratios -- it would certainly make transformer design easier. We could easily and accurately match 50 to 75 ohms, for example: a 1.5:1 impedance ratio. But we cannot: a 1.22474... turns ratio is needed. We can only get arbitrarily close. (The continued fraction representation of this ratio goes [1; 4, 2, 4, 2, ...]; large numbers in the continued fraction are desirable as they represent points of especially good (but still not perfect!) fit, but repeating sequences like this don't give any especially good stopping points.)
Tim
Maybe I was not clear enough.
If you have orders of magnitude larger capacity than inductance then you will not even be able to measure the signal (and I'm referring to amplitude not frequency).
So a sine wave with 0.001 Vi amplitude and a 0.707 Vi DC offset. An 8 bit scope will not be able to measure that and even with a 12bit scope the signal will be in the noise so you will read a DC voltage of abut 0.707 Vi
The setup that you build was something like 0.9 Vi waveform amplitude so an almost perfectly tuned LC oscillator due to using almost equal L and C in therms of energy storage.
Also all this is irrelevant as energy is conserved in LC vs the RLC where half of the energy ends up as heat when energy is transferred from one identical capacitor to another. This is a fact both if you do the calculations using appropriate equations and if you measure the temperature rise of the conductors.
Now that I think about you can not get a perfect capacitor as there is no perfect dielectric (as far as I know) so there will be losses there and energy from this super conductor made capacitor will slowly dissipate over time as heat due to leakage in dielectric.
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I mean, how would you know? Given the above information, can you solve for the resistance (if any) in my circuit?
Tim
Any resistance value will result in half of the energy lost as heat on the resistance.
A smaller resistance will mean wires will heat faster and a larger resistance value means wires will heat slower but the exact same total amount will be wasted as heat.
With 20nF capacitors energy transferred is small so that half energy ending as heat is to small for you to be able to measure.
But use larger capacitors and a large value resistor in series then you will be able to measure that lost energy.
You know that half the energy was wasted by the end of the experiment
1F cap with 3V charging a discharged 1F capacitor
Start energy 4.5Ws end energy 2.25Ws
Current was limited by resistance so current trough a resistor means wasted heat.
You can add a 1Ohm, 100Ohm and 10kOhm between the two capacitors to increase the resistance and you will see that no matter what resistance you use you end up with the same 0.5 Vi on the two capacitors.
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Actually, power peaks for R = Zo. If it were true that smaller resistance heats faster -- what would superconductors do? Just explode? :D
Tim
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Actually, power peaks for R = Zo. If it were true that smaller resistance heats faster -- what would superconductors do? Just explode? :D
Tim
Of course the speed of electron wave is not infinite so there is a limit to how fast the energy will be transferred from one capacitor to the other.
Yes a superconductor will explode if the temperature increases and it gets out of the super-conduction zone while high current is passing trough it.
Superconductor has no resistance not a small resistance but absolute zero.
Any resistance value no matter how small will dissipate the same amount of energy as heat in this two capacitor example assuming proper ratio of inductance and capacitance with inductance much, much smaller than capacitance.
If you have this resonant circuit as it was in your example energy will flow in multiple directions multiple times significantly increasing the losses then you no longer end up at 0.5Vi.
That is why industrial users of electricity are charged for apparent power so that they will try and correct the power factor.
If we use this same example with two parallel capacitors but the wires are super long so switch is super far from the two capacitors.
How long to you think it will take for the first unit of energy to get to the discharged capacitor?
Will energy get to the discharged capacitor based on the short distance between the two capacitors or it will be based on the long distance between capacitors and switch ?
This should demonstrate if energy travels inside or outside the wires.
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If we use this same example with two parallel capacitors but the wires are super long so switch is super far from the two capacitors.
How long to you think it will take for the first unit of energy to get to the discharged capacitor?
Will energy get to the discharged capacitor based on the short distance between the two capacitors or it will be based on the long distance between capacitors and switch ?
This should demonstrate if energy travels inside or outside the wires.
We already know the answer to that from the video - some energy will travel a shorter distance, via the EM fields, due to the changes in that fields. The bulk of the energy will start once the a stable field has built up around the whole length of the wire.
It doesn't answer if energy is transported inside or outside of the wires, just that energy can travel outside of wires (through the EM field), and that energy can travel along wires.
For me the interesting observation is that only the tiniest bit of EM energy travels the shortest distance, and after that the energy is then being supplied by changes in the EM field that are happening further and further away on the wires.
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If so: HOW MUCH ENERGY does travel along the path?
If we assume that potential energy is located in the rock, and arbitrarily say sea level is zero energy, then zero energy travels along the surface path, and negative energy travels back along the underground path.
Between the two paths (passing through a plane perpendicular to the paths), there is a net energy flow from A to B. The amount of energy is the energy converted to heat. So it depends on which hole it goes down.
This is equivalent to say that the path the energy follows depends on where we arbitrarily choose to set the zero reference. We have seen it with the electrical system as well: for a given convention we can have all energy traveling along the top wire, for another half in the top and half in the bottom wire, and for another one yet all energy travels along the bottom wire. They can't be all true, can they?
Now, in this particular example you elected the return path because it allows you to remove the chance element, but what if I put a second cusp on the return path? Do I have to wait till the weight has fallen in the return hole to know how much energy has traveled?
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The relativistic answer (convention) is that a momentum density p of matter corresponds to an energy flux of pc².
If you want general relativity then the answer is, well, complicated. https://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html
Classical physics is simple and says you have a potential energy of -GMm/r, you're not moving 0 energy, you're just converting: potential->kinetic->heat.
In the quasistatic idealization I had in mind, the mass takes an infinite time to travel from point A to point B. So, velocity is basically zero along the path and only changes during 'generation' (the machine raises the weight) and 'dissipation' (the weight does work against the gravitational field and it all becomes heat).
In my view the energy is in the gravitational field of the system planet + weight. It is being added to the system during 'generation' and it is extracted from the system during 'dissipation'. The system occupies all space, so does it even make sense to ask if the energy is traveling?
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We already know the answer to that from the video - some energy will travel a shorter distance, via the EM fields, due to the changes in that fields. The bulk of the energy will start once the a stable field has built up around the whole length of the wire.
It doesn't answer if energy is transported inside or outside of the wires, just that energy can travel outside of wires (through the EM field), and that energy can travel along wires.
For me the interesting observation is that only the tiniest bit of EM energy travels the shortest distance, and after that the energy is then being supplied by changes in the EM field that are happening further and further away on the wires.
What video ? The one with the wrong info made by Derek ?
The energy is transported trough wires that is why there is no energy transfer from battery to load until electron wave has the time to travel from source to load along the wire and not trough air.
It is also important I think to talk about the electric and magnetic field separately in this context.
Before you close the switch all you have is a constant electric field so no magnetic field.
Only the exact moment when the two conductors of the switch get close enough so that an electron can jump to the other conductor that has a different distribution of electrons the energy starts to be transferred or do any work.
If you just move the switch but not close enough for any electrons to move you have some local small variation in the electric field but no magnetic field and no work done so no transfer of energy.
But if you think you understand better the magnetic field think about at permanent magnet that has fields extending quite some distance outside the magnet but they can not do any work or transfer any energy.
if you move a conductive loop in a magnetic field then the energy to the system is provided by the one that moves the loop and as electrons move they create an equal and opposite field. It is not the magnetic field that moves the electrons is the person that moves the loop trough a magnetic field and that opposite field is the result of electrons moving.
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We already know the answer to that from the video - some energy will travel a shorter distance, via the EM fields, due to the changes in that fields. The bulk of the energy will start once the a stable field has built up around the whole length of the wire.
It doesn't answer if energy is transported inside or outside of the wires, just that energy can travel outside of wires (through the EM field), and that energy can travel along wires.
For me the interesting observation is that only the tiniest bit of EM energy travels the shortest distance, and after that the energy is then being supplied by changes in the EM field that are happening further and further away on the wires.
What video ? The one with the wrong info made by Derek ?
The energy is transported trough wires that is why there is no energy transfer from battery to load until electron wave has the time to travel from source to load along the wire and not trough air.
It is also important I think to talk about the electric and magnetic field separately in this context.
Before you close the switch all you have is a constant electric field so no magnetic field.
Only the exact moment when the two conductors of the switch get close enough so that an electron can jump to the other conductor that has a different distribution of electrons the energy starts to be transferred or do any work.
If you just move the switch but not close enough for any electrons to move you have some local small variation in the electric field but no magnetic field and no work done so no transfer of energy.
But if you think you understand better the magnetic field think about at permanent magnet that has fields extending quite some distance outside the magnet but they can not do any work or transfer any energy.
if you move a conductive loop in a magnetic field then the energy to the system is provided by the one that moves the loop and as electrons move they create an equal and opposite field. It is not the magnetic field that moves the electrons is the person that moves the loop trough a magnetic field and that opposite field is the result of electrons moving.
There will be a potential difference across the switch (otherwise no current will flow when the switch is closed). And as you point out it is a small capacitor - different charges separated by distance. There is then a force across the gap - and with simple geometries it can be calculated - https://physicstasks.eu/1535/force-acting-on-capacitor-plates. Force x distance = something or other... You can listen to this on electrostatic speakers if you want.
For a constant magnetic field what you say is true. Likewise a transformer with DC over the primary windings has 0V over the secondary winding.
But if the magnetic field is changing, then you also have an electric field. How else would transformers work - the wires are not moving, yet energy is transferred?
As captured in the Faraday-Maxwell equation:
The Maxwell–Faraday equation states that a time-varying magnetic field always accompanies a spatially varying (also possibly time-varying), non-conservative electric field, and vice versa.
Connecting the battery causes a time-varying magnetic and electric field, and this transfers (a limited amount of) power across the 1m gap between wires.
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There will be a potential difference across the switch (otherwise no current will flow when the switch is closed). And as you point out it is a small capacitor - different charges separated by distance. There is then a force across the gap - and with simple geometries it can be calculated - https://physicstasks.eu/1535/force-acting-on-capacitor-plates. Force x distance = something or other... You can listen to this on electrostatic speakers if you want.
For a constant magnetic field what you say is true. Likewise a transformer with DC over the primary windings has 0V over the secondary winding.
But if the magnetic field is changing, then you also have an electric field. How else would transformers work - the wires are not moving, yet energy is transferred?
As captured in the Faraday-Maxwell equation:
The Maxwell–Faraday equation states that a time-varying magnetic field always accompanies a spatially varying (also possibly time-varying), non-conservative electric field, and vice versa.
Connecting the battery causes a time-varying magnetic and electric field, and this transfers (a limited amount of) power across the 1m gap between wires.
The potential difference is due to imbalance of charges and the electric field is the effect of that.
You mentioned electrostatic speakers but those same as electrostatic microphones/condenser microphones require a DC bias voltage / phantom voltage in order to work unlike dynamic microphones / speakers that generate energy by moving the coil installed on the membrane in a constant magnetic field.
(piezoelectric effect is a different story and plays no role here).
So you can not produce energy by moving the plates of a capacitor like example condenser microphone. You push the energy stored in the capacitor in and out in that DC bias (usually you measure the voltage drop on a resistor).
So you charge or discharge the capacitor and that phantom voltage / DC bias is there to cover any losses and keep the capacitor charged.
But even if there was a piezoelectric effect you will have energy generated just when there is motion so say air was piezoelectric (not the case) and when you moved the switch you produced some energy then as soon as you close the switch (relevant moment) no more energy is generated.
That is why I mentioned a few times that nothing moves in this circuit.
The transformer will not work with DC not that energy is transferred trough the field it is still transferred trough wires.
If there are no electrons to move in primary then you do not have any electron movement in secondary.
If you apply DC to a transformer you are using that coil as a heater. The transient at connection is not DC.
Yes the transient is due to electron wave that starts to form electrons just moving in random direction canceling out to starting to have a defined direction.
You are charging the line witch is a capacitor thus you have a current flow due to creating an electric field by moving electrons on the capacitor plate that results in an electron vacating the opposite plate and that happens at light speed so depends on the gap between capacitor plates.
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The transformer will not work with DC not that energy is transferred trough the field it is still transferred trough wires.
If there are no electrons to move in primary then you do not have any electron movement in secondary.
You might have missed a word in there somewhere.
Are you saying if I have 1A DC in the primary I will have current flowing in the secondary?
Or just that with no current in the primary windings there will be no current in the secondary?
The transient at connection [of a DC source to a transformer] is not DC.
Agreed - during the transient there is a changing magnetic field, and that changing field will causes a voltage across the transformer's secondary windings during the transient.
Once the magnetic field in the primary becomes stable then no more energy will be transferred to the secondary windings (apart from heat). That is until the DC source is disconnected. Then more as second pulse of energy will be transferred into the secondary (even though the DC source is disconnected!)
A resistor connected across the transformer's secondary windings will get (slightly) warmer during each of these transients - energy has been transferred from DC source to the resistor without them being electrically connected. That energy has definitely flowed between the two insulated wires that make up the transformer's winding (yes, through the insulation!).
The original Veritasium experiment has very long wires and so has a very long transient, during which a small amount of energy is transferred into the load through the space between the wires. But where along the wires this energy transfer is happening continuously changes depending on where the transient has got to. Only a tiniest initial amount is will be transferred directly over the 1m gap.
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Only a tiniest initial amount is will be transferred directly over the 1m gap.
Yes. Which is basically what most of us had been considering all along, and that was later shown by a couple experiments. So, well, this is one of those topics that are kinda running in circles.
Has the whole thing shown that many people, including engineers, had misconceptions about "electricity" and that the model they frequently use, while working well enough in a large number of situations, is flawed? Yes. Has it really fully explained "how electricity works"? There are still some quirky corners there. I'm afraid that quite a few people, after having followed all this, will now understand that they had misconceptions, but will embrace a new concept that might itself lead to more misconceptions. Whichever is a better misconception among all those might not be so obvious in the end.
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You might have missed a word in there somewhere.
Are you saying if I have 1A DC in the primary I will have current flowing in the secondary?
Or just that with no current in the primary windings there will be no current in the secondary?
No there will be no current in the secondary but there will be a current in the primary so basically a heating element.
Say you have a 600VA 120Vac 5A input transformer the output is irrelevant it can be anything as it will not be connected to anything.
So with primary connected to 120Vac and nothing on the output you maybe have a little bit of losses in the magnetic core so a bit of heat from there and the small associated IR losses in the primary copper.
With AC you just push energy in to the transformer to store as magnetic field in the ferromagnetic core but then that gets returned with just a bit of loss as the magnetic core is not perfect. So all energy is transmitted trough wires then converted to a magnetic field then due to ferromagnetic materials not being perfect you have some small loss then energy is converted back form magnetic to electric (electron flow) and sent back to your source so you have two way losses trough copper wires and a bit of loss in the ferromagnetic core.
You will sure not want to apply 120Vdc to that primary as you will then see that energy travels trough wires and not outside of them.
The transient at connection [of a DC source to a transformer] is not DC.
Agreed - during the transient there is a changing magnetic field, and that changing field will causes a voltage across the transformer's secondary windings during the transient.
Once the magnetic field in the primary becomes stable then no more energy will be transferred to the secondary windings (apart from heat). That is until the DC source is disconnected. Then more as second pulse of energy will be transferred into the secondary (even though the DC source is disconnected!)
A resistor connected across the transformer's secondary windings will get (slightly) warmer during each of these transients - energy has been transferred from DC source to the resistor without them being electrically connected. That energy has definitely flowed between the two insulated wires that make up the transformer's winding (yes, through the insulation!).
The original Veritasium experiment has very long wires and so has a very long transient, during which a small amount of energy is transferred into the load through the space between the wires. But where along the wires this energy transfer is happening continuously changes depending on where the transient has got to. Only a tiniest initial amount is will be transferred directly over the 1m gap.
Sorry I answered first part without reading this second part that shows you understand how a transformer works not quite sure why you do not realize that energy is delivered trough wires to the load.
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Yes. Which is basically what most of us had been considering all along, and that was later shown by a couple experiments. So, well, this is one of those topics that are kinda running in circles.
Has the whole thing shown that many people, including engineers, had misconceptions about "electricity" and that the model they frequently use, while working well enough in a large number of situations, is flawed? Yes. Has it really fully explained "how electricity works"? There are still some quirky corners there. I'm afraid that quite a few people, after having followed all this, will now understand that they had misconceptions, but will embrace a new concept that might itself lead to more misconceptions. Whichever is a better misconception among all those might not be so obvious in the end.
I will say current models that engineers use are very accurate. Of course a transmission line modeled as finite number of LCR components will provide an approximation but even if you use analog computers you still need to read the result and that may be even less accurate due to measurement (reading) precision.
I still think that understanding what energy is and how energy storage is everywhere will be helpful for a better understanding.
Many including Derek just do not understand what energy is and it gets confused with all sorts of other things.
I guess I have the advantage that I work with energy generation and storage as a hobby.
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I find it funny and at the same time annoying that these people who are reluctant to properly study math and physics, and end up struggling especially with electromagnetism, like to talk in the name of engineers, as if their misconceptions were the general mindset of our class so as to justify their position.
The hydraulic analogy, the origin of the energy in the wire idea, was dismissed right from the start by Maxwell himself in the 19th century. Derek only made this incontestable knowledge available to the masses.
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I find it funny and at the same time annoying that these people who are reluctant to properly study math and physics, and end up struggling especially with electromagnetism, like to talk in the name of engineers, as if their misconceptions were the general mindset of our class so as to justify their position.
The hydraulic analogy, the origin of the energy in the wire idea, was dismissed right from the start by Maxwell himself in the 19th century. Derek only made this incontestable knowledge available to the masses.
Point to me where I have posted a wrong equation. All equations that I used to make the demonstration that energy travels trough wires and not outside the wires as Derek claims are used by engineers and physicist.
There are many posts as I try to refine/distill my explanation to the simplest form possible.
The two parallel capacitors are the simplest example I can give that demonstrate without any doubt that energy transfer is done trough wires and not outside of it.
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I will try to make a resume of my demonstration that energy travels trough wires and not outside.
(https://upload.wikimedia.org/wikipedia/commons/8/86/Two_capacitor_paradox.svg)
1) We have a capacitor on the right that is 1F and charged at 3V and one capacitor on the right also 1F but with no potential across 0V
Before the switch is closed there is a constant electric field inside the charged capacitor but there is no transfer of energy from the charged capacitor to the discharged capacitor.
By shorting the discharged capacitor at the beginning of the experiment we can ensure that there is no energy stored in that capacitor on the right and if the switch has any capacity then that switch capacity is at the same Vi potential as the capacitor on the left.
This proves that the constant electric field inside the charged capacitor can not transfer the energy to the discharged capacitor as long as the switch is open and alone should be sufficient proof that for energy to flow it requires a close loop.
2) If capacitors are 1m apart and the switch is 10m from both capacitors the time it takes from the moment the switch is closed to the moment any amount of energy is transferred to the discharged capacitor is the time it takes the electron wave to travel 10m.
This again is a proof that energy transfer from the source (charged capacitor) to the discharged capacitor travels trough wires.
3) The initial energy in this isolated system is:
0.5 * 1F * 3V2 = 4.5Ws (I prefer Ws as opposed to Joules as a unit but they are the same thing).
With a real circuit where resistance is higher than zero the voltage after the switch was closed and circuit reached steady state will be:
0.5 * 1F * 1.5V2 = 1.125Ws and the same amount in the other capacitor so total 2.25W
So only half of the initial energy is now found in the system the other half was dissipated as heat inside the wires and this can be verified with a thermal camera.
This is again another proof that energy has traveled inside the wire else it will not make sense to have lost half of the initial energy as heat and also see the wires heat up by the exact amount of energy missing.
I may add more points but I think this 3 points should be enough conclusive evidence that energy flows trough wires and not outside.
I'm open to any criticism just mention the point out of this 3 you do not agree with and what I did wrong.
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2) If capacitors are 1m apart and the switch is 10m from both capacitors the time it takes from the moment the switch is closed to the moment any amount of energy is transferred to the discharged capacitor is the time it takes the electron wave to travel 10m.
This again is a proof that energy transfer from the source (charged capacitor) to the discharged capacitor travels trough wires.
If you put the switch at the end of the line then yes it will take the delay of the length of the line to start charging, exactly the same as Derek's original question would if the switch was at the end of the line. No difference at all, so that doesn't prove in any way that the energy flows inside the wire.
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2) If capacitors are 1m apart and the switch is 10m from both capacitors the time it takes from the moment the switch is closed to the moment any amount of energy is transferred to the discharged capacitor is the time it takes the electron wave to travel 10m.
This again is a proof that energy transfer from the source (charged capacitor) to the discharged capacitor travels trough wires.
If you put the switch at the end of the line then yes it will take the delay of the length of the line to start charging, exactly the same as Derek's original question would if the switch was at the end of the line. No difference at all, so that doesn't prove in any way that the energy flows inside the wire.
Yes it proves that since energy is not flowing from the switch to the load but from the source to the load. The position of the switch should not be relevant if the energy from the charged capacitor will not need to travel trough wire.
If the switch is closer to charged capacitor / source then the wire after the switch forms another capacitor with the wire of the load thus effectively the small current Derek's is seeing is the one needed to charge that extra capacitor (the transmission line).
With the switch far away you also have a capacitor made by the transmission wires (that is fully charged before starting the test) and what the switch is doing is shorting that capacitor but from the far end of the capacitor thus electron wave that move the energy trough wires needs time to travel that distance.
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Re Energy not transferred in the wires.
1 The light bulb will not light without the wires. Not even with the tiniest part of wire snipped out.
2 Any material not directly touching the wires may surround the wires and not affect the light bulb function in any way.
Energy is flowing in the wires.
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Alternative wording: current flows through the wires into the light bulb.
The current flowing through the light bulb heats the filament to the point where it emits energy in the form of light and heat radiation.
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If so: HOW MUCH ENERGY does travel along the path?
If we assume that potential energy is located in the rock, and arbitrarily say sea level is zero energy, then zero energy travels along the surface path, and negative energy travels back along the underground path.
Between the two paths (passing through a plane perpendicular to the paths), there is a net energy flow from A to B. The amount of energy is the energy converted to heat. So it depends on which hole it goes down.
This is equivalent to say that the path the energy follows depends on where we arbitrarily choose to set the zero reference.
I don't think this is a problem if you consider the energy is relative to a reference. So zero relative energy may still be nonzero "absolute" energy.
Instead of assuming the energy is located in the rock, you can eliminate the problem perhaps by using the gravitational potential: https://en.wikipedia.org/wiki/Gravitational_potential (https://en.wikipedia.org/wiki/Gravitational_potential)
If the gravitational potential is a property of location, then when you move the mass you are not "transporting energy". As Naej said, you are only converting energy at the two end points.
We have heard the same point of view applied to electromagnetic systems.
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The unasked question so far is:
Those pipes are hollow. What if you put a thick wire inside those pipes, electrically connected only at both ends of the pipe.
What if you only connected pipe at one end. The wire is connected at both ends. Then the current only flows through the wire.
But since the pipe is at the same voltage as the wire, you have no electric field around the wire. You moved the electric field to the outside of the pipe.
Now according to the Poynting vector, the energy flow is on the outside of the pipe, not even at the surface of the wire.
The current flows through the wire, but the energy flows on the outside of the pipe. You could make the pipe 1 meter in diameter and the energy flow would be half a meter away from the current flow.
Yet another odd result of the Poynting vector applied to DC.
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2) If capacitors are 1m apart and the switch is 10m from both capacitors the time it takes from the moment the switch is closed to the moment any amount of energy is transferred to the discharged capacitor is the time it takes the electron wave to travel 10m.
This again is a proof that energy transfer from the source (charged capacitor) to the discharged capacitor travels trough wires.
If you put the switch at the end of the line then yes it will take the delay of the length of the line to start charging, exactly the same as Derek's original question would if the switch was at the end of the line. No difference at all, so that doesn't prove in any way that the energy flows inside the wire.
Yes it proves that since energy is not flowing from the switch to the load but from the source to the load. The position of the switch should not be relevant if the energy from the charged capacitor will not need to travel trough wire.
If the switch is closer to charged capacitor / source then the wire after the switch forms another capacitor with the wire of the load thus effectively the small current Derek's is seeing is the one needed to charge that extra capacitor (the transmission line).
With the switch far away you also have a capacitor made by the transmission wires (that is fully charged before starting the test) and what the switch is doing is shorting that capacitor but from the far end of the capacitor thus electron wave that move the energy trough wires needs time to travel that distance.
Dave is correct(ish).
If the open switch is at the far end of a loop then when the switch is suddenly closed there is an immediate sudden change in voltage amps etc at the switch.
This change will eventually propagate to the bulb. There will be an early weak fast signal at the bulb (via the em field in the air), & then later stronger but slower signals (via the wires mainly).
The early (air) signal will need to cross the direct clear distance from switch to bulb, through the air, at the speed of light.
This distance is say 10 m (or praps 10.1 m)(if this switch sits in the Veritasium Pt2-X), in which case the early (air) delay is say 34 ns.
Or, alternatively, we could claim that there is a strong (wire) signal that propagates along the wire (on the wire)(or in the wire) at the speed of electricity.
If the wire is not insulated then this too propagates at the speed of light, & the delay for say 10.5 m is say 35 ns, nearly the same as for the early (air) signal.
If the wire is insulated then this 35 ns would be 50% greater, ie 52.5 ns. But the 34 ns for the early signal would remain at 34 ns.
Much of what electrodacus says is ok i think (if i could understand it).
But, my memory of Veritasium's 2 youtubes is that Veritasium ignored the pozzy of the switch in the first youtube, & he acknowledged the importance of the pozzy of the switch in the second youtube.
I will have to have another look to check.
Yes, at [11:55] Derek says that if u mooved the switch then this would change the delay.
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Dave is correct(ish).
If the open switch is at the far end of a loop then when the switch is suddenly closed there is an immediate sudden change in voltage amps etc at the switch.
This change will eventually propagate to the bulb. There will be an early weak fast signal at the bulb (via the em field in the air), & then later stronger but slower signals (via the wires mainly).
The early (air) signal will need to cross the direct clear distance from switch to bulb, through the air, at the speed of light. This distance is say 10 m (or praps 10.1 m)(if this switch sits in the Veritasium Pt2-X), in which case the early (air) delay is say 34 ns.
Or, alternatively, we could claim that there is a strong (wire) signal that propagates along the wire (on the wire)(or in the wire) at the speed of electricity. If the wire is not insulated then this too propagates at the speed of light, & the delay for say 10.5 m is say 36 ns, nearly the same as for the early (air) signal.
Much of what electrodacus says is ok i think (if i could understand it).
But, my memory of Veritasium's 2 youtubes is that Veritasium ignored the pozzy of the switch in the first youtube, & he acknowledged the importance of the pozzy of the switch in the second youtube.
I will have to have another look to check.
Yes, at [11:55] Derek says that if u mooved the switch then this would change the delay.
There will be no "early weak fast signal" with the switch 10m away from the source and from the load as mentioned.
The small constant electric field already exists on the transmission line going to the switch as that is just a low value capacitor and it is charged at Vi
The electric field will be converted to magnetic field and some IR heat loss as the electron wave travels along the wire and the electron wave will arrive first then you will have the electric field thus the energy is transferred by the electron wave moving inside the wire and not by any field outside the wire.
Derek (Veritasium) has no understanding of what energy is else he will not make such a claim that energy is transferred by a field.
Any of the 3 points I made are sufficient to disprove Derek's claim and all 3 of them are correct unless you have proof that they are not since I only used tested and accepted equations and theories to make those points.
Link to that post: https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/msg4156462/#msg4156462 (https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/msg4156462/#msg4156462)
Internet is also fool of java type animations and explanations of capacitors and most of them are incorrectly made as the author did not understand the physics and made wrong assumptions.
The most common mistake is to think that charge is conserved if you move the plates of an isolated capacitor when energy is the one that is conserved.
The explanation they have for this is also wrong as they think that since there is a force from the electric field thus you put energy in to move the plates further apart that is the reason charge is conserved thus they show final voltage as 2x initial voltage if you double the distance between plates when the correct answer is 1.414x and conserved energy not conserved charge.
The energy you put in to move the plates will be recovered as soon as you let go of the plates so it is potential kinetic energy storage.
It is like pulling two permanent magnets apart. You do not do any work you store some kinetic energy that you can get back by letting go of the magnets.
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Dave is correct(ish).
If the open switch is at the far end of a loop then when the switch is suddenly closed there is an immediate sudden change in voltage amps etc at the switch.
This change will eventually propagate to the bulb. There will be an early weak fast signal at the bulb (via the em field in the air), & then later stronger but slower signals (via the wires mainly).
The early (air) signal will need to cross the direct clear distance from switch to bulb, through the air, at the speed of light. This distance is say 10 m (or praps 10.1 m)(if this switch sits in the Veritasium Pt2-X), in which case the early (air) delay is say 34 ns.
Or, alternatively, we could claim that there is a strong (wire) signal that propagates along the wire (on the wire)(or in the wire) at the speed of electricity. If the wire is not insulated then this too propagates at the speed of light, & the delay for say 10.5 m is say 36 ns, nearly the same as for the early (air) signal.
Much of what electrodacus says is ok i think (if i could understand it).
But, my memory of Veritasium's 2 youtubes is that Veritasium ignored the pozzy of the switch in the first youtube, & he acknowledged the importance of the pozzy of the switch in the second youtube.
I will have to have another look to check.
Yes, at [11:55] Derek says that if u mooved the switch then this would change the delay.
There will be no "early weak fast signal" with the switch 10m away from the source and from the load as mentioned.
The small constant electric field already exists on the transmission line going to the switch as that is just a low value capacitor and it is charged at Vi
The electric field will be converted to magnetic field and some IR heat loss as the electron wave travels along the wire and the electron wave will arrive first then you will have the electric field thus the energy is transferred by the electron wave moving inside the wire and not by any field outside the wire.
Derek (Veritasium) has no understanding of what energy is else he will not make such a claim that energy is transferred by a field.
Any of the 3 points I made are sufficient to disprove Derek's claim and all 3 of them are correct unless you have proof that they are not since I only used tested and accepted equations and theories to make those points.
Link to that post: https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/msg4156462/#msg4156462 (https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/msg4156462/#msg4156462)
Internet is also fool of java type animations and explanations of capacitors and most of them are incorrectly made as the author did not understand the physics and made wrong assumptions.
The most common mistake is to think that charge is conserved if you move the plates of an isolated capacitor when energy is the one that is conserved.
The explanation they have for this is also wrong as they think that since there is a force from the electric field thus you put energy in to move the plates further apart that is the reason charge is conserved thus they show final voltage as 2x initial voltage if you double the distance between plates when the correct answer is 1.414x and conserved energy not conserved charge.
The energy you put in to move the plates will be recovered as soon as you let go of the plates so it is potential kinetic energy storage.
It is like pulling two permanent magnets apart. You do not do any work you store some kinetic energy that you can get back by letting go of the magnets.
Yes i agree that the wire is a capacitor & has an initial field. But this duznt affect the early signal.
I insist that in theory there will be an early weak fast signal, but i grant that praps at 10 m (instead of the 1 m) i should have called it an early (very very) weak fast signal.
It would be easy to show that u are wrong. Lets place the switch at 5 m from the battery, then the distance switch to bulb on an angle would be say 5.1 m, & the early weak delay would be 5.1 m by 3.4 ns/m which is 17.0 ns. And your late strong delay would be 16 m along the wire at 3.4 ns/m which is 54.4 ns, or, if insulated, add 50% which makes it 81.6 ns.
I have been looking at the arguments re capacitors, & everyone has been wrong in every way.
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Yes i agree that the wire is a capacitor & has an initial field. But this duznt affect the early signal.
I insist that in theory there will be an early weak fast signal, but i grant that praps at 10 m (instead of the 1 m) i should have called it an early (very very) weak fast signal.
It would be easy to show that u are wrong. Lets place the switch at 5 m from the battery, then the distance switch to bulb on an angle would be say 5.1 m, & the early weak delay would be 5.1 m by 3.4 ns/m which is 17.0 ns. And your late strong delay would be 16 m along the wire at 3.4 ns/m which is 54.4 ns, or, if insulated, add 50% which makes it 81.6 ns.
I have been looking at the arguments re capacitors, & everyone has been wrong in every way.
I specifically mentioned that circuit is symmetrical. In this case both the charged and discharged capacitors are at the example same distance from the switch.
The so called small current when circuit is not symmetrical is the current to charge the capacitor made by the transmission line. In that case as it was in Derek's test your lamp is just an indicator to show the current used to charge that capacitor made from the transmission line wires.
As another example you can have the two capacitors circuit that I mentioned and add a lamp in series with the switch just next to the switch and you will see a current trough that lamp as soon as you close the switch but that current is provided by the energy stored in the transmission line so if you open the switch immediately after you closed it the charged in the transmission line will redistribute and the energy lamp used will be provided some nano seconds later by the main energy storage so that charged capacitor in this case.
Thus energy travels trough wires.
As long as your lamp is next to the switch it does not matter how far the battery is the lamp will see a small current immediately but that energy traveled trough wires and from wires (capacitor).
There is no evidence of energy traveling outside the wire just a wrong interpretation of the test result.
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Yes i agree that the wire is a capacitor & has an initial field. But this duznt affect the early signal.
I insist that in theory there will be an early weak fast signal, but i grant that praps at 10 m (instead of the 1 m) i should have called it an early (very very) weak fast signal.
It would be easy to show that u are wrong. Lets place the switch at 5 m from the battery, then the distance switch to bulb on an angle would be say 5.1 m, & the early weak delay would be 5.1 m by 3.4 ns/m which is 17.0 ns. And your late strong delay would be 16 m along the wire at 3.4 ns/m which is 54.4 ns, or, if insulated, add 50% which makes it 81.6 ns.
I have been looking at the arguments re capacitors, & everyone has been wrong in every way.
I specifically mentioned that circuit is symmetrical. In this case both the charged and discharged capacitors are at the example same distance from the switch.
The so called small current when circuit is not symmetrical is the current to charge the capacitor made by the transmission line. In that case as it was in Derek's test your lamp is just an indicator to show the current used to charge that capacitor made from the transmission line wires.
As another example you can have the two capacitors circuit that I mentioned and add a lamp in series with the switch just next to the switch and you will see a current trough that lamp as soon as you close the switch but that current is provided by the energy stored in the transmission line so if you open the switch immediately after you closed it the charged in the transmission line will redistribute and the energy lamp used will be provided some nano seconds later by the main energy storage so that charged capacitor in this case.
Thus energy travels trough wires.
As long as your lamp is next to the switch it does not matter how far the battery is the lamp will see a small current immediately but that energy traveled trough wires and from wires (capacitor).
There is no evidence of energy traveling outside the wire just a wrong interpretation of the test result.
Capacitance/induction & induction/induction & radio/induction are all inductions.
Veritasium's 3.3 ns is due to induction (ie through the air), no matter whether u say that it is due to capacitance/induction or some other kind of induction.
However, Veritasium is of course wrong re his silly Poynting Vector having or transferring energy.
U & i agree that the main source of electric energy comes later via the wire (not via the silly Poynting field). But i don’t see how capacitance plays a part in that.
In any case the descriptions that i have seen re what happens with 2 capacitors in a circuit is all wrong.
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Capacitance/induction & induction/induction & radio/induction are all inductions.
Veritasium's 3.3 ns is due to induction (ie through the air), no matter whether u say that it is due to capacitance/induction or some other kind of induction.
However, Veritasium is of course wrong re his silly Poynting Vector having or transferring energy.
U & i agree that the main source of electric energy comes later via the wire (not via the silly Poynting field). But i don’t see how capacitance plays a part in that.
In any case the descriptions that i have seen re what happens with 2 capacitors in a circuit is all wrong.
All energy transfer from source to load/lamp is done done trough wires and none of it radiated unless you want to take in consideration that some IR radiated photon from the wires heated the lamp filament.
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Capacitance/induction & induction/induction & radio/induction are all inductions.
Veritasium's 3.3 ns is due to induction (ie through the air), no matter whether u say that it is due to capacitance/induction or some other kind of induction.
However, Veritasium is of course wrong re his silly Poynting Vector having or transferring energy.
U & i agree that the main source of electric energy comes later via the wire (not via the silly Poynting field). But i don’t see how capacitance plays a part in that.
In any case the descriptions that i have seen re what happens with 2 capacitors in a circuit is all wrong.
All energy transfer from source to load/lamp is done done trough wires and none of it radiated unless you want to take in consideration that some IR radiated photon from the wires heated the lamp filament.
The energy is in the wire(s). Then some of the energy transfers from wire to wire throo the air via the em field, the delay being 3.3 ns, as proven by every X that i have ever seen.
We agree that the full electricity arrives via the wires (ie we agree not via the silly Poynting field).
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The relativistic answer (convention) is that a momentum density p of matter corresponds to an energy flux of pc².
If you want general relativity then the answer is, well, complicated. https://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html
Classical physics is simple and says you have a potential energy of -GMm/r, you're not moving 0 energy, you're just converting: potential->kinetic->heat.
In the quasistatic idealization I had in mind, the mass takes an infinite time to travel from point A to point B. So, velocity is basically zero along the path and only changes during 'generation' (the machine raises the weight) and 'dissipation' (the weight does work against the gravitational field and it all becomes heat).
In my view the energy is in the gravitational field of the system planet + weight. It is being added to the system during 'generation' and it is extracted from the system during 'dissipation'. The system occupies all space, so does it even make sense to ask if the energy is traveling?
It does not occupy all space, because potential energy is the sum of -Gm1m2/r.[
quote author=bsfeechannel link=topic=322795.msg4156429#msg4156429 date=1651721925]
I find it funny and at the same time annoying that these people who are reluctant to properly study math and physics, and end up struggling especially with electromagnetism, like to talk in the name of engineers, as if their misconceptions were the general mindset of our class so as to justify their position.
The hydraulic analogy, the origin of the energy in the wire idea, was dismissed right from the start by Maxwell himself in the 19th century. Derek only made this incontestable knowledge available to the masses.
[/quote]
Isn't it interesting how every argument for 'energy in vacuum' is an argument from authority? :o
Also, Maxwell wrote:
https://en.wikisource.org/wiki/Page:A_Treatise_on_Electricity_and_Magnetism_-_Volume_1.djvu/102
" We are thus led to a very remarkable consequence of the theory which we are examining, namely, that the motions of electricity are like those of an incompressible fluid, so that the total quantity within an imaginary fixed closed surface remains always the same."
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The energy is in the wire(s). Then some of the energy transfers from wire to wire throo the air via the em field, the delay being 3.3 ns, as proven by every X that i have ever seen.
We agree that the full electricity arrives via the wires (ie we agree not via the silly Poynting field).
What do you mean by "full electricity"?
Do you understand what energy is ? And do you understand the conservation of energy ?
Main claim made by Darek and that is false is that "Energy doesn't flow in wires"
All 3 points that I made using the science accepted and used by everyone shows that is not true. Even just one of those points alone is enough to dismiss the claim Derek made.
And unless you are referring to IR from the wires no energy from the source travels outside the wire.
Electrons moving trough the wire did all the work. I think the two parallel capacitors explain best what happens and that all energy flows in the wire.
All is required is to understand what energy is how energy in a capacitors is calculated (the discharged capacitor is like a energy recording device) then all energy will be accounted for and the results shows very clearly that no energy flowed outside the wire other than in this example half of the total energy at the start that ended as heat due to wire resistance.
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The energy is in the wire(s). Then some of the energy transfers from wire to wire throo the air via the em field, the delay being 3.3 ns, as proven by every X that i have ever seen.
We agree that the full electricity arrives via the wires (ie we agree not via the silly Poynting field).
What do you mean by "full electricity"?
Do you understand what energy is ? And do you understand the conservation of energy ?
Main claim made by Darek and that is false is that "Energy doesn't flow in wires"
All 3 points that I made using the science accepted and used by everyone shows that is not true. Even just one of those points alone is enough to dismiss the claim Derek made.
And unless you are referring to IR from the wires no energy from the source travels outside the wire.
Electrons moving trough the wire did all the work. I think the two parallel capacitors explain best what happens and that all energy flows in the wire.
All is required is to understand what energy is how energy in a capacitors is calculated (the discharged capacitor is like a energy recording device) then all energy will be accounted for and the results shows very clearly that no energy flowed outside the wire other than in this example half of the total energy at the start that ended as heat due to wire resistance.
Me myself i reckon that the energy flows on the surface of the wires (electons)(& electrons), not in the wires (electrons).
Plus i reckon that there is a small amount of energy in the em field around a wire, but not enuff to justify some silly kind of Poynting field source.
In addition the em field around a wire transmits (can transmit) the energy of the electricity on the wire.
If electrons drift inside a wire then that is merely a drag effect, robbing energy. Energy duznt flow in wires.
The workings of capacitors can be explained by my electricity, but can't be explained by silly drifting electrons.
The full electricity arrives when the electons arrive.
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Me myself i reckon that the energy flows on the surface of the wires (electons)(& electrons), not in the wires (electrons).
Plus i reckon that there is a small amount of energy in the em field around a wire, but not enuff to justify some silly kind of Poynting field source.
In addition the em field around a wire transmits (can transmit) the energy of the electricity on the wire.
If electrons drift inside a wire then that is merely a drag effect, robbing energy. Energy duznt flow in wires.
The workings of capacitors can be explained by my electricity, but can't be explained by silly drifting electrons.
The full electricity arrives when the electons arrive.
You did not provide an answer to what "full electricity" means.
You seems to have guesses about the subject but no concrete mathematical or experimental proof.
This is not unknown science. Is just Derek that comes with ridiculous false claims thanks in large part due to his inability to understand what energy is and he exposes how many have the same gap in understanding (he has millions of views on his videos and multiple other creators make videos mostly agreeing with his points).
Energy flows in to a capacitor not trough a capacitor.
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Me myself i reckon that the energy flows on the surface of the wires (electons)(& electrons), not in the wires (electrons).
Plus i reckon that there is a small amount of energy in the em field around a wire, but not enuff to justify some silly kind of Poynting field source.
In addition the em field around a wire transmits (can transmit) the energy of the electricity on the wire.
If electrons drift inside a wire then that is merely a drag effect, robbing energy. Energy duznt flow in wires.
The workings of capacitors can be explained by my electricity, but can't be explained by silly drifting electrons.
The full electricity arrives when the electons arrive.
You did not provide an answer to what "full electricity" means.
You seems to have guesses about the subject but no concrete mathematical or experimental proof.
This is not unknown science. Is just Derek that comes with ridiculous false claims thanks in large part due to his inability to understand what energy is and he exposes how many have the same gap in understanding (he has millions of views on his videos and multiple other creators make videos mostly agreeing with his points).
Energy flows in to a capacitor not trough a capacitor.
Full electricity is the arrival of my electons (photons hugging the surface of the wire). If u prefer to believe in old electricity (ie drifting electrons) then it is the arrival of the wavefront of the electron to electron push-wave.
I agree that energy flows in to a capacitor when it is charging.
I reckon that energy duz flow through a capacitor – by virtue of induction across the gap.
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Point to me where I have posted a wrong equation.
Ex falso quodlibet, after a contradiction you can conclude whatever.
However, equations are not the question, here. Maxwell debunked the hydraulic analogy for the flow of energy based on experimental data. He noticed that a stretch of wire through which a current was flowing would tend to show a spike of voltage if interrupted at the point of the interruption, just like a pipe experiences sudden pressure at the point where a valve has interrupted the flow. However what hinted him that the energy didn't flow like a fluid in a pipe is that if you coil up the wire the voltage increases, while with a coiled up pipe you do not have this effect. Also, if you put a piece of iron inside the wire coil, the voltage will increase even more, but not the pressure for the pipe coil. Electricity flowing in wires affect other wires nearby not connected to each other either by attracting or repelling them if they too are conducting current, or inducing voltages if the currents vary.
Derek repeated the experiment Maxwell observed, only that, instead of opening the switch, he closed it. And he observed that, unlike a pipe, where the energy would never arrive at 1m/c, the electric energy does, and that the hydraulic analogy is what it is, a (should I say poor) analogy, not a description of How Electricity Actually Works.
So the experimental data debunks whatever theory you may have to sustain your misconception.
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Isn't it interesting how every argument for 'energy in vacuum' is an argument from authority? :o
Also, Maxwell wrote:
Contradiction, thy name is Naej.
https://en.wikisource.org/wiki/Page:A_Treatise_on_Electricity_and_Magnetism_-_Volume_1.djvu/102
" We are thus led to a very remarkable consequence of the theory which we are examining, namely, that the motions of electricity are like those of an incompressible fluid, so that the total quantity within an imaginary fixed closed surface remains always the same."
If you didn't take snippets of texts out of context like every pseudo-sciencer you would turn the page and find the following snippet:
https://en.wikisource.org/wiki/Page:A_Treatise_on_Electricity_and_Magnetism_-_Volume_1.djvu/103
The peculiar features of the theory as we have now developed them are: -
That the energy of electrification resides in the dielectric medium, whether that medium be solid, liquid, or gaseous, dense or rare, or even deprived of ordinary gross matter, provided it be still capable of transmitting electrical action.
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Point to me where I have posted a wrong equation.
Ex falso quodlibet, after a contradiction you can conclude whatever.
However, equations are not the question, here. Maxwell debunked the hydraulic analogy for the flow of energy based on experimental data. He noticed that a stretch of wire through which a current was flowing would tend to show a spike of voltage if interrupted at the point of the interruption, just like a pipe experiences sudden pressure at the point where a valve has interrupted the flow. However what hinted him that the energy didn't flow like a fluid in a pipe is that if you coil up the wire the voltage increases, while with a coiled up pipe you do not have this effect. Also, if you put a piece of iron inside the wire coil, the voltage will increase even more, but not the pressure for the pipe coil. Electricity flowing in wires affect other wires nearby not connected to each other either by attracting or repelling them if they too are conducting current, or inducing voltages if the currents vary.
Derek repeated the experiment Maxwell observed, only that, instead of opening the switch, he closed it. And he observed that, unlike a pipe, where the energy would never arrive at 1m/c, the electric energy does, and that the hydraulic analogy is what it is, a (should I say poor) analogy, not a description of How Electricity Actually Works.
So the experimental data debunks whatever theory you may have to sustain your misconception.
As any analogy there will be limitations but I was not mentioning the hydraulic model in my demonstration. https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/msg4156462/#msg4156462 (https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/msg4156462/#msg4156462)
Derek clearly has no understanding of what energy is and is not just based on this two videos related to "how electricity actually works".
Do you agree with this simple statement:
Energy flows in or out of a capacitor and not trough a capacitor.
I say and is easy to demonstrate that energy flows in and out of a capacitor while Derek that has no clue what energy is basically saying that energy flows trough a capacitor.
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Why not both?
Stick a 10mF capacitor in series with an AC load and tell me it doesn't transmit energy. :)
Tim
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Why not both?
Stick a 10mF capacitor in series with an AC load and tell me it doesn't transmit energy. :)
Tim
It will not. You will have a lot of heat loss in the conductors due to repeatably charging and discharging that capacitor.
No energy will get across the dielectric.
That is why there is no current with DC after the capacitor is charged and gets to same voltage as your source and if you disconnect the source and short the capacitor you will get all that energy you put in back as heat.
Edit: And if you attempt to do this test make sure to use a non polarized capacitor (likely much smaller value than 10mF as you do not need that). Use a resistor to limit the current.
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Whelp... time for another experiment I guess.
Ed: You're not just thinking parallel (shunt), right? You know what I mean by "series"?
Tim
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Whelp... time for another experiment I guess.
Ed: You're not just thinking parallel (shunt), right? You know what I mean by "series"?
Tim
If you want to do the experiment then in order to make things simpler for you just select the R and C so that capacitor is fully charged or almost fully charged with just half sine. That way you much easier can see what happens as measurements will be more accurate.
Yes of course resistor will be in series with capacitor in order to limit the current and be able to calculate the current by measuring the voltage drop across the resistor.
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2) If capacitors are 1m apart and the switch is 10m from both capacitors the time it takes from the moment the switch is closed to the moment any amount of energy is transferred to the discharged capacitor is the time it takes the electron wave to travel 10m.
This again is a proof that energy transfer from the source (charged capacitor) to the discharged capacitor travels trough wires.
If you put the switch at the end of the line then yes it will take the delay of the length of the line to start charging, exactly the same as Derek's original question would if the switch was at the end of the line. No difference at all, so that doesn't prove in any way that the energy flows inside the wire.
Yes it proves that since energy is not flowing from the switch to the load but from the source to the load. The position of the switch should not be relevant if the energy from the charged capacitor will not need to travel trough wire.
The position of the switch matters, otherwise that would break speed of light causality.
If anyone can't get their head around that then put the switch at a Derek inspired moon length. Flipping the switch at the moon can't instantly make something happen on earth, to do so obviously requires the speed of light to be infinite. This is wacko land.
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However, equations are not the question, here. Maxwell debunked the hydraulic analogy for the flow of energy based on experimental data. He noticed that a stretch of wire through which a current was flowing would tend to show a spike of voltage if interrupted at the point of the interruption, just like a pipe experiences sudden pressure at the point where a valve has interrupted the flow. However what hinted him that the energy didn't flow like a fluid in a pipe is that if you coil up the wire the voltage increases, while with a coiled up pipe you do not have this effect. Also, if you put a piece of iron inside the wire coil, the voltage will increase even more, but not the pressure for the pipe coil. Electricity flowing in wires affect other wires nearby not connected to each other either by attracting or repelling them if they too are conducting current, or inducing voltages if the currents vary.
Yes, this is how the hydraulic analogy is inaccurate. Which is why the word 'analogy' is used: it's very similar, not identical.
Derek repeated the experiment Maxwell observed, only that, instead of opening the switch, he closed it. And he observed that, unlike a pipe, where the energy would never arrive at 1m/c, the electric energy does, and that the hydraulic analogy is what it is, a (should I say poor) analogy, not a description of How Electricity Actually Works.
So the experimental data debunks whatever theory you may have to sustain your misconception.
The energy arrives 1m/c where c is the speed of sound in air. For the same reason: opening the valve emits a sound wave, because you're violently compressing/accelerating water.
Just as in the experiment proposed, it's a tiny amount of energy.
Also there's a pressure wave propagating inside the pipe, much like there's a potential wave propagating along the wire (telegrapher's equations also work for the pipe).
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Why not both?
Stick a 10mF capacitor in series with an AC load and tell me it doesn't transmit energy. :)
Tim
It will not. You will have a lot of heat loss in the conductors due to repeatably charging and discharging that capacitor.
No energy will get across the dielectric.
Yes it will. It's called the displacement current.
Displacement current is "flowing" anywhere , even in an insulator , where the electric field is varying with time.
https://en.wikipedia.org/wiki/Displacement_current
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The position of the switch matters, otherwise that would break speed of light causality.
If anyone can't get their head around that then put the switch at a Derek inspired moon length. Flipping the switch at the moon can't instantly make something happen on earth, to do so obviously requires the speed of light to be infinite. This is wacko land.
Dave, I do understand that. The point is that source (battery or charged capacitor) is 1m from the lamp/discharged capacitor on earth while switch is at the moon.
The transmission line is just a very long capacitor and it is charged as same potential like the the battery/charged capacitor. So when you connect a multimeter across the open switch at the moon you will not wait the time it takes electron wave propagation from earth to moon to measure the voltage it will be basically instant as the transmission line/capacitor is fully charged.
At the moment you close the switch first electron from one side of the switch will move on the other side collapsing the electric field associated with that electron imbalance. This will continue at the speed of light so if you where able to visualize the electric field line you will see how they collapse starting from the switch on the moon all the way down to earth at basically the speed of light and all that energy contained in the transmission line is radiated as infrared photos in the vacuum of space so ends up as heat.
Up to this point no energy left the battery/charged capacitor and no energy was delivered to bulb/discharged capacitor.
So when switch is closed the transmission line is a long capacitor with significant inductance and resistance.
No energy will flow trough the capacitor / transmission line but it will flow in or out. So all energy delivered to load flows in the wire and not outside of it unless you count the radiated heat in the form of infrared photons due to wire resistance and it is not even that if you are using a superconductor.
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Yes it will. It's called the displacement current.
Displacement current is "flowing" anywhere , even in an insulator , where the electric field is varying with time.
https://en.wikipedia.org/wiki/Displacement_current
This is the part you are referring to
"In physical materials (as opposed to vacuum), there is also a contribution from the slight motion of charges bound in atoms, called dielectric polarization."
We do not consider the limitations of dielectrics here as there is no such thing as a perfect dielectric but I'm fairly certain that Tim or you do not have the equipment to measure such small losses due to dielectric polarization.
And all examples where 3 to 20V with distance between the plates of 1m in air. Plus that energy is not flowing from one plate to the other of the capacitor. That is lost energy in the form of heat in the dielectric (air in case of a transmission line). This is insignificant compared to IR radiation from the wires and that IR also will not transfer any energy to the load unless you say that an IR photon will help a bit the filament of a lamp but will sure not charge a discharged capacitor.
When you charge a capacitor the energy remains in the capacitor (except for the one ending up as heat due to capacitor plate resistance) so energy did not flow trough the capacitor so trough dielectric on the other side but remained stored so that if you disconnect the capacitor and apply a load (say a resistor or lamp) you can get that energy out.
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Yes, this is how the hydraulic analogy is inaccurate. Which is why the word 'analogy' is used: it's very similar, not identical.
But it is brain-damaging engineers. It's time to get rid of it.
The energy arrives 1m/c where c is the speed of sound in air.
But not in vacuum.
For the same reason: opening the valve emits a sound wave, because you're violently compressing/accelerating water.
Just as in the experiment proposed, it's a tiny amount of energy.
The amount of energy is irrelevant. What you are not paying attention is that the energy that arrives first through space is not a spike of energy. It is a sustained continuous step and it does not disappear. When the rest of the energy flowing through space guided by the wires finally arrives, it is just added to the initial step.
Comparing Derek's and Alpha Phoenix's setup, you'll see that the step "duration" is proportional to the length of the line. If it were 300km, it would "last" 1s. If it were infinite, it would last as much as you'd want. So, it is not a "transient", it is "standient", that doesn't go away.
Another problem is that some people who do not accept that the energy flows through space, are prepared to understand that the Poynting vector, perpendicular to the surface of the load, really conveys the energy that it will dissipate. But they don't understand that the wires are also resistors, and that the Poynting vector, perpendicular to whatever surface of the wire, be it radial or axial, will cause that energy to be dissipated by the wire.
In other words, you cannot hand energy to a wire in one side and expect that this energy will reappear at the other side. The wire will dissipate it entirely.
You can prove that by connecting any piece of wire to the poles of a battery. The energy will not be transferred from the negative pole to the positive pole. It'll simply be dissipated. Completely.
Sounds ridiculous, but people do not connect the dots.
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dielectric polarization is not about losses. It is about the atoms in the dielectric having a polar electrostatic potential and in the presence of an electric field they mostly rotate to line up with
that field thereby strengthening the field.
Energy absolutely does flow through a capacitor.
https://en.wikipedia.org/wiki/Capacitive_power_supply
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Just to annoy those who dislike complex values, the dielectric "constant" or permittivity of material due to internal polarization with applied electric field is represented by a complex frequency-dependent value to include dielectric loss.
See: https://en.wikipedia.org/wiki/Permittivity for a detailed explanation.
The relationship with dielectric loss is discussed in https://en.wikipedia.org/wiki/Dielectric_loss
As is well known, the dielectric loss of typical insulators at reasonable frequencies is measurable: good plastics such as polypropylene have dielectric loss around 0.1%, but polyester (Mylar) is roughly 1%. PVC is much worse.
My anecdote (already posted here several times) about the late Professor U Fano (at the University of Chicago) lecturing on dielectric phenomena was about a quantum-mechanical calculation of the polarization in a medium with harmonically-bound electrons.
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dielectric polarization is not about losses. It is about the atoms in the dielectric having a polar electrostatic potential and in the presence of an electric field they mostly rotate to line up with
that field thereby strengthening the field.
Energy absolutely does flow through a capacitor.
https://en.wikipedia.org/wiki/Capacitive_power_supply
If energy does not flow trough a capacitor then it can only flow trough wire.
So what are we disagreeing on ?
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Another problem is that some people who do not accept that the energy flows through space, are prepared to understand that the Poynting vector, perpendicular to the surface of the load, really conveys the energy that it will dissipate. But they don't understand that the wires are also resistors, and that the Poynting vector, perpendicular to whatever surface of the wire, be it radial or axial, will cause that energy to be dissipated by the wire.
In other words, you cannot hand energy to a wire in one side and expect that this energy will reappear at the other side. The wire will dissipate it entirely.
You can prove that by connecting any piece of wire to the poles of a battery. The energy will not be transferred from the negative pole to the positive pole. It'll simply be dissipated. Completely.
Sounds ridiculous, but people do not connect the dots.
How will you charge a discharged capacitor from a charged one if you can not transfer energy trough a wire ?
Do you just keep the capacitors close and wait for energy to be transferred or do you connect them with wires ?
And yes part of the energy (not all) will end up as heat due to wire resistance. In the case of two identical capacitors one charged and one discharged exactly half of the energy will end up as heat with the other half of the energy remaining in the two capacitors in equal quantity so a quarter of the initial energy in each capacitor.
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How will you charge a discharged capacitor from a charged one if you can not transfer energy trough a wire ?
If you are charging or discharging, a wire will come in handy for the transfer of the charges.
Do you just keep the capacitors close and wait for energy to be transferred or do you connect them with wires ?
Have in mind that wires, capacitors, inductors, antennas, etc. are all conductors and that nature does not distinguish between them.
If you manage to put the plates of both capacitors in contact without the use of wires, the charges will redistribute and you'll have a loss of energy due to the current through the plates. And part of the rest, that was in the air between the plates of the first capacitor, will be transferred in between the plates of the second capacitor through the air.
Nice try, though.
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Capacitance/induction & induction/induction & radio/induction are all inductions.
Veritasium's 3.3 ns is due to induction (ie through the air), no matter whether u say that it is due to capacitance/induction or some other kind of induction.
However, Veritasium is of course wrong re his silly Poynting Vector having or transferring energy.
U & i agree that the main source of electric energy comes later via the wire (not via the silly Poynting field). But i don’t see how capacitance plays a part in that.
In any case the descriptions that i have seen re what happens with 2 capacitors in a circuit is all wrong.
All energy transfer from source to load/lamp is done trough wires and none of it radiated unless you want to take in consideration that some IR radiated photon from the wires heated the lamp filament.
It seems to me that u have defined electricity as being energy carried by wires. And then u say that electric energy can only be carried by wires.
It seems to me that u are ignoring wireless energy transfer.
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The very fact that in a given wire the electrons drift the same speed for the same current should be enough to conclusively prove that electrons are not carrying the energy, and the energy is in the fields.
1A @ 1V (1 J/s)through a 2mm diameter wire? anywhere along the wire 6.24 x 10^18 electrons will drift past, drifting at an average of 23 μm/s.
1A @ 1000V, (1000 J/s), the same 2mm diameter wire? anywhere along the wire 6.24 x 10^18 electrons will drift past, drifting at an average of 23 μm/s.
The motion of electrons is not transferring the energy - it does not change with energy being transferred.
Also, how much does that number of electrons weigh? (you can do the math if you like) - so little weight moving at such a slow speed transfers very close to zero kinetic energy. There is very little 'electron hammer' effect (analogous to water hammer).
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The very fact that in a given wire the electrons drift the same speed for the same current should be enough to conclusively prove that electrons are not carrying the energy, and the energy is in the fields.
1A @ 1V (1 J/s)through a 2mm diameter wire? anywhere along the wire 6.24 x 10^18 electrons will drift past, drifting at an average of 23 μm/s.
1A @ 1000V, (1000 J/s), the same 2mm diameter wire? anywhere along the wire 6.24 x 10^18 electrons will drift past, drifting at an average of 23 μm/s.
The motion of electrons is not transferring the energy - it does not change with energy being transferred.
Also, how much does that number of electrons weigh? (you can do the math if you like) - so little weight moving at such a slow speed transfers very close to zero kinetic energy. There is very little 'electron hammer' effect (analogous to water hammer).
Drifting electrons always weigh 1.7 kg per cubic m of Cu.
If drifting exists.
If electrons exist.
And i have my suspicions about wire.
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It will make no sense for me to answer to you individually (last 4 posts).
You all lack significant understanding about what energy is and what energy storage is.
Today is late and I do not feel like but tomorrow I will take that oscilloscope screenshot from Derek's last video and I will explain exactly what each part means and what it represents. Maybe that will be helpful but I doubt that.
Derek I think mentioned the same thing is his last video, that he can not see how if same number of electrons exit the battery as they enter then how any energy was transferred by them.
The answer is very simple. The electrons could only exit one side and enter the other because there was an imbalance between the two sides.
Analogies are never good but imagine a compressed air cylinder with two chambers divided by an elastic membrane (dielectric equivalent in a capacitor).
When this cylinder has the same amount of gas particle (any gas say Nitrogen) on each side of the membrane it contains no energy but if you use say mechanical energy to move with a pump molecules from one side to the other side then you have a device that stored energy equivalent to a charged capacitor or battery.
Now if you connect a pipe connecting the two sides/chambers of this cylinder you have the equivalent to connecting a wire between the terminals of a charged capacitor. In both cases the stored energy will end up as heat.
You can use that pressure differential to convert that stored energy in to something more useful than heat say mechanical energy by spinning a turbine and still the same number of molecules will exit one side of the cylinder and enter the other side.
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The very fact that in a given wire the electrons drift the same speed for the same current should be enough to conclusively prove that electrons are not carrying the energy, and the energy is in the fields.
1A @ 1V (1 J/s)through a 2mm diameter wire? anywhere along the wire 6.24 x 10^18 electrons will drift past, drifting at an average of 23 μm/s.
1A @ 1000V, (1000 J/s), the same 2mm diameter wire? anywhere along the wire 6.24 x 10^18 electrons will drift past, drifting at an average of 23 μm/s.
The motion of electrons is not transferring the energy - it does not change with energy being transferred.
Also, how much does that number of electrons weigh? (you can do the math if you like) - so little weight moving at such a slow speed transfers very close to zero kinetic energy. There is very little 'electron hammer' effect (analogous to water hammer).
Drifting electrons always weigh 1.7 kg per cubic m of Cu.
If drifting exists.
If electrons exist.
And i have my suspicions about wire.
So for 1m length of 2mm diameter copper wire (~30g of copper) that's about 5mg of drifting electrons.
At 1A those electrons are drifting at the average speed of about 23 μm/s, regardless of the if that wire is transferring 0.1W or 1000W...
The kinetic energy of those electrons is half of stuff all (or more formally 0.5 * 0.005g * 0.000023 m/s * 0.000023 m/s = 0.00000000000000132 J)
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It will make no sense for me to answer to you individually (last 4 posts).
You all lack significant understanding about what energy is and what energy storage is.
Today is late and I do not feel like but tomorrow I will take that oscilloscope screenshot from Derek's last video and I will explain exactly what each part means and what it represents. Maybe that will be helpful but I doubt that.
Derek I think mentioned the same thing is his last video, that he can not see how if same number of electrons exit the battery as they enter then how any energy was transferred by them.
The answer is very simple. The electrons could only exit one side and enter the other because there was an imbalance between the two sides.
Analogies are never good but imagine a compressed air cylinder with two chambers divided by an elastic membrane (dielectric equivalent in a capacitor).
When this cylinder has the same amount of gas particle (any gas say Nitrogen) on each side of the membrane it contains no energy but if you use say mechanical energy to move with a pump molecules from one side to the other side then you have a device that stored energy equivalent to a charged capacitor or battery.
Now if you connect a pipe connecting the two sides/chambers of this cylinder you have the equivalent to connecting a wire between the terminals of a charged capacitor. In both cases the stored energy will end up as heat.
You can use that pressure differential to convert that stored energy in to something more useful than heat say mechanical energy by spinning a turbine and still the same number of molecules will exit one side of the cylinder and enter the other side.
Ok, i am keen to see what u have to say about the scope screenshot. In the meantime allow me to soften u up re energy.
(1) Energy is a source of force. The creation of a force needs mass (2). And it (3) needs a medium & a process to transmit the energy or force & (4) it needs a medium & process to tap into that source.
There are only 4 kinds of force, (5) gravitational force, & (6) inertial force, & (7) electric (charge) force, & (8 ) magnetic force.
There is only one medium, it is (9) the aether.
(10) There are at least 8 kinds of aether process, we need 2 ovem here today, (10a) the bulk macro flow of aether (giving 5 & 6), & (10b) the micro excitation of aether (giving 7 & 8 ).
(11) Everything we feel & see is made of photons, the fundamental (quasi) particle. (12) Photons have mass. (13) Photons emit photaenos (14). (15) Photaenos have mass. (16) Photaenos give us the em field (giving us 7 & 8 ).
(17) Energy is stored in photons & (18 ) photaenos.
(19 ) Energy is stored in the kinetic movement of mass.
(20) Energy is stored in the position of mass (potential energy).
(21) Electric energy is stored in electons, ie photons that hug the surface of the wire. (22) Plus it is stored in the photaenos that are emitted as a part of every photon & electron (ie the em field). (23) Plus it is stored in free-ish surface electrons on a wire. (24) Plus it is stored in drifting electrons.
(25) Re the other 6 processes not mentioned in (10), 5 of these 6 are (26) the creation of aether, (27) the annihilation of aether, (28) the creation of photons, (29) the annihilation of photons. (30) the annihilation of photaenos. The remaining process is my secret.
(31) Energy duznt exist, what we have is (32) force. (30) Force duznt exist, what we have is mass & position.
(31) Mass (gravitational mass) duznt exist, what we have is the desire to change position.
(32) Mass (inertial mass) duznt exist, what we have is the resistance to any change of position.
(33) Everything is a process (of the aether).
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The very fact that in a given wire the electrons drift the same speed for the same current should be enough to conclusively prove that electrons are not carrying the energy, and the energy is in the fields.
1A @ 1V (1 J/s)through a 2mm diameter wire? anywhere along the wire 6.24 x 10^18 electrons will drift past, drifting at an average of 23 μm/s.
1A @ 1000V, (1000 J/s), the same 2mm diameter wire? anywhere along the wire 6.24 x 10^18 electrons will drift past, drifting at an average of 23 μm/s.
The motion of electrons is not transferring the energy - it does not change with energy being transferred.
Also, how much does that number of electrons weigh? (you can do the math if you like) - so little weight moving at such a slow speed transfers very close to zero kinetic energy. There is very little 'electron hammer' effect (analogous to water hammer).
Drifting electrons always weigh 1.7 kg per cubic m of Cu.
If drifting exists.
If electrons exist.
And i have my suspicions about wire.
So for 1m length of 2mm diameter copper wire (~30g of copper) that's about 5mg of drifting electrons.
At 1A those electrons are drifting at the average speed of about 23 μm/s, regardless of the if that wire is transferring 0.1W or 1000W...
The kinetic energy of those electrons is half of stuff all (or more formally 0.5 * 0.005g * 0.000023 m/s * 0.000023 m/s = 0.00000000000000132 J)
They might not carry much energy, but they can transmit lots of energy (from the source).
If drifting electrons exist.
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Ok, i am keen to see what u have to say about the scope screenshot. In the meantime allow me to soften u up re energy.
(1) Energy is a source of force. The creation of a force needs mass (2). And it (3) needs a medium & a process to transmit the energy or force & (4) it needs a medium & process to tap into that source.
There are only 4 kinds of force, (5) gravitational force, & (6) inertial force, & (7) electric (charge) force, & (8 ) magnetic force.
There is only one medium, it is (9) the aether.
I need to ask what is your qualification.
I will not bother to read the rest of what you wrote.
There is absolutely zero evidence that energy in this particular case can travel outside the wire.
I guess my only way to explain is some sort of mechanical analogies but you did not mentioned anything about my last analogy with compressed gas in a cylinder with two chambers separated by an elastic membrane.
If there are more molecules in one chamber then if they have a path they will like to move in the other chamber in order to get to the lowest energy state so equal amount of molecules in both chambers.
While as many molecules will enter the low pressure chamber as they will exit form the high pressure one there will be energy delivered by them.
Why will you think there will be any difference for electrons stored in a capacitor ?
Also why when transferring energy between two identical capacitors half of the energy is lost and using a thermal camera you can see where all that energy went and it is in conductors making it quite obvious that energy traveled trough wires and due to resistance half of transferred energy was lost.
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Ok, i am keen to see what u have to say about the scope screenshot. In the meantime allow me to soften u up re energy.
(1) Energy is a source of force. The creation of a force needs mass (2). And it (3) needs a medium & a process to transmit the energy or force & (4) it needs a medium & process to tap into that source.
There are only 4 kinds of force, (5) gravitational force, & (6) inertial force, & (7) electric (charge) force, & (8 ) magnetic force.
There is only one medium, it is (9) the aether.
I need to ask what is your qualification.
I will not bother to read the rest of what you wrote.
There is absolutely zero evidence that energy in this particular case can travel outside the wire.
I guess my only way to explain is some sort of mechanical analogies but you did not mentioned anything about my last analogy with compressed gas in a cylinder with two chambers separated by an elastic membrane.
If there are more molecules in one chamber then if they have a path they will like to move in the other chamber in order to get to the lowest energy state so equal amount of molecules in both chambers.
While as many molecules will enter the low pressure chamber as they will exit form the high pressure one there will be energy delivered by them.
Why will you think there will be any difference for electrons stored in a capacitor ?
Also why when transferring energy between two identical capacitors half of the energy is lost and using a thermal camera you can see where all that energy went and it is in conductors making it quite obvious that energy traveled trough wires and due to resistance half of transferred energy was lost.
I am a retired civil engineer.
That a half of the energy is lost during transfer is interesting, i don’t have a view re that, but it smells fishy.
I have kept out of capacitor to capacitor stuff here, except that i did say that almost everything said/written here has looked wrong to me. Remember i suggested u needed a 2nd switch.
I have agreed with u that the Poynting Field carries zero energy – u must be confusing me with someone else.
The gas cylinder with a membrane is a standard analogy for capacitors – something to do with waving away the need for displacement current or something.
Electrons can't cross from one plate to the other, they would have to jump through the air or whatever.
[G] If we shorted across the gap from one plate to the other then electrons would flow through the short, & would deliver energy (whatever that means). But this is never done.
[C] If we shorted around the circuit from one plate to the other then electrons would flow through the wire, & would deliver energy (whatever that means). This is always done.
In both cases no energy is transferred via the air.
But the amount of energy transfer will usually not be the same in [G] & [C].
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Let's return to Veritasium's original topic.
Consider the steady-state (DC) case where all the voltages and currents are constant. Also assume that the wires are perfect conductors. The electronics only carry a small amount of kinetic energy as they travel through the wires. Once they reach the resistor (lightbulb) they are accelerated by the electric field associated with the voltage across the resistor. In the process, electric potential energy of the electrons is converted to kinetic energy. This kinetic energy is, in turn, converted to thermal energy through collisions with atoms in the resistor.
Now let's consider the analogy of a spaceship returning to earth. It has both kinetic energy and gravitational potential energy. Both are converted to thermal energy as it reenters the atmosphere. So what carries the gravitational potential energy? Is it the spaceship or the Earth's gravitational field?
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Let's return to Veritasium's original topic.
Consider the steady-state (DC) case where all the voltages and currents are constant. Also assume that the wires are perfect conductors. The electronics only carry a small amount of kinetic energy as they travel through the wires. Once they reach the resistor (lightbulb) they are accelerated by the electric field associated with the voltage across the resistor. In the process, electric potential energy of the electrons is converted to kinetic energy. This kinetic energy is, in turn, converted to thermal energy through collisions with atoms in the resistor.
Why should electrons drift faster when they reach a resistor?
Why can't electrons go slower when they reach a resistor, but have more collisions or resistance than when going faster in the wire?
If electrons have zero resistance when in a perfect conductor then they must drift at almost c/1. If so then how could they accelerate when they get to the resistor?
Now let's consider the analogy of a spaceship returning to earth. It has both kinetic energy and gravitational potential energy. Both are converted to thermal energy as it reenters the atmosphere. So what carries the gravitational potential energy? Is it the spaceship or the Earth's gravitational field?
The gravitational potential energy is not carried by anything.
The existence of the spaceship has no effect on any gravitational field.
The spaceship's gravitational potential energy is available, & it has a certain value, but it is due to the spaceship's relative position to Earth [caveat] in the surrounding gravitational field including Earth's gravitational field.
In any case, if Earth's escape velocity is 11.8 km/s, & if the Sun's escape velocity at Earth's radius is 42 km/s, then the spaceship's direction of approach to Earth, & the spaceship's velocity would affect whether the Earth or the Sun were the major factor.
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I will reply to your post after you received the Noble prize.
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Remember that the drift of electrons (or holes in a semiconductor) can be measured using the Hall effect.
In solid-state physics, the solid structure imposes a mobility factor (in m2V s) on the charge carriers.
The conductivity of the medium is proportional to the product of mobility and carrier density.
In the article cited below, quite a few properties of the material contribute to the mobility: drifting electrons encounter an obstacle course as they try to follow the E field.
Note that the Hall effect can discern the polarity of the charge carriers: if both polarities be present, then there is a net effect.
For a summary, see https://en.wikipedia.org/wiki/Electron_mobility
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I posted this comment on Derek's video, slighly expanded:
Riddle me this, a new thought experiment: Let's go extreme and say you have a 100mm diameter copper conductor at pure steady state DC delivering a small amount of power to a pure resitive load, say 1W, but go as low as you want. No transients, no skin effect, no nothing, just pure steady state DC into a resistive load.
Is there NO energy WITHIN this comicly large wire? None? It's all on the OUTSIDE of the wire in the fields at DC? Really? REALLY?
The classicial field theory math might work at DC, but I just can't get over the feeling that it doesn't pass the sniff test at DC. I don't get The Vibe I get with AC and transients. Quantum Electrodynamics and probability theory in the electron fields within the wire better passes the sniff test at DC.
Can someone please convince me that there is no energy flow within this 100mm diameter wire at all, and that all the energy flows outside the wire at DC.
I was thinking about this in terms of a water analogy.
Let's suppose we have a large lake, with a weir on the far side, and the lake is full to the top of the weir. Now let's suppose we turn on a hose, and start adding a small stream of water on the opposite side of the lake. By and by, water will start flowing over the weir at the same rate it is being added from the hose.
Clearly water is flowing across the lake from the hose to the weir. However, if we try to measure any gradient in the surface level of the lake, we will likely fail--it will be quite level from one side to the other to the precision of our instruments. Similarly, if we try to measure any change in pressure throughout the lake we will also fail, it will once again be the same everywhere to the precision of our instruments.
Yet for the water to flow across the lake, there must be a driving force. There must be a pressure/level gradient, no matter how infinitesimal, for the water to flow.
Perhaps the analogy here with the surface electric charge would be the gravitational field. The level of the lake surface must be slightly higher on the hose side than on the weir side, and this change in level in the presence of a gravitational potential field leads to a driving force that moves the water across the lake. The slightly higher level will simultaneously lead to a slightly higher hydrodynamic pressure below the surface, which in this analogy would correspond to a small electric field inside the wire.
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I am a retired civil engineer.
That a half of the energy is lost during transfer is interesting, i don’t have a view re that, but it smells fishy.
I have kept out of capacitor to capacitor stuff here, except that i did say that almost everything said/written here has looked wrong to me. Remember i suggested u needed a 2nd switch.
I have agreed with u that the Poynting Field carries zero energy – u must be confusing me with someone else.
The gas cylinder with a membrane is a standard analogy for capacitors – something to do with waving away the need for displacement current or something.
Electrons can't cross from one plate to the other, they would have to jump through the air or whatever.
[G] If we shorted across the gap from one plate to the other then electrons would flow through the short, & would deliver energy (whatever that means). But this is never done.
[C] If we shorted around the circuit from one plate to the other then electrons would flow through the wire, & would deliver energy (whatever that means). This is always done.
In both cases no energy is transferred via the air.
But the amount of energy transfer will usually not be the same in [G] & [C].
There is no strange smell :) This is a particular case where both capacitors are identical and you charge one form another. You can imagine that a real resistor has a DC series resistance and for transient/AC impedance.
Since this are equal (identical capacitors) you will build a divider when you parallel them so if you measure symmetrical you will always measure 1/2 of the charged capacitor voltage from the time you parallel them to when they are in steady state so no more current flow.
If capacitors are not identical then more or less than half of the energy will be lost as heat the exact half is for the identical capacitor case.
Both G and C will convert the same amount of energy to heat.
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"That a half of the energy is lost during transfer is interesting, i don’t have a view re that, but it smells fishy."
You can do a very simple experiment about energy loss when connecting two capacitors.
Obtain (and measure) two 10 uF polypropylene capacitors. Make sure that each is discharged before proceeding.
Wire any available switch between the two capacitors. It's your choice: DPST or SPST and a wire.
Connect a 10 megohm voltmeter across one of them (time constant = 100 seconds), called C1.
Then, connect a reasonable and safe (say, 9 V battery) DC supply to that capacitor C1 and wait until the voltage stabilizes at voltage V1.
Disconnect the DC supply, note the voltage, close the switch, and note the voltage V2 before it decays due to the 10 megohm resistance.
The only reason to measure the capacitors before the test is to improve the accuracy.
Before closing the switch, the total energy in the system is (1/2) x C1 x V12.
After closing the switch, the total energy is (1/2) x (C1 + C2) x V22.
I don't need to repeat this test, since I already know the answer.
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I promised an explanation of what happens in the waveform captured by Derek.
Below with red is the voltage across the 1KOhm resistor with green is the supply voltage just after that R4 but it is an ideal voltage source so is basically the 20V supplied.
You can see that from simulation "LTspice" you get the same small power to resistor that is byproduct of charging the transmission line capacitance (I only simulated half the circuit as it is symmetrical).
Then at around 32ms current electron wave will travel half the line so at this time all line capacity is charged tho the part that charged first has already discharged not fully but significantly. Then as the electron travels on the other side of the transmission line the capacitance is being discharged starting with that far end continuing to maintain that limited amount of power for the lamp/resistor until about 65ns when the electron wave that left the source has finally arrived and rump up the lamp at full power (in fact slightly over as the remaining energy in capacitors is now basically fully discharged).
The inductors are also charged but at steady state they still contain significant amount of energy that will be delivered when switch is turned off.
[attach=1]
Below is the power graph (area under the graph represents energy).
Here you can see that significantly more power is provided by the source than what it ends up as heat on the lamp/resistor in the first 65ns and that difference is stored energy. The wires are much lower resistance so while there is some thermal energy wasted there it is insignificant compared to the 1K resistor.
[attach=2]
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I am a retired civil engineer.
That a half of the energy is lost during transfer is interesting, i don’t have a view re that, but it smells fishy.
I have kept out of capacitor to capacitor stuff here, except that i did say that almost everything said/written here has looked wrong to me. Remember i suggested u needed a 2nd switch.
I have agreed with u that the Poynting Field carries zero energy – u must be confusing me with someone else.
The gas cylinder with a membrane is a standard analogy for capacitors – something to do with waving away the need for displacement current or something.
Electrons can't cross from one plate to the other, they would have to jump through the air or whatever.
[G] If we shorted across the gap from one plate to the other then electrons would flow through the short, & would deliver energy (whatever that means). But this is never done.
[C] If we shorted around the circuit from one plate to the other then electrons would flow through the wire, & would deliver energy (whatever that means). This is always done.
In both cases no energy is transferred via the air.
But the amount of energy transfer will usually not be the same in [G] & [C].
There is no strange smell :) This is a particular case where both capacitors are identical and you charge one form another. You can imagine that a real resistor has a DC series resistance and for transient/AC impedance.
Since this are equal (identical capacitors) you will build a divider when you parallel them so if you measure symmetrical you will always measure 1/2 of the charged capacitor voltage from the time you parallel them to when they are in steady state so no more current flow.
If capacitors are not identical then more or less than half of the energy will be lost as heat the exact half is for the identical capacitor case.
Both G and C will convert the same amount of energy to heat.
Heat loss in G & C will depend on initial potentials. In the simplest case the heat loss would be equal, but in most cases it would not be equal (but this is a side issue)(not important).
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For the same reason: opening the valve emits a sound wave, because you're violently compressing/accelerating water.
Just as in the experiment proposed, it's a tiny amount of energy.
The amount of energy is irrelevant. What you are not paying attention is that the energy that arrives first through space is not a spike of energy. It is a sustained continuous step and it does not disappear. When the rest of the energy flowing through space guided by the wires finally arrives, it is just added to the initial step.
Comparing Derek's and Alpha Phoenix's setup, you'll see that the step "duration" is proportional to the length of the line. If it were 300km, it would "last" 1s. If it were infinite, it would last as much as you'd want. So, it is not a "transient", it is "standient", that doesn't go away.
Derek clearly talked about the earliest event and not what happened next, so I talked about it.
If you want to know what happens next, it's in the line after:
Also there's a pressure wave propagating inside the pipe, much like there's a potential wave propagating along the wire (telegrapher's equations also work for the pipe).
It certainly deserves to be separated because the physical behavior is quite different: in the first part, you emit/receive radio waves, and it is why you get the d/c delay, but quickly decays; after, you have capacitive coupling, with a much higher power/duration and no radio waves emitted.
In both cases, you observe the exact same thing with pipes (given the impedance of air/metal/water, there's usually a very good shielding) https://www.youtube.com/watch?v=oZbguheiVs4 (https://www.youtube.com/watch?v=oZbguheiVs4)
Another problem is that some people who do not accept that the energy flows through space, are prepared to understand that the Poynting vector, perpendicular to the surface of the load, really conveys the energy that it will dissipate. But they don't understand that the wires are also resistors, and that the Poynting vector, perpendicular to whatever surface of the wire, be it radial or axial, will cause that energy to be dissipated by the wire.
In other words, you cannot hand energy to a wire in one side and expect that this energy will reappear at the other side. The wire will dissipate it entirely.
Yes I can. Energy flows in wires and reappear in a lightbulb/engine/LED.
You can prove that by connecting any piece of wire to the poles of a battery. The energy will not be transferred from the negative pole to the positive pole. It'll simply be dissipated. Completely.
Sounds ridiculous, but people do not connect the dots.
Yes energy is transferred from the battery to the short, through the wire.
Saying it flows from the battery to the vacuum and then from the vacuum into the wire is not wrong, but it definitely sounds ridiculous.
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I promised an explanation of what happens in the waveform captured by Derek.
Below with red is the voltage across the 1KOhm resistor with green is the supply voltage just after that R4 but it is an ideal voltage source so is basically the 20V supplied.
You can see that from simulation "LTspice" you get the same small power to resistor that is byproduct of charging the transmission line capacitance (I only simulated half the circuit as it is symmetrical).
Then at around 32ms current electron wave will travel half the line so at this time all line capacity is charged tho the part that charged first has already discharged not fully but significantly. Then as the electron travels on the other side of the transmission line the capacitance is being discharged starting with that far end continuing to maintain that limited amount of power for the lamp/resistor until about 65ns when the electron wave that left the source has finally arrived and rump up the lamp at full power (in fact slightly over as the remaining energy in capacitors is now basically fully discharged).
The inductors are also charged but at steady state they still contain significant amount of energy that will be delivered when switch is turned off.
(Attachment Link)
Below is the power graph (area under the graph represents energy).
Here you can see that significantly more power is provided by the source than what it ends up as heat on the lamp/resistor in the first 65ns and that difference is stored energy. The wires are much lower resistance so while there is some thermal energy wasted there it is insignificant compared to the 1K resistor.
(Attachment Link)
Nice.
The full current bit is of course a no-brainer – ie the rise starting at 60.0/65.5 ns & finishing at 80.0 ns – plus the reflexion at 136.1 ns is standard stuff.
But i am very interested in the rise from 4.1 ns to 21.1/24.8 ns -- & the plateau of 3.7V at 24.8 ns, falling to 3.1V at 63.0/65.5 ns.
The 4.1 ns is only a ruff estimate, & it confirms Veritasium's claim of 3.3 ns (ie 1/c) from his first youtube gedanken.
So, can u get LTSpice to confirm the delay & rise & plateau in the first 65.5 ns?
In particular the weird angle of the initial rise from 4.1 ns to 21.1 ns.
AlphaPhoenix also got that there falling plateau in his AlphaPhoenix Xpt1 (pt2 yet to come)(if ever).
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Nice.
The full current bit is of course a no-brainer – ie the rise starting at 60.0/65.5 ns & finishing at 80.0 ns – plus the reflexion at 136.1 ns is standard stuff.
But i am very interested in the rise from 4.1 ns to 21.1/24.8 ns -- & the plateau of 3.7V at 24.8 ns, falling to 3.1V at 63.0/65.5 ns.
The 4.1 ns is only a ruff estimate, & it confirms Veritasium's claim of 3.3 ns (ie 1/c) from his first youtube gedanken.
So, can u get LTSpice to confirm the delay & rise & plateau in the first 65.5 ns?
In particular the weird angle of the initial rise from 4.1 ns to 21.1 ns.
AlphaPhoenix also got that there falling plateau in his AlphaPhoenix Xpt1 (pt2 yet to come)(if ever).
You have a capacitor with air as dielectric and very long plates (the wires).
That initial rise in voltage is due to capacitance of the transmission line not being charged all at once as the line is long and the electron wave needs time to get from one side to the other.
The transmission line is simulated as multiple inductors and capacitors but they are ideal so it is consider the distance between plates is infinitely small so there no 3.3ns or so delay to correspond with the 1m distance between the plates (wires) in the real example.
That can be added to simulation but is not the relevant part. If you bring the wires closer you will get rid of those 3.3ns (reduce that time significantly) but the energy transfer will be done in the exact same way.
All energy travels trough wires and that initial energy is due to capacitive coupling so you are charging the capacitor (energy flows in the capacitor being stored there for later used and does not flow trough capacitor).
If that was the case then you will not need to close the switch to transfer energy as the switch is also a capacitor.
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Nice.
The full current bit is of course a no-brainer – ie the rise starting at 60.0/65.5 ns & finishing at 80.0 ns – plus the reflexion at 136.1 ns is standard stuff.
But i am very interested in the rise from 4.1 ns to 21.1/24.8 ns -- & the plateau of 3.7V at 24.8 ns, falling to 3.1V at 63.0/65.5 ns.
The 4.1 ns is only a ruff estimate, & it confirms Veritasium's claim of 3.3 ns (ie 1/c) from his first youtube gedanken.
So, can u get LTSpice to confirm the delay & rise & plateau in the first 65.5 ns?
In particular the weird angle of the initial rise from 4.1 ns to 21.1 ns.
AlphaPhoenix also got that there falling plateau in his AlphaPhoenix Xpt1 (pt2 yet to come)(if ever).
You have a capacitor with air as dielectric and very long plates (the wires).
That initial rise in voltage is due to capacitance of the transmission line not being charged all at once as the line is long and the electron wave needs time to get from one side to the other.
The transmission line is simulated as multiple inductors and capacitors but they are ideal so it is consider the distance between plates is infinitely small so there no 3.3ns or so delay to correspond with the 1m distance between the plates (wires) in the real example.
That can be added to simulation but is not the relevant part. If you bring the wires closer you will get rid of those 3.3ns (reduce that time significantly) but the energy transfer will be done in the exact same way.
All energy travels trough wires and that initial energy is due to capacitive coupling so you are charging the capacitor (energy flows in the capacitor being stored there for later used and does not flow trough capacitor).
If that was the case then you will not need to close the switch to transfer energy as the switch is also a capacitor.
If the gap is small then nonetheless the capacitor sits in the middle of a 1000 mm long short, & the speed of electricity is c/1 in the short, hence LTSpice should still give at least 3.3 ns.
And, if modelled correctly, LTSpice should give the funny rise & the funny falling plateau.
Re the switch, this might be a capacitor, but before it is closed the circuit is in steady state.
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If the gap is small then nonetheless the capacitor sits in the middle of a 1000 mm long short, & the speed of electricity is c/1 in the short, hence LTSpice should still give at least 3.3 ns.
And, if modelled correctly, LTSpice should give the funny rise & the funny falling plateau.
Re the switch, this might be a capacitor, but before it is closed the circuit is in steady state.
A capacitor with a 1000mm gap between plates will require 3.3ns and one with 1mm gap will require 3.3ps
LTspice as far as I know has no option to specify the gap between plates and all comercial capacitors even the very high voltage ones will typically have much less than 1mm gap as the smaller the gap the larger the capacity.
So there is nothing strange or not understood about that 3.3ns.
When you close the switch exactly at that contact interface one electron will move from one side of the switch to the other and if a capacitor is in series in the loop as is the case in Derek's experiment then it takes 3.3ns for an electron on the other side of the 1m thick dielectric (air) to feel the effect and vacate a space.
In the battery the space is much smaller so when switch is closed and one electron moves on the other side of the switch the electron wave will travel in to battery say maybe 100mm from the switch then inside the battery the gap say is 1mm and so that electron that left from battery to switch will cause an electron from the wire connected on the other side of the battery to be accepted in the battery and so if total distance is 101mm in a straight line it may take as little as 0.33ns but likely it can not capacitively couple in straight line so it will be an ark maybe say 150mm so around 0.50ns then both sides are coupled at the same time with the wire above that is at about 1m thus 3.3nm + 0.5ns will be the total time from closing the switch until the first packet of energy is transferred from battery to the capacitor (transmission line is a capacitor) and since the lamp is in series between this two 1m capacitors all current used to charge the capacitor also will go through the lamp and be lost as heat and maybe photos.
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If the gap is small then nonetheless the capacitor sits in the middle of a 1000 mm long short, & the speed of electricity is c/1 in the short, hence LTSpice should still give at least 3.3 ns.
And, if modelled correctly, LTSpice should give the funny rise & the funny falling plateau.
Re the switch, this might be a capacitor, but before it is closed the circuit is in steady state.
A capacitor with a 1000mm gap between plates will require 3.3ns and one with 1mm gap will require 3.3ps
LTspice as far as I know has no option to specify the gap between plates and all comercial capacitors even the very high voltage ones will typically have much less than 1mm gap as the smaller the gap the larger the capacity.
So there is nothing strange or not understood about that 3.3ns.
When you close the switch exactly at that contact interface one electron will move from one side of the switch to the other and if a capacitor is in series in the loop as is the case in Derek's experiment then it takes 3.3ns for an electron on the other side of the 1m thick dielectric (air) to feel the effect and vacate a space.
In the battery the space is much smaller so when switch is closed and one electron moves on the other side of the switch the electron wave will travel in to battery say maybe 100mm from the switch then inside the battery the gap say is 1mm and so that electron that left from battery to switch will cause an electron from the wire connected on the other side of the battery to be accepted in the battery and so if total distance is 101mm in a straight line it may take as little as 0.33ns but likely it can not capacitively couple in straight line so it will be an ark maybe say 150mm so around 0.50ns then both sides are coupled at the same time with the wire above that is at about 1m thus 3.3nm + 0.5ns will be the total time from closing the switch until the first packet of energy is transferred from battery to the capacitor (transmission line is a capacitor) and since the lamp is in series between this two 1m capacitors all current used to charge the capacitor also will go through the lamp and be lost as heat and maybe photos.
Surely LTSpice is told the distance tween wires, ie 1000 mm. And the small gap tween plates can be ignored.
In which case LTSpice should be able to model the initial transient delay & rise & plateau.
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If the gap is small then nonetheless the capacitor sits in the middle of a 1000 mm long short, & the speed of electricity is c/1 in the short, hence LTSpice should still give at least 3.3 ns.
And, if modelled correctly, LTSpice should give the funny rise & the funny falling plateau.
Re the switch, this might be a capacitor, but before it is closed the circuit is in steady state.
A capacitor with a 1000mm gap between plates will require 3.3ns and one with 1mm gap will require 3.3ps
LTspice as far as I know has no option to specify the gap between plates and all comercial capacitors even the very high voltage ones will typically have much less than 1mm gap as the smaller the gap the larger the capacity.
So there is nothing strange or not understood about that 3.3ns.
When you close the switch exactly at that contact interface one electron will move from one side of the switch to the other and if a capacitor is in series in the loop as is the case in Derek's experiment then it takes 3.3ns for an electron on the other side of the 1m thick dielectric (air) to feel the effect and vacate a space.
In the battery the space is much smaller so when switch is closed and one electron moves on the other side of the switch the electron wave will travel in to battery say maybe 100mm from the switch then inside the battery the gap say is 1mm and so that electron that left from battery to switch will cause an electron from the wire connected on the other side of the battery to be accepted in the battery and so if total distance is 101mm in a straight line it may take as little as 0.33ns but likely it can not capacitively couple in straight line so it will be an ark maybe say 150mm so around 0.50ns then both sides are coupled at the same time with the wire above that is at about 1m thus 3.3nm + 0.5ns will be the total time from closing the switch until the first packet of energy is transferred from battery to the capacitor (transmission line is a capacitor) and since the lamp is in series between this two 1m capacitors all current used to charge the capacitor also will go through the lamp and be lost as heat and maybe photos.
Surely LTSpice is told the distance tween wires, ie 1000 mm. And the small gap tween plates can be ignored.
In which case LTSpice should be able to model the initial transient delay & rise & plateau.
Spice is not the sort of simulation you are imagining. The transmission line is done by adding a bunch of inductors and capacitors together to simulate those two wires.
And as mentioned that 3.3 or so ns delay at the beginning is irrelevant. I can add that delay but it will make no difference.
What you need to understand is that two parallel wires are a capacitor thus an energy storage device.
If you leave the ends open (the 1m pipes that connects the two parallel 10m pipes) the you will still have this first 65ns current through the lamp/resistor as the capacitor's one on each side will be charged but once they are charged you no longer have any current unless you reverse the battery polarity so that you discharge those capacitors and charge them in the other direction then it will again stop.
So energy flows in and out of capacitors and not through capacitors. Energy flows in to capacitor is the key as that energy is not doing work it is stored and just because wires have resistance and a lamp is in series it means that charging is not efficient so not all energy from the source will end up in capacitor as some will be lost to power the lamp/heat the resistor.
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Now let's consider the analogy of a spaceship returning to earth. It has both kinetic energy and gravitational potential energy. Both are converted to thermal energy as it reenters the atmosphere. So what carries the gravitational potential energy? Is it the spaceship or the Earth's gravitational field?
Gravitational potential energy is the additional energy the spaceship would gain if it were to travel a path between the two points used as the references.
It is not 'carried', it is transferred from the field as it makes the journey though the field.
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Now let's consider the analogy of a spaceship returning to earth. It has both kinetic energy and gravitational potential energy. Both are converted to thermal energy as it reenters the atmosphere. So what carries the gravitational potential energy? Is it the spaceship or the Earth's gravitational field?
Gravitational potential energy is the additional energy the spaceship would gain if it were to travel a path between the two points used as the references.
It is not 'carried', it is transferred from the field as it makes the journey though the field.
This seems to be getting out of subject but here is an explanation:
Have you seen the Einstein's explanation of gravity ?
It shows a straight fabric and when an object with mass is added it creates a valley
(https://media-cldnry.s-nbcnews.com/image/upload/newscms/2019_31/2951706/190729-earth-gravity-fabric-space-time-ac-1036p.jpg)
Now any other object with mass near the planet like the spaceship will also bend that fabric based on spaceship weight thus both planet and spaceship will slide closer to each other.
If there is no mass there is no gravitational field and similarly if there is no electron imbalance there is no electric field.
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This picture is widely understood to be an analogy. It does not represent what really happens, it is only a pictorial simplification.
The complication here is that general relativity indicates that you can have gravity without mass, and an internal observer experiencing such gravity cannot tell whether the experienced gravitational field is due to mass or otherwise.
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If the gap is small then nonetheless the capacitor sits in the middle of a 1000 mm long short, & the speed of electricity is c/1 in the short, hence LTSpice should still give at least 3.3 ns.
And, if modelled correctly, LTSpice should give the funny rise & the funny falling plateau.
Re the switch, this might be a capacitor, but before it is closed the circuit is in steady state.
A capacitor with a 1000mm gap between plates will require 3.3ns and one with 1mm gap will require 3.3ps
LTspice as far as I know has no option to specify the gap between plates and all comercial capacitors even the very high voltage ones will typically have much less than 1mm gap as the smaller the gap the larger the capacity.
So there is nothing strange or not understood about that 3.3ns.
When you close the switch exactly at that contact interface one electron will move from one side of the switch to the other and if a capacitor is in series in the loop as is the case in Derek's experiment then it takes 3.3ns for an electron on the other side of the 1m thick dielectric (air) to feel the effect and vacate a space.
In the battery the space is much smaller so when switch is closed and one electron moves on the other side of the switch the electron wave will travel in to battery say maybe 100mm from the switch then inside the battery the gap say is 1mm and so that electron that left from battery to switch will cause an electron from the wire connected on the other side of the battery to be accepted in the battery and so if total distance is 101mm in a straight line it may take as little as 0.33ns but likely it can not capacitively couple in straight line so it will be an ark maybe say 150mm so around 0.50ns then both sides are coupled at the same time with the wire above that is at about 1m thus 3.3nm + 0.5ns will be the total time from closing the switch until the first packet of energy is transferred from battery to the capacitor (transmission line is a capacitor) and since the lamp is in series between this two 1m capacitors all current used to charge the capacitor also will go through the lamp and be lost as heat and maybe photos.
Surely LTSpice is told the distance tween wires, ie 1000 mm. And the small gap tween plates can be ignored.
In which case LTSpice should be able to model the initial transient delay & rise & plateau.
Spice is not the sort of simulation you are imagining. The transmission line is done by adding a bunch of inductors and capacitors together to simulate those two wires.
And as mentioned that 3.3 or so ns delay at the beginning is irrelevant. I can add that delay but it will make no difference.
What you need to understand is that two parallel wires are a capacitor thus an energy storage device.
If you leave the ends open (the 1m pipes that connects the two parallel 10m pipes) the you will still have this first 65ns current through the lamp/resistor as the capacitor's one on each side will be charged but once they are charged you no longer have any current unless you reverse the battery polarity so that you discharge those capacitors and charge them in the other direction then it will again stop.
So energy flows in and out of capacitors and not through capacitors. Energy flows in to capacitor is the key as that energy is not doing work it is stored and just because wires have resistance and a lamp is in series it means that charging is not efficient so not all energy from the source will end up in capacitor as some will be lost to power the lamp/heat the resistor.
I agree that the 1000 mm end connection is irrelevant to the initial transients.
And i agree that parallel wires are a capacitor.
Old electricity i think says that electrons passing the switch will repel electrons from the other wire. Some of these will go left (towards the bulb) & some will go right.
Hence electrons will start to pass through the bulb at 3.3 ns.
The rise ends at 21.1/24.8 ns. This means that the electron wavefront passing the switch has propagated say 7.3 m at 3.3 ns/m
That propagation includes say 1 m crossing tween the wires, which leaves 6.3 m in the wires.
That 6.3 m is say 3.2 m along the primary wire, plus 3.1 m to the bulb along the secondary wire.
Hence the plateauing started when the wavefront was 3.2 m along past the switch. This is 1/3rd of the way to the end of that 10 m stretch of wire/tube.
At 24 ns the induction process of squeezing of electrons from the secondary wire reaches a peak value of 3.7V, after which the voltage across the bulb drops from 3.7v to 3.1V.
I think that the same kind of rise & plateau can be seen in the AlphaPhoenix-X.
I had a look. Brian's rise takes about 75 ns, for a gap of say 250 mm, ie ¼ of Veritasium's gap of 1000 mm.
AlphaPhoenix's L was say 250 m, & Veritasium's L was say 10 m.
AlphaPhoenix's plateaux went up not down, ie it went from say 1.8V up to 2.2V at say 1600 ns.
AlphaPhoenix's 75 ns indicates that the end of his rise was when the wavefront was 11.2 m past the switch.
Add 1 ns for the 250 mm gap, plus 11.2 m going back to the bulb, & we have our 75 ns.
I suspect that the height of the wires & tubes above the dirt would affect the rises etc.
Anyhow, any lumped element transmission line model worth its salt will surely give these kinds of numbers.
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This picture is widely understood to be an analogy. It does not represent what really happens, it is only a pictorial simplification.
The complication here is that general relativity indicates that you can have gravity without mass, and an internal observer experiencing such gravity cannot tell whether the experienced gravitational field is due to mass or otherwise.
As any analogy it has limitations.
The fact remains that energy cannot be transferred by the electric field else capacitors will not exist.
It all reducess to whatever you think energy flows through a capacitor (demonstrably untrue) or flows in and out of a capacitor.
All you need to check this is take a battery a lamp and a capacitor and connect all of them in series. Depending how large is the capacitor (in therms of energy storage capacity) you may see the lamp light up as energy is transferred from the battery to the capacitor to be charged but once the capacitor is at same potential as the battery meaning it can not accept any more energy the current flow will stop and so no energy transfer.
Then by removing the battery and only connecting the lamp and capacitor in a loop the energy that was stored can be recovered and lamp will be illuminated for the same amount of time as it was when capacitor was charged.
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It all reducess to whatever you think energy flows through a capacitor (demonstrably untrue) or flows in and out of a capacitor.
Oh? What will you say about that?
All you need to check this is take a battery a lamp and a capacitor and connect all of them in series. Depending how large is the capacitor (in therms of energy storage capacity) you may see the lamp light up as energy is transferred from the battery to the capacitor to be charged but once the capacitor is at same potential as the battery meaning it can not accept any more energy the current flow will stop and so no energy transfer.
Then by removing the battery and only connecting the lamp and capacitor in a loop the energy that was stored can be recovered and lamp will be illuminated for the same amount of time as it was when capacitor was charged.
Congratulations. You just demonstrated that energy flows through the capacitor.
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The fact remains that energy cannot be transferred by the electric field else capacitors will not exist.
It all reducess to whatever you think energy flows through a capacitor (demonstrably untrue) or flows in and out of a capacitor.
All you need to check this is take a battery a lamp and a capacitor and connect all of them in series. Depending how large is the capacitor (in therms of energy storage capacity) you may see the lamp light up as energy is transferred from the battery to the capacitor to be charged but once the capacitor is at same potential as the battery meaning it can not accept any more energy the current flow will stop and so no energy transfer.
Then by removing the battery and only connecting the lamp and capacitor in a loop the energy that was stored can be recovered and lamp will be illuminated for the same amount of time as it was when capacitor was charged.
Are you saying that if you have a lamp between two capacitors (so all three components are in series) that it is impossible to get the lamp to glow?
Because that would be "energy passing through the capacitors" to me. If not, where did the energy that makes the lamp glow come from?
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Congratulations. You just demonstrated that energy flows through the capacitor.
You will need to understand what energy and energy storage is to make that claim.
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Are you saying that if you have a lamp between two capacitors (so all three components are in series) that it is impossible to get the lamp to glow?
Because that would be "energy passing through the capacitors" to me. If not, where did the energy that makes the lamp glow come from?
I was mentioning a battery but yes if one of the capacitors is charged then yes.
It will only glow for a few moments and stop as soon as the two capacitors become equally charged. There is still energy to make the lamp glow in the two capacitors but they can not do that as energy will not flow through them.
There is no current flow through a dielectric (if there is then something is wrong and that is no longer a dielectric).
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Are you saying that if you have a lamp between two capacitors (so all three components are in series) that it is impossible to get the lamp to glow?
Because that would be "energy passing through the capacitors" to me. If not, where did the energy that makes the lamp glow come from?
I was mentioning a battery but yes if one of the capacitors is charged then yes.
It will only glow for a few moments and stop as soon as the two capacitors become equally charged. There is still energy to make the lamp glow in the two capacitors but they can not do that as energy will not flow through them.
There is no current flow through a dielectric (if there is then something is wrong and that is no longer a dielectric).
I've put a pair of LEDs between two caps, with a current limiting resistor. Where is the energy coming from that is making these LED glow?
I've used 2 LEDs so you can be sure that I am not playing any funny games with AC. One glows when DC voltage is applied, the other when the caps are discharged after DC is removed.
https://www.youtube.com/watch?v=5qdDqOwGomY (https://www.youtube.com/watch?v=5qdDqOwGomY)
I can repeat the cycle over and over, so more energy is getting to the LEDs somehow, and charges are moving through the LEDs... but you say this energy is not going through the capacitors?
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It certainly deserves to be separated because the physical behavior is quite different:
Separated.
Since the 19th century, everybody thought that everything electric and magnetic, from DC to cosmic rays, through radio waves, heat, light, ultraviolet, X-rays and whatnot, is the manifestation of the same freaking physical phenomenon.
Now you're saying that they are different. I wonder why the Nobel Committee has not noticed you yet.
Yes I can. Energy flows in wires and reappear in a lightbulb/engine/LED.
Yes energy is transferred from the battery to the short, through the wire.
If you say so, it must be true.
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Considering capacitors is sure interesting, considering inductive coupling as well.
We know we can transfer energy without "wires". That's for sure. But what would the fundamental difference be? If there is any?
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I've put a pair of LEDs between two caps, with a current limiting resistor. Where is the energy coming from that is making these LED glow?
I've used 2 LEDs so you can be sure that I am not playing any funny games with AC. One glows when DC voltage is applied, the other when the caps are discharged after DC is removed.
I can repeat the cycle over and over, so more energy is getting to the LEDs somehow, and charges are moving through the LEDs... but you say this energy is not going through the capacitors?
Use a battery. You have a low quality power supply that has a lot of ripple thus your capacitors are charged and discharged continuously.
Is basically an AC supply with a DC offset.
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It certainly deserves to be separated because the physical behavior is quite different:
Separated.
Since the 19th century, everybody thought that everything electric and magnetic, from DC to cosmic rays, through radio waves, heat, light, ultraviolet, X-rays and whatnot, is the manifestation of the same freaking physical phenomenon.
Now you're saying that they are different. I wonder why the Nobel Committee has not noticed you yet.Yes I can. Energy flows in wires and reappear in a lightbulb/engine/LED.
Yes energy is transferred from the battery to the short, through the wire.
If you say so, it must be true.
I am not sure about the gist of this argument, but, radio waves are not photons, & photons are not radio waves.
Radio waves are em radiation.
Photons are photons, they emit em radiation.
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I've put a pair of LEDs between two caps, with a current limiting resistor. Where is the energy coming from that is making these LED glow?
I've used 2 LEDs so you can be sure that I am not playing any funny games with AC. One glows when DC voltage is applied, the other when the caps are discharged after DC is removed.
I can repeat the cycle over and over, so more energy is getting to the LEDs somehow, and charges are moving through the LEDs... but you say this energy is not going through the capacitors?
Use a battery. You have a low quality power supply that has a lot of ripple thus your capacitors are charged and discharged continuously.
Is basically an AC supply with a DC offset.
My admittedly low quality PSU, replaced with a 9V battery. Same result.
https://www.youtube.com/watch?v=KYGzFRjagkY (https://www.youtube.com/watch?v=KYGzFRjagkY)
Also if it was an AC supply, both LEDs would glow.
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My admittedly low quality PSU, replaced with a 9V battery. Same result.
Also if it was an AC supply, both LEDs would glow.
Not sure what you are doing there. Where is the black cable connected ? I see just what looks like a red banana plug that you touch on either positive or negative of a 9V battery.
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My admittedly low quality PSU, replaced with a 9V battery. Same result.
Also if it was an AC supply, both LEDs would glow.
Not sure what you are doing there. Where is the black cable connected ? I see just what looks like a red banana plug that you touch on either positive or negative of a 9V battery.
Black cable is held against the negative of the 9V battery.
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Black cable is held against the negative of the 9V battery.
I see so you where charging and discharging the capacitors. LED's can glow at a few uA so it will take quite some time for them to dim as capacitors charge and those capacitors may have leakage current in that region.
I just tested with a large 4700uF 35Vdc new capacitor and my small white LED had a small glow even after 30 to 50 seconds (did not count) the same level of glow it has if it conducts through my finger so I suspect a few nA
I never used electrolytic capacitors in any of my projects and did a quick search and it seems they do have some leakage current due to the type of electrolyte they use. Water based ones have the largest leakage but all have even a few uA much more than nA modern LED's can glow at.
So I learned something new and that is that electrolytics are not just capacitors but also a parallel resistor.
There are good reason I never consider electrolytics for my projects (I do not like things with finite life).
Edit: This is the exact one that I just used https://www.mouser.ca/ProductDetail/United-Chemi-Con/EGPA350ELL472MM40S?qs=beQ1fBGcmj2M%2Fui1vdONPg%3D%3D (https://www.mouser.ca/ProductDetail/United-Chemi-Con/EGPA350ELL472MM40S?qs=beQ1fBGcmj2M%2Fui1vdONPg%3D%3D)
Looking at the spec it has 4uA of leakage current so fairly significant and this is not a non brand and purchased from mouser.
Is just there on first page and that 4uA is best case can bemore https://www.mouser.ca/datasheet/2/420/GPALL_e-2509122.pdf (https://www.mouser.ca/datasheet/2/420/GPALL_e-2509122.pdf)
In my particular case should be about 705uA based on use case 8V battery 3V on LED so 5V on the capacitor 5V * 4700uF * 0.03 but mine was no where near close to that probably is worse case and was more like 10uA max based on almost invisible glow.
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Black cable is held against the negative of the 9V battery.
I see so you where charging and discharging the capacitors. LED's can glow at a few uA so it will take quite some time for them to dim as capacitors charge and those capacitors may have leakage current in that region.
I just tested with a large 4700uF 35Vdc new capacitor and my small white LED had a small glow even after 30 to 50 seconds (did not count) the same level of glow it has if it conducts through my finger so I suspect a few nA
I never used electrolytic capacitors in any of my projects and did a quick search and it seems they do have some leakage current due to the type of electrolyte they use. Water based ones have the largest leakage but all have even a few uA much more than nA modern LED's can glow at.
So I learned something new and that is that electrolytics are not just capacitors but also a parallel resistor.
There are good reason I never consider electrolytics for my projects (I do not like things with finite life).
Edit: This is the exact one that I just used https://www.mouser.ca/ProductDetail/United-Chemi-Con/EGPA350ELL472MM40S?qs=beQ1fBGcmj2M%2Fui1vdONPg%3D%3D (https://www.mouser.ca/ProductDetail/United-Chemi-Con/EGPA350ELL472MM40S?qs=beQ1fBGcmj2M%2Fui1vdONPg%3D%3D)
Looking at the spec it has 4uA of leakage current so fairly significant and this is not a non brand and purchased from mouser.
Is just there on first page and that 4uA is best case can bemore https://www.mouser.ca/datasheet/2/420/GPALL_e-2509122.pdf (https://www.mouser.ca/datasheet/2/420/GPALL_e-2509122.pdf)
In my particular case should be about 705uA based on use case 8V battery 3V on LED so 5V on the capacitor 5V * 4700uF * 0.03 but mine was no where near close to that probably is worse case and was more like 10uA max based on almost invisible glow.
It is a good glow - I put a meter in line with the 10k resistor - current starts out at well over half a mA, falling off over time (as expected). I just left it to settle down to steady-state, and there is no discernable glow, at least during the day, so it isn't leakage.
Actually... I'll just go measure the leakage then update this post... leakage is less than a microamp.
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Actually... I'll just go measure the leakage then update this post.
Please do as there is sure a leakage. Those red LED's may not be as efficient as the modern white ones.
The white one I have is visible at 1uA and that is about where it settles with that 4700uF cap after about one minute.
You have two capacitors in series and seems like lower capacity not sure what voltage rating but there will be a leakage with any electrolytic.
In any case they are perfectly fine for testing things like charge one cap with the other directly or with a small DC-DC with constant current control.
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Yea nah, that's a mA or so. Clearly not leakage.
It also clearly fades over time, as the capacitor(s) charge; exactly as conventional theory would predict. How odd... :)
Tim
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Circuit theory isn't sufficient to solve the two-capacitor problem. We need to take the electrodynamic behavior of the system into account.
So let's do the thought experiment of connecting a charged and a discharged capacitor in parallel at t=0s. We assume that all the conductors are perfect and that the two capacitors are of the parallel plate type with vacuum between the plates. The two capacitors can be modelled as two lossless transmission lines (see for instance House and Melcher example 14.2.1). After the two capacitors are connected in parallel, an electromagnetic wave will bounce back and forth between the two capacitors. The amount of energy stored in each capacitor will fluctuate, but the total amount of stored energy in the two capacitors will remain the same.
If we go to the next level and take electromagnetic radiation into account, then al the energy will be radiated into space over time.
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So let's do the thought experiment of connecting a charged and a discharged capacitor in parallel at t=0s.
May I suggest an alternative thought experiment?
Assume lossless capacitors and conductors. Assume two capacitors of equal capacitance, C. Assume one of the capacitors is discharged. Assume that the other capacitor is charged to the exact value 1.602176634×10−19 Coulombs (the charge on an electron). Assume that the charged capacitor has a voltage of 1V, equivalent to 1eV Joules of energy. When the switch is closed, the electron migrates from the initially-charged capacitor to the initially-discharged capacitor. The electron then migrates back to the initially-charged capacitor. Oscillation occurs. Is my alternative thought experiment valid?
If those two capacitors are identical (same capacity) the voltage after switch is closed will be 0.707 * Vi
I was thinking further with this idea of 70.7% of Vi per capacitor after the switch is closed and noted a disparity regarding the issue of time relating to current flow between the two capacitors and the voltages across the capacitors. If both capacitors are 1F and the charged capacitor is charged to 3V then the initial charge is 3 Coulombs.
If the initially-charged capacitor discharges from 3V to 2.121V then the change in voltage is -29.3% of Vi. If the initially-discharged capacitor charges from 0V to 2.121V then the change in voltage is +70.7% of Vi.
If both capacitors have 70.7% of Vi after the same time period has past then how did the initially-discharged capacitor charge up faster from 0V to 70.7% of Vi than the initially-charged up capacitor discharge from 100% to 70.7% of Vi? This would violate Kirchhoff's Current Law where the current flowing into the node is not equal to the current flowing out of the node. In this case the node is located between the capacitors.
The current flowing out of the initially-charged capacitor must be equal in magnitude to the current flowing into the initially-discharged capacitor. The initially-discharged capacitor would need to charge from 29.3% of Vi to 70.7% of Vi in a time period of zero seconds. This is impossible.
I have seen a similiar problem in a YouTube video.
https://youtu.be/q398AqtTEL8
\$I = C \frac{dV}{dt}\$
\$I.dt = C.dV\$
Time taken for capacitor to charge/dischage to target voltage:
\$dt = C\frac{dV}{I}\$
For reference:
+70.7% \$V_i \approx \frac{V_i}{\sqrt{2}}\$
-29.3% \$V_i \approx \frac{-V_i{(\sqrt{2}} - 1)}{\sqrt{2}}\$
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Circuit theory isn't sufficient to solve the two-capacitor problem. We need to take the electrodynamic behavior of the system into account.
So let's do the thought experiment of connecting a charged and a discharged capacitor in parallel at t=0s. We assume that all the conductors are perfect and that the two capacitors are of the parallel plate type with vacuum between the plates. The two capacitors can be modelled as two lossless transmission lines (see for instance House and Melcher example 14.2.1). After the two capacitors are connected in parallel, an electromagnetic wave will bounce back and forth between the two capacitors. The amount of energy stored in each capacitor will fluctuate, but the total amount of stored energy in the two capacitors will remain the same.
If we go to the next level and take electromagnetic radiation into account, then al the energy will be radiated into space over time.
The circuit theory is perfectly capable in solving the two parallel capacitor problem.
With ideal conductors so no resistance energy stored in the two capacitors at the end of the experiment will be the same as at the beginning just split between the two capacitors.
With the example I used where capacitors are 1F and charged capacitor starts with 3V so 4.5Ws of stored energy the end result will be that each capacitor will contain 2.25Ws meaning 2.121V across each capacitor.
That is in contrast to normal wires where you have resistance and in that case energy at the end of the experiment in the two capacitors will be just half of the energy at the beginning so just 1.125Ws in each capacitor the other half of the energy 2.25Ws will be lost in the wires as heat because energy travels through wires and wires have resistance.
While yes there will be some bounce both in the experiment with resistance and the one with superconductors the speed of electron wave is finite and there will be reflexions at the open ends so voltage will stabilize and not forever slush around as you imagine as reflected waves will interact and in time cancel each other.
You also probably imagine that inductance and capacitance will match but that is not the case in a capacitor where capacitance is much higher than inductance same ways as for an inductor inductance is much higher than capacitance.
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If those two capacitors are identical (same capacity) the voltage after switch is closed will be 0.707 * Vi
I was thinking further with this idea of 70.7% of Vi per capacitor after the switch is closed and noted a disparity regarding the issue of time relating to current flow between the two capacitors and the voltages across the capacitors. If both capacitors are 1F and the charged capacitor is charged to 3V then the initial charge is 3 Coulombs.
If the initially-charged capacitor discharges from 3V to 2.121V then the change in voltage is -29.3% of Vi. If the initially-discharged capacitor charges from 0V to 2.121V then the change in voltage is +70.7% of Vi.
If both capacitors have 70.7% of Vi after the same time period has past then how did the initially-discharged capacitor charge up faster from 0V to 70.7% of Vi than the initially-charged up capacitor discharge from 100% to 70.7% of Vi? This would violate Kirchhoff's Current Law where the current flowing into the node is not equal to the current flowing out of the node. In this case the node is located between the capacitors.
The current flowing out of the initially-charged capacitor must be equal in magnitude to the current flowing into the initially-discharged capacitor. The initially-discharged capacitor would need to charge from 29.3% of Vi to 70.7% of Vi in a time period of zero seconds. This is impossible.
I have seen a similiar problem in a YouTube video.
https://youtu.be/q398AqtTEL8
\$I = C \frac{dV}{dt}\$
\$I.dt = C.dV\$
Time taken for capacitor to charge/dischage to target voltage:
\$dt = C\frac{dV}{I}\$
For reference:
+70.7% \$V_i \approx \frac{V_i}{\sqrt{2}}\$
-29.3% \$V_i \approx \frac{-V_i{(\sqrt{2}} - 1)}{\sqrt{2}}\$
You are confusing charge with energy and charge is not energy.
A capacitor is the same as a transmission line and vice versa so there is both inductance and capacitance just a different rates. You can consider each electron and hole pair a capacitor so that first electron that will move from the charged capacitor forms another small capacitor on the other side of you perfect middle point where you have the ideal meter. And while that is fast is not infinitely fast and it is just the beginning as the electron wave needs to travel down to the end of the capacitor.
If you better understand rechargeable batteries then better convert Coulombs to mAh by dividing to 3600 seconds.
Now if you say that the capacitor contains 3C/3600 = 0.833mAh the you will not call this energy instead you will need to multiply this with the average voltage or integrate if voltage will not drop linear like on a capacitor.
So in this case 3V/2 = 1.5V * 0.833mAh = 1.25mWh and this is the energy contained in the capacitor.
1.25mWh * 3600 = 4.5Ws
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Circuit theory isn't sufficient to solve the two-capacitor problem. We need to take the electrodynamic behavior of the system into account.
Actually, we don't. It is only necessary to look at the starting and ending states and apply basic principles of physics (no circuit theory required).
At the start we have one charged capacitor with a charge of \$Q = C_a V_0\$ and a second capacitor \$C_b\$ with a charge of zero. The system is fully insulated and no charge can leak, meaning we can apply conservation of charge between any two states.
After we connect the two capacitors in parallel the charge is distributed between them until in the end both have equal voltage. At this point we have \$Q = C_a V + C_b V\$
If both capacitors are identical (\$C_a = C_b\$) we can write \$CV_0 = 2CV\$ and the \$C\$ cancels giving
$$V = \frac{V_0}{2}$$
Thus the final voltage is half the initial voltage.
It is not necessary to consider what path is taken between the initial state and the final state if it is given that any path followed will satisfy the conservation of charge.
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Circuit theory isn't sufficient to solve the two-capacitor problem. We need to take the electrodynamic behavior of the system into account.
Actually, we don't. It is only necessary to look at the starting and ending states and apply basic principles of physics (no circuit theory required).
At the start we have one charged capacitor with a charge of \$Q = C_a V_0\$ and a second capacitor \$C_b\$ with a charge of zero. The system is fully insulated and no charge can leak, meaning we can apply conservation of charge between any two states.
After we connect the two capacitors in parallel the charge is distributed between them until in the end both have equal voltage. At this point we have \$Q = C_a V + C_b V\$
If both capacitors are identical (\$C_a = C_b\$) we can write \$CV_0 = 2CV\$ and the \$C\$ cancels giving
$$V = \frac{V_0}{2}$$
Thus the final voltage is half the initial voltage.
It is not necessary to consider what path is taken between the initial state and the final state if it is given that any path followed will satisfy the conservation of charge.
You also seems not to understand conservation of energy.
Charge is not conserved energy is.
This is just a special case where you selected two identical capacitors and you have resistance that is why you get exactly half the voltage at the end but that is just a quarter of initial energy in each capacitor so just half the energy you started with. The other half of the energy ended up as heat.
If you add an inductor as an intermediary energy storage you can move the energy from one capacitor to another at much higher efficiency than 50% maybe as high as 90% then the final voltage on those two capacitors will be much closer to 0.707 of the charged capacitor and so much less energy will be wasted as heat.
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When applying circuit theory to the problem, either the law of conservation of charge or the law of conservation of energy is violated.
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When applying circuit theory to the problem, either the law of conservation of charge or the law of conservation of energy is violated.
There is no such thing as conservation of charge. The conservation of energy is a law.
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When applying circuit theory to the problem, either the law of conservation of charge or the law of conservation of energy is violated.
There is no such thing as conservation of charge. The conservation of energy is a law.
Look at section 1.5 of Haus and Melcher.
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When applying circuit theory to the problem, either the law of conservation of charge or the law of conservation of energy is violated.
If the two capacitors form a closed system then charge is conserved within that system. The model then indicates that electrical energy is not conserved, but this says nothing about total energy. Energy could be converted to other forms such as heat, or it could cross the system boundaries and leave. Neither of these possibilities violate causality.
If you look at the situation from a physics perspective, the connecting together of the two capacitors is an irreversible process, which therefore causes an increase in entropy. This increase in entropy translates to a decrease in the free energy of the system, which means you can get less useful work out of the two capacitors after you have joined them than you could get out of the one capacitor at the start.
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Look at section 1.5 of Haus and Melcher.
??? Never heard of them before now and likely for a good reason.
People seems to confuse charge with energy.
Energy is the one that can not be created or destroyed just converted from one form to another.
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When applying circuit theory to the problem, either the law of conservation of charge or the law of conservation of energy is violated.
If the two capacitors form a closed system then charge is conserved within that system. The model then indicates that electrical energy is not conserved, but this says nothing about total energy. Energy could be converted to other forms such as heat, or it could cross the system boundaries and leave. Neither of these possibilities violate causality.
If you look at the situation from a physics perspective, the connecting together of the two capacitors is an irreversible process, which therefore causes an increase in entropy. This increase in entropy translates to a decrease in the free energy of the system, which means you can get less useful work out of the two capacitors after you have joined them than you could get out of the one capacitor at the start.
That's very true. The problem is that circuit theory, on its own, cannot explain how this energy is converted from one form to another. It is problems like these that indicate that we need a deeper understanding, i.e. electrodynamics.
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Look at section 1.5 of Haus and Melcher.
??? Never heard of them before now and likely for a good reason.
People seems to confuse charge with energy.
Energy is the one that can not be created or destroyed just converted from one form to another.
https://web.mit.edu/6.013_book/www/
One of the best books on Electromagnetics.
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https://web.mit.edu/6.013_book/www/
One of the best books on Electromagnetics.
Is either that you misinterpreted what you read in the book or the book is just not as good as you say.
For a single capacitor that say it is charged so there is an imbalance of electrons on the two plates. What do you think it will happen with charge if you change the size of the plates?
I guess you realized that Q=C*V and Ec=0.5 * C * V2 are both true and nobody ever disproved the conservation of energy so it seems clear that charge will not be conserved in this example.
Q = 3C at the beginning and for case where there is no resistance so no energy lost as heat end will be Q = 4.242C
It just happens that due to the perfect symmetry of the two identical capacitors you end up with half of the energy lost as heat and so charge is conserved but this is just thanks to sysmentry you can call this a coincidence.
Try to calculate for two capacitors that are not identical and see then what you will get even with resistance and heat loss. You will be surprised.
I guess those two are professors at MIT and same as the other professor at University of California Alex have a poor understanding of energy storage.
Alex kusenko knew that energy conservation can not be broken yet he lost due to his ability to understand energy storage.
He knew Derek was wrong he was just unable to understand himself how vehicle works thus he lost the bet.
I try to explain this first as it is simpler to explain than the "faster than wind direct down wind vehicle" but there the exact same issue exist and that is understanding what energy is.
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While yes there will be some bounce both in the experiment with resistance and the one with superconductors the speed of electron wave is finite and there will be reflexions at the open ends so voltage will stabilize and not forever slush around as you imagine as reflected waves will interact and in time cancel each other.
Consequent does not match the antecedent.
Think: if you didn't know a damn thing about waves, you just heard someone say, "this thing is reflective therefore energy is absorbed", would you be not the least bit concerned?
The definition of something being reflective, is that it does not absorb energy, at least not predominantly so.
Indeed, some very excellent mirrors can be constructed, whether using superconductors (at radio frequencies), or stacks of dielectrics (at optical frequencies).
If the mirrors are indeed perfect, where pray tell is the energy going to go?
Indeed it is exactly the reflection which causes it to oscillate. If the energy could be absorbed, it would, but there is insufficient resistance in the circuit to absorb it, at least before it's reflected around a few times. There is a direct equivalence between a lumped-equivalent circuit (like the present CLC network) and a transmission line construction, and further to a full-fields (mirrors and light beams) model, assuming of course the geometry keeps the waves confined. So it is perfectly consistent to say that energy is reflecting in this network, and that it is precisely that reflection, at each end, that causes the energy to "slosh" back and forth.
Tim
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Is either that you misinterpreted what you read in the book or the book is just not as good as you say.
For a single capacitor that say it is charged so there is an imbalance of electrons on the two plates. What do you think it will happen with charge if you change the size of the plates?
I guess you realized that Q=C*V and Ec=0.5 * C * V2 are both true and nobody ever disproved the conservation of energy so it seems clear that charge will not be conserved in this example.
Pray tell, how much force is required to "change the size of the plates"?
Also, what meaning does "change size" have? Surely you aren't changing just one, that accomplishes [almost] nothing.
But I would be inclined to understand this as "separation of plates", which is perfectly meaningful, a standard experiment.
And indeed, if there is a force, and changing the separation of the plates implies a change in distance... then does that not also imply.......? ;)
I guess those two are professors at MIT and same as the other professor at University of California Alex have a poor understanding of energy storage.
:-DD
Tim
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Consequent does not match the antecedent.
Think: if you didn't know a damn thing about waves, you just heard someone say, "this thing is reflective therefore energy is absorbed", would you be not the least bit concerned?
The definition of something being reflective, is that it does not absorb energy, at least not predominantly so.
Indeed, some very excellent mirrors can be constructed, whether using superconductors (at radio frequencies), or stacks of dielectrics (at optical frequencies).
If the mirrors are indeed perfect, where pray tell is the energy going to go?
Indeed it is exactly the reflection which causes it to oscillate. If the energy could be absorbed, it would, but there is insufficient resistance in the circuit to absorb it, at least before it's reflected around a few times. There is a direct equivalence between a lumped-equivalent circuit (like the present CLC network) and a transmission line construction, and further to a full-fields (mirrors and light beams) model, assuming of course the geometry keeps the waves confined. So it is perfectly consistent to say that energy is reflecting in this network, and that it is precisely that reflection, at each end, that causes the energy to "slosh" back and forth.
Tim
There is not just a single wave that moves from one side to the other. and they interact with each other.
Not sure why we discus so much the ideal superconductor when you can get very similar result by just adding an inductor as intermediary energy storage to move the energy from one capacitor to another.
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Pray tell, how much force is required to "change the size of the plates"?
Also, what meaning does "change size" have? Surely you aren't changing just one, that accomplishes [almost] nothing.
But I would be inclined to understand this as "separation of plates", which is perfectly meaningful, a standard experiment.
And indeed, if there is a force, and changing the separation of the plates implies a change in distance... then does that not also imply.......? ;)
No energy is needed if there is no friction. Variable capacitors are build this way but they do not care about friction. In any case energy from the capacitor will always be lost as heat due to plate resistance when you move the plates in any direction.
That force * speed * time is energy and it will have nothing to do with electrical energy stored in a capacitor.
There is an attraction force between the two plates of the capacitor so it takes energy to move the plates apart but that will not translate in any increase in electrical energy stored in the capacitor.
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I guess those two are professors at MIT and same as the other professor at University of California Alex have a poor understanding of energy storage.
:-DD
Tim
I will need to correct this as it is unfair since I have no read the book written by them. In case of Alex I have seen his inability to understand energy by being unable to explain how that vehicle works and win the bet with Derek.
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TIL this machine doesn't exist! :-DD
https://www.youtube.com/watch?v=Zilvl9tS0Og (https://www.youtube.com/watch?v=Zilvl9tS0Og)
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TIL this machine doesn't exist! :-DD
That is not what you think. It is an electrostatic energy generator. There are also piezoelectric capacitors that can convert mechanical energy to electrical energy.
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I guess those two are professors at MIT and same as the other professor at University of California Alex have a poor understanding of energy storage.
:-DD
Tim
I will need to correct this as it is unfair since I have no read the book written by them. In case of Alex I have seen his inability to understand energy by being unable to explain how that vehicle works and win the bet with Derek.
As you asked someone else a few messages back, "I need to ask what is your qualification." Seems that literally everyone is wrong except for you, so what is your qualification for having the definitive knowledge about energy?
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As you asked someone else a few messages back, "I need to ask what is your qualification." Seems that literally everyone is wrong except for you, so what is your qualification for having the definitive knowledge about energy?
Fair turnabout; ah, but qualifications only matter if anyone was taking him seriously. :popcorn:
Tim
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I guess those two are professors at MIT and same as the other professor at University of California Alex have a poor understanding of energy storage.
:-DD
Tim
I will need to correct this as it is unfair since I have no read the book written by them. In case of Alex I have seen his inability to understand energy by being unable to explain how that vehicle works and win the bet with Derek.
As you asked someone else a few messages back, "I need to ask what is your qualification." Seems that literally everyone is wrong except for you, so what is your qualification for having the definitive knowledge about energy?
I'm an electrical engineer but work with energy storage and energy generation (is both my job but more importantly my hobby).
I looked probably at all forms of energy generation and energy storage so maybe that gives me a better understanding of energy.
It seems in this particular problem people confuse charge with energy. I do not think I seen the energy equation in Derek's video yet the main claim he makes is that "energy doesn't travel through wires" with is completely wrong.
He made the same mistake with the "faster than wind direct downwind vehicle with is powered by the wind only". That is more than a perpetuum mobile is an overunity device claim (obviously not true).
Alex the profesor knew that since such a machine will violate the conservation of energy but was unable to figure out that energy storage was involved so that vehicle energy storage device was charged while below wind speed using wind energy and that stored energy is what allowed the vehicle to temporarily exceed wind speed (not indefinitely as claimed).
To make the claim fit they used wrong equations (equation Derek or the guy that "invented" the machine came up just to fit their silly explanation).
So while Alex did not made any false claim on camera he as a physics professor should have understood how the vehicle works and debunk the claim instead of offer it even more credibility.
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Circuit theory isn't sufficient to solve the two-capacitor problem. We need to take the electrodynamic behavior of the system into account.
So let's do the thought experiment of connecting a charged and a discharged capacitor in parallel at t=0s. We assume that all the conductors are perfect and that the two capacitors are of the parallel plate type with vacuum between the plates. The two capacitors can be modelled as two lossless transmission lines (see for instance House and Melcher example 14.2.1). After the two capacitors are connected in parallel, an electromagnetic wave will bounce back and forth between the two capacitors. The amount of energy stored in each capacitor will fluctuate, but the total amount of stored energy in the two capacitors will remain the same.
If we go to the next level and take electromagnetic radiation into account, then al the energy will be radiated into space over time.
U suggest a gedanken where a charged capacitor & a discharged capacitor are suddenly connected in parallel.
One way to connect the 2 capacitors in parallel is to have 2 switches, which is what i asked for earlier.
Yes, a capacitor can be modelled as a parallel pair of transmission lines.
I don’t agree that an em wave will bounce back & forth tween the 2 capacitors. Electricity will bounce back & forth. And if electricity is an em wave then yes an em wave will bounce back & forth.
But i suspect that your em wave is Veritasium's Poynting Vector or Poynting Field. The Poynting Vector or Field are not electricity. I don’t know what they are. Praps some kind of description of what exists. But certainly not a description of what makes things happen.
No, circuit theory is not sufficient to solve the 2 capacitor problem, & no, the electrodynamic behaviour of the system is not sufficient. Both fail. Both are wrong. It comes down to what is electricity.
If we use a bad theory for electricity then we are unlikely to solve an electricity problem. If we do solve an electricity problem then that would be due to good luck (the scientific term is i think equivalence). But, we know that circuit theory plus electrodynamic behaviour plus luck fail every time.
If lossless, then all of the energy will not be radiated into space over time.
There will be no heat losses, there will be no radiation losses.
Listen. Photons do not looz energy over time (ignoring things which rob energy from a photon). Photons are eternal (ignoring things). Electricity consists of photons (not electrons)(not em waves)(not silly Poynting stuff), hence in your gedanken the electricity will be eternal.
Hey, just noticed, this is my 400th posting/reply/comment on this forum. :-+
I have cast 400 pearls.
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"Photons do not looz energy over time (ignoring things which rob energy from a photon)."
Tautological, yet mis-spelled.
Photons are emitted and absorbed all the time. In Compton scattering of x-ray photons, the photon loses a fraction of its energy.
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"Photons do not looz energy over time (ignoring things which rob energy from a photon)."
Tautological, yet mis-spelled.
Photons are emitted and absorbed all the time. In Compton scattering of x-ray photons, the photon loses a fraction of its energy.
A photon left well alone duznt change in any way.
I will spell proper when u & all talk proper (ie i will use archaic spelling when u & all use archaic pronunciation).
Photons are eternal (almost). Once created they exist forever (until they are killed)(there are at least 2 ways, which i wont go into today).
Yes, photons can be emitted, & absorbed, & praps cut up (Compton), & stretched, & bent (& in the end annihilated).
But until they are annihilated they are eternal.
So too electrons. But, electrons are photons. Hence in a way old (electron) electricity is photonic.
My new (electon) electricity is of course purely photonic (my electons propagating at the speed of light).
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When radioactivity was first discovered, around 1900, three types were identified: alpha, beta, and gamma.
They were initially named in order of their penetrating power through matter.
Since their first discovery, it was found and demonstrated experimentally that beta particles were negatively-charged electrons, and gamma rays were uncharged photons.
In 1900, Becquerel determined that the charge/mass ratio of beta particles equaled that of electrons in cathode rays.
Around 1914, it was demonstrated that gammas are electromagnetic radiation, i.e., photons.
How, then, can electrons be photons and vice-versa? Electrons have charge and mass, photons have neither charge nor mass.
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When radioactivity was first discovered, around 1900, three types were identified: alpha, beta, and gamma.
They were initially named in order of their penetrating power through matter.
Since their first discovery, it was found and demonstrated experimentally that beta particles were negatively-charged electrons, and gamma rays were uncharged photons.
In 1900, Becquerel determined that the charge/mass ratio of beta particles equalled that of electrons in cathode rays.
Around 1914, it was demonstrated that gammas are electromagnetic radiation, i.e., photons.
How, then, can electrons be photons and vice-versa? Electrons have charge and mass, photons have neither charge nor mass.
Good questions.
Gammas are photons. Radio waves are em radiation. Radio waves are not photons. Gammas are not em radiation. EM radiation is emitted by photons, it is a part of every photon.
An electron is a photon that has formed a loop by biting its own tail. At which time the em radiation splits, some going out (negative charge), some going in (which is then annihilated)(Williamson explains)(he says that the loop has a twist, hence all of the negative em radiation always goes outwards)(or most of it).
Jeans called an electron "bottled light".
Catt i think called an electron "a rolled up photon".
A positron has the positive charge going out, negative charge going in (where it is annihilated).
Electrons do not orbit a nucleus. Photons orbit a nucleus. In that sense we have 2 kinds of electron. One kind orbits nothing (& has the form of a loop). The other kind orbits a nucleus (& need not be a complete loop)(ie it need not bite its own tail).
Mass is a bit of a mystery. Firstly photons do have mass. The question arises -- how come a photon gains lots of mass when it becomes an electron. Williamson mentions a possible way.
Mass is the ability to annihilate aether. Something that has more mass annihilates more aether.
However, i have my own theory. A photon propagates along a line, hence the inflow of aether giving us what we call mass flows in perpendicular to the line, ie the inflow streamlines converge in 2 dimensions, while the guilty photon leaves the scene at the speed of light.
Meanwhile back at the ranch, an electron is a photon that is constrained in 3 dimension, the inflow streamlines converge in 3 dimensions, to a (stationary) point rather than to a line (which is fleeing at c).
I think that the convergence of streamlines, & the stationary versus fleeing stuff, might explain the difference in gravitational mass tween a photon & an electron.
My new (electon) electricity, ie my electons, consist of photons that hug the surface of a wire, rather than hugging a nucleus. Electons are photons that are constrained in 2 dimensions. Free photons (light) are constrained in 1 dimension. Electrons are photons that are constrained in 3 dimension (usually called confined photons).
Everything that we feel & see is made of photons. Photons are the fundamental (quasi) particle.
When photons form loops they give us every other kind of elementary particle (electrons quarks etc).
There are 3 forms of electricity. My electons (on a wire). Free surface electrons (on a wire). Possibly drifting electrons (inside a wire).
The charge on a capacitor consists of my electons on the negative plate, & induced free-surface-electrons on the other plate. However, my new (electon) electricity is a work in progress. I need to tick all of the boxes.
My electons immediately & simply explain why the discharge of a capacitor takes twice the time predicted by the flawed standard circuit theory. In the meantime everyone around here is talking rubbish about capacitors.
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So photons form a loop, and gain mass to become electrons.
Whence comes the charge?
I'm glad to see you admit the possibility of electrons drifting: TI and the other semiconductor manufacturers can now continue their processes.
Electrons accelerating emit electromagnetic radiation: see antenna theory and synchrotron radiation (both of which work).
Photons interacting with electrons can increase the energy of the electrons: see atomic structure theory (not the archaic "orbits" you keep harping on) and spectroscopy (both of which work).
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It certainly deserves to be separated because the physical behavior is quite different:
Separated.
Since the 19th century, everybody thought that everything electric and magnetic, from DC to cosmic rays, through radio waves, heat, light, ultraviolet, X-rays and whatnot, is the manifestation of the same freaking physical phenomenon.
Now you're saying that they are different. I wonder why the Nobel Committee has not noticed you yet.
Oh they are both implied by Maxwell's equations.
But because their physical behavior is quite different, they are called differently. Much like DC current and gamma rays are called differently. Duh.
You don't call wires, capacitors, coils and antennae "Maxwell stuff" do you?
Yes I can. Energy flows in wires and reappear in a lightbulb/engine/LED.
Yes energy is transferred from the battery to the short, through the wire.
If you say so, it must be true.
Well you say energy flows through vacuum just to mock engineers, why can't I say it flows through wires?
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The circuit theory is perfectly capable in solving the two parallel capacitor problem.
There is no such thing as conservation of charge. The conservation of energy is a law.
Kirchhoff's circuital laws (KCL/KVL) are an extension of the principle of conservation of charge. So your assertions above are contradictory.
I looked probably at all forms of energy generation and energy storage so maybe that gives me a better understanding of energy.
You lack fundamental understanding though.
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You lack fundamental understanding though.
I proved that energy travels through wires in multiple ways sumarized in this post https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/msg4156462/#msg4156462 (https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/msg4156462/#msg4156462)
Let me know what part of that is wrong.
The entire discussion is about energy traveling through wires or outside of wires. Same can be summarized as energy traveling through a capacitor (not the leakage through dielectric) that is basically what Derek will say vs energy flows in and out of the capacitor as capacitor is an energy storage device and that is what I and everyone that has correct understanding of reality is saying.
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So photons form a loop, and gain mass to become electrons.
Whence comes the charge?
I'm glad to see you admit the possibility of electrons drifting: TI and the other semiconductor manufacturers can now continue their processes.
Electrons accelerating emit electromagnetic radiation: see antenna theory and synchrotron radiation (both of which work).
Photons interacting with electrons can increase the energy of the electrons: see atomic structure theory (not the archaic "orbits" you keep harping on) and spectroscopy (both of which work).
I like the ideas (re charge) coming from Williamson, & also from Conrad Ranzan. Here are some links to Williamson.
Is the electron a photon with toroidal topology. Williamson & van der Mark. 1997.
http://home.claranet.nl/users/benschop/electron.pdf (http://home.claranet.nl/users/benschop/electron.pdf)
On the nature of the photon and the electron. J.G.Williamson 2015?
https://www.researchgate.net/publication/281749668_On_the_nature_of_the_photon_and_the_electron (https://www.researchgate.net/publication/281749668_On_the_nature_of_the_photon_and_the_electron)
A new theory of light and matter. J.G.Williamson. (Dated: July 18, 2014)
https://www.researchgate.net/publication/323873485_A_new_theory_of_light_and_matter (https://www.researchgate.net/publication/323873485_A_new_theory_of_light_and_matter)
I think i am ok with that electrons accelerating stuff. And i am ok with that photons interacting with electrons stuff.
Re electricity, one can in many cases just change electron to electon.
In antennas it is the electons sloshing up&back, not electrons. And the electons emit em radiation.
In synchrotrons it is i suppose electrons going round & round, emitting em radiation.
I am ok with that.
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The entire discussion is about energy traveling through wires or outside of wires. Same can be summarized as energy traveling through a capacitor (not the leakage through dielectric) that is basically what Derek will say vs energy flows in and out of the capacitor as capacitor is an energy storage device and that is what I and everyone that has correct understanding of reality is saying.
Is that statement compatible with an LED + resistor between two physically small capacitors emitting light for a few seconds - a physical demonstration that energy can flow through the dielectric of the capacitors, even though electric charge can't.
If a glowing LED doesn't suffice, how about you put a third identical capacitor in the middle (so three capacitors in series). apply 12V. Then, with out switching off the 12V, remove the middle capacitor, walk it to the other side of the room and measure the voltage on it with a DMM. It will read 4V, and will have 1/2 C V^2 Joules of energy in it. That energy is far higher than a microamp or so of leakage current count account for.
Where did it this energy come from? How does that energy get into the middle capacitor if energy can only flow in wires and not through capacitors?
Maybe your definition of "energy traveling through [a capacitor]" doesn't match mine?
Acutally, what is your definition of "energy traveling through" something?
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Oh they are both implied by Maxwell's equations.
But because their physical behavior is quite different, they are called differently. Much like DC current and gamma rays are called differently. Duh.
Precisely. Theories like the Newton's law of gravitation or Maxwell's equations are breakthroughs because they show that things that on the surface appear to be different are in fact just aspects of the same crap.
So, to the untrained eye, radiation, capacitive coupling and "transmission of energy through wires" look like different and uncorrelated things.
But those in the know know that they're the same thing.
You don't call wires, capacitors, coils and antennae "Maxwell stuff" do you?
Why not? Will they get offended?
Well you say energy flows through vacuum just to mock engineers, why can't I say it flows through wires?
You don't get it. I AM an engineer. If you want to contradict the experimental data, knock yourself out.
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I proved that energy travels through wires in multiple ways sumarized in this post https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/msg4156462/#msg4156462 (https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/msg4156462/#msg4156462)
Let me know what part of that is wrong.
I just did it. From a contradiction you can prove whatever, remember? So the flaw of your entire reasoning lies in the fact that you don't understand the most fundamental tenets of electricity. Go back to the books and if you have any question, we're here to help.
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The entire discussion is about energy traveling through wires or outside of wires. Same can be summarized as energy traveling through a capacitor (not the leakage through dielectric) that is basically what Derek will say vs energy flows in and out of the capacitor as capacitor is an energy storage device and that is what I and everyone that has correct understanding of reality is saying.
Is that statement compatible with an LED + resistor between two physically small capacitors emitting light for a few seconds - a physical demonstration that energy can flow through the dielectric of the capacitors, even though electric charge can't.
If a glowing LED doesn't suffice, how about you put a third identical capacitor in the middle (so three capacitors in series). apply 12V. Then remove the middle capacitor, walk it to the other side of the room and measure the voltage on it with a DMM. It will read 4V, and will have 1/2 C V^2 Joules of energy in it. That energy is far higher than a microamp or so of leakage current count account for.
Where did it this energy come from? How does that energy get into the middle capacitor if energy can only flow in wires and not through capacitors?
Maybe your definition of "energy traveling through [a capacitor]" doesn't match mine?
Acutally, what is your definition of "energy traveling through" something?
That was energy through a leaky capacitor (as all electrolytic capacitors are). But if you are referring to initial high current through LED as you seems you do that is due to capacitors charging but no current passes through capacitor. Current flows in or out of the capacitor.
I may actually post a problem with 3 different case and provide the solution also to see if you agree with the results.
Yes the capacitor in the middle will charge as all 3 capacitors in series are viewed as a single capacitor.
But the fact that you measure 4V means that all energy that flowed in that circuit with 3 identical capacitors in series supplied by 12V power supply flowed into the capacitors and not through them.
Maybe is a matter of understanding the definition of flowing in or flowing through.
Like again an analogy with limited scope just so we batter define flowing through vs flowing in/out of.
Water flows in a bucket and what will mean for water to flow through a bucket will be a bucket with a hole on the bottom so water can flow through.
Those electrolytic capacitors have a leakage current same as a bucket with a very small leak (the bucket is still useful to transfer liquid from some part to another but no long storage).
The capacitor made by the long transmission wires are more like a small coffee cup so low capacity but very well build with no measurable leakage.
The small current in the first 65ns or so Derek observes in his test is due to current flowing in to the transmission line capacitance (as I showed in spice simulation) and is not due to leakage with for that transmission line with 1m of air is not measurable (way to small).
Also back to those two identical parallel capacitors. If you add an incandescent lamp instead of the switch to transfer the energy from one capacitor to the other you will get the exact same half voltage and half the energy remaining in the capacitors while you get some visible photons (maybe depends of how large the capacitors are vs the energy need for filament to glow) but the fact remains that you can add anything in between the capacitors to do some work and the energy left in the capacitors at the end of the experiment is the same as directly paralleling the capacitors to transfer the energy.
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I just did it. From a contradiction you can prove whatever, remember? So the flaw of your entire reasoning lies in the fact that you don't understand the most fundamental tenets of electricity. Go back to the books and if you have any question, we're here to help.
OK here is a problem and let me know if it the results are correct as I will provide the answer to the problem also.
A) Two identical 1F capacitors one fully discharged so 0V and one charged at 3V that will be paralleled.
Initial energy 0.5 * 1F * 32 = 4.5Ws
End energy 1.125Ws in each capacitors so just half of the total initial energy the rest ended as heat.
You get the same result if you insert at resistor or an incandescent lamp instead of the switch.
B) Three identical 1F capacitors two of them charged at 3V and one discharged 0V that will all be paralleled
Initial energy 9Ws
End energy 6Ws so all capacitors will be at 2V (0.5 * 3F * 22) = 6Ws
There is just 2Ws in the capacitor that started empty so that came from the two charged ones 1Ws from each.
They started with 4.5W each and ended up with 2Ws each so the 1Ws that each provided to discharged capacitor resulted in 1.5W of loss so worse transfer energy than first case.
Again you can add any resistor or incandescent bulb and result will be the same.
C) Three identical 1F capacitors just one of them charged at 3V the other two empty.
Initial energy 4.5Ws
End energy 1.5Ws (just 1V across each capacitor). So 3Ws ended up as heat double compared to what is left stored.
If you agree that the calculations above are correct and you also agree that all that wasted heat is in the wires/capacitor plates and lamp or resistor plus radiated as IR especially with the lamp as filament has low thermal mass and it is in vacuum then you agree that all energy traveled through wires/conductors.
I do not see how any engineer will be able to contest the above. You can measure and see that is what will happen including the wasted energy ending up as heat.
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That was energy through a leaky capacitor (as all electrolytic capacitors are). But if you are referring to initial high current through LED as you seems you do that is due to capacitors charging but no current passes through capacitor. Current flows in or out of the capacitor.
"Energy through a leaky capacitor"? What does "leaky" mean in the context of energy?
How long after removing from the circuit do you want me to leave the middle capacitor alone before I measure it? If it is 'leaky' and the energy is there because of such leaks, then it should self-discharge in a short period of time.
But the fact that you measure 4V means that all energy that flowed in that circuit with 3 identical capacitors in series supplied by 12V power supply flowed into the capacitors and not through them.
So how did the energy get into the middle capacitor? If it didn't go through the other two capacitors then it must have gone around them? Those are really the only two options. Pick one (or both even). Energy not being able to pass through capacitors is demonstratable false.
Maybe is a matter of understanding the definition of flowing in or flowing through.
Like again an analogy with limited scope just so we batter define flowing through vs flowing in/out of.
Water flows in a bucket and what will mean for water to flow through a bucket will be a bucket with a hole on the bottom so water can flow through.
Those electrolytic capacitors have a leakage current same as a bucket with a very small leak (the bucket is still useful to transfer liquid from some part to another but no long storage).
The capacitor made by the long transmission wires are more like a small coffee cup so low capacity but very well build with no measurable leakage.
In this case I think you are confusing the model as being reality, not being representative of reality.
If I made this test out of three transmission lines (say 100m rolls of coax, would my results be different?
If I went out and sourced the lowest possible leakage capacitors in the known universe, would my results be different?
Also back to those two identical parallel capacitors.
I'ld rather not - exploring how you explain that energy gets into the middle capacitor is far more interesting and enlightening.
I just left a charged capacitor on the bench for 10 minutes before I put the meter over it. Still has energy in it, so leakage doesn't seem to be an major issue.
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"Energy through a leaky capacitor"? What does "leaky" mean in the context of energy?
How long after removing from the circuit do you want me to leave the middle capacitor alone before I measure it? If it is 'leaky' and the energy is there because of such leaks, then it should self-discharge in a short period of time.
Leakage is an undesired effect of real capacitors and has nothing to do with anything related to the main question.
Energy flows in capacitors not through capacitors except for that pesky small current of few uA typical.
When you have a few mA or A of current in a loop with a capacitor in series then you have current going in to the capacitor not through it.
I sort of get frustrated and annoyed of explaining what in/out and trough means.
Se the problem above post #316 as that is proof energy only travels through wires not through the space around the wire or capacitors.
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If current "goes into" a capacitor and not "through" it then would you agree Vr = 0 for an initially discharged capacitor and voltage step Vs?
C
+---||---+--o Vr
|+ |
Vs R
|- |
+--------+
_|_
Tim
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I do not see how any engineer will be able to contest the above.
Then see. The energy resides in the dielectric between the plates of any of the charged capacitors and will happily migrate through space to the uncharged ones. The energy that goes into the wires and plates will be immediately dissipated.
Wires transfer charge. Space transfers energy.
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"Energy through a leaky capacitor"? What does "leaky" mean in the context of energy?
How long after removing from the circuit do you want me to leave the middle capacitor alone before I measure it? If it is 'leaky' and the energy is there because of such leaks, then it should self-discharge in a short period of time.
Leakage is an undesired effect of real capacitors and has nothing to do with anything related to the main question.
Energy flows in capacitors not through capacitors except for that pesky small current of few uA typical.
When you have a few mA or A of current in a loop with a capacitor in series then you have current going in to the capacitor not through it.
I sort of get frustrated and annoyed of explaining what in/out and trough means.
Se the problem above post #316 as that is proof energy only travels through wires not through the space around the wire or capacitors.
I too get frustrated when we talk about "energy flowing through" and then you reply talking about current. If you said something like "proof that charges only travels through wires not through the space around the wire or capacitors" I would agree with you (as long as it is at low voltages).
It is easily proven that we are able to move energy - actual Joules of measurable energy, that can do real work - from a battery into the middle capacitor of three in series. That energy will persists in the capacitor after it is removed from the circuit, and can be moved to the other side of the room. The rest of the original circuit could be disassembled, or even destroyed, and yet that energy still persists in what was the middle capacitor.
It should be undeniable proof that energy has been transferred from the battery into that middle capacitor, even though it has no other connection to the source except through the two capacitors. The only way for energy to get into that capacitor is either through the other two capacitors, or through the air around it. Both of which are things you say can't happen.
And the more 'ideal' the capacitors are the better this can be shown - an ideal capacitor with zero leakage could be left for months, and that energy would still be in there. Even with my less-than-ideal AliExpress components that energy sits there for minutes, if not hours.
It isn't as if my caps suddenly go all leaky when connected to a DC source - I have put a uA meter inline (once the caps are charged), and have measured exactly how little leakage is - under a uA at 10V.
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Wires transfer charge. Space transfers energy.
Oh.. that's good. I like that!
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If current "goes into" a capacitor and not "through" it then would you agree Vr = 0 for an initially discharged capacitor and voltage step Vs?
C
+---||---+--o Vr
|+ |
Vs R
|- |
+--------+
_|_
Tim
Electrons entering on one side of the capacitor to increase the density of free electrons means that electrons will need to leave the other plate.
But this energy is not used to do any work but stored and can be retrieved later to do work.
Say Vs is 3V and capacitor 1F the value of the resistor is irrelevant.
The capacitor will have a voltage potential of zero at the start so all 3V will drop across the resistor (will say resistor value is large enough that voltage source and capacitor DC resistance is low enough to to be of importance).
The charge current will start high then drop as the capacitor voltage increases until it gets to zero when voltage across the capacitor equal Vs (3V) then no current will flow so no energy will travel through the circuit (we ignore the capacitor small leakage current).
So now no energy can be transferred from Vs to resistor as current is basically zero.
Charging was inefficient but energy is stored in capacitor and can be used.
At all time no current (except that annoying leakage) has flown through the capacitor.
The way you charge a capacitor is by moving electrons from one plate to another but that is not done through dielectric but externally through wires.
There are now 4.5Ws of energy stored in the capacitor even if the power supply needed to deliver 9Ws as the other 4.5W was lost as heat on the resistor.
If you had a constant current power supply then no current resistor will have been needed for current limiting and power supply will have delivered 4.5Ws and all of that 4.5Ws will have been stored thus no work done at all / no heat loss.
Is like charging the capacitor with a linear regulator vs a DC-DC charger.
If you agree with the above (current flows in to capacitor as it is being charged and current flow stops when capacitor is fully charged) then you agree that energy is not flowing through the capacitor but in to capacitor.
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If current "goes into" a capacitor and not "through" it then would you agree Vr = 0 for an initially discharged capacitor and voltage step Vs?
C
+---||---+--o Vr
|+ |
Vs R
|- |
+--------+
_|_
Tim
Electrons entering on one side of the capacitor to increase the density of free electrons means that electrons will need to leave the other plate.
But this energy is not used to do any work but stored and can be retrieved later to do work.
Say Vs is 3V and capacitor 1F the value of the resistor is irrelevant.
The capacitor will have a voltage potential of zero at the start so all 3V will drop across the resistor (will say resistor value is large enough that voltage source and capacitor DC resistance is low enough to to be of importance).
The charge current will start high then drop as the capacitor voltage increases until it gets to zero when voltage across the capacitor equal Vs (3V) then no current will flow so no energy will travel through the circuit (we ignore the capacitor small leakage current).
So now no energy can be transferred from Vs to resistor as current is basically zero.
Charging was inefficient but energy is stored in capacitor and can be used.
At all time no current (except that annoying leakage) has flown through the capacitor.
The way you charge a capacitor is by moving electrons from one plate to another but that is not done through dielectric but externally through wires.
There are now 4.5Ws of energy stored in the capacitor even if the power supply needed to deliver 9Ws as the other 4.5W was lost as heat on the resistor.
If you had a constant current power supply then no current resistor will have been needed for current limiting and power supply will have delivered 4.5Ws and all of that 4.5Ws will have been stored thus no work done at all / no heat loss.
Is like charging the capacitor with a linear regulator vs a DC-DC charger.
If you agree with the above (current flows in to capacitor as it is being charged and current flow stops when capacitor is fully charged) then you agree that energy is not flowing through the capacitor but in to capacitor.
If you agree with the above (current flows into the left hand side of the capacitor, and that current is flowing out of the right hand side of the capacitor, as it is being charged and both current flows stop when capacitor is fully charged) then you agree that during that time energy was flowing through the capacitor.
FTFY.
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If you agree with the above (current flows into the left hand side of the capacitor, and that current is flowing out of the right hand side of the capacitor, as it is being charged and both current flows stop when capacitor is fully charged) then you agree that during that time energy was flowing through the capacitor.
FTFY.
I assure you that energy is flowing in to capacitor and that distinction is very important.
It is irrelevant what else is in series with the capacitor and of course that thing in this case a resistor will see that charge current.
In case of Derek's experiment those first 65ns where used to charge the long transmission line and that charge current passed trough the lamp/resistor then after that the next 65ns the energy from the transmission line was discharged also in the lamp so nothing was lost. After those first 65ns the electron wave got to lamp through wires and continued to supply all the energy.
At no point in time energy was flowing to lamp/resistor outside the wires. Each unit of energy no matter how insignificant was delivered through wires and the wire resistance heated up from caring that energy.
You can not charge a capacitor by just connecting one of the terminals. Charging means moving electrons from one side of the capacitor to the other creating an excess of free electrons on one side and a deficit on the other side.
The dielectric can be anything including air or even vacuum and that is not what has the stored energy. The energy is contained in the capacitor plate same as transmission wires.
I'm sorry I fail at explaining things.
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I sort of get frustrated and annoyed of explaining what in/out and trough means.
So current is going into, but it's not? And it's going through, but it's not? And energy is going into, but it's also not?
It seems you can't produce a coherent description of any of the particulars in these matters, in addition to use of the above words... at least not in terms that anyone else can make any meaning of. If it's so frustrating and annoying, then... why bother?
You know the phrase, "better to be thought a fool..."?
Tim
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You can not charge a capacitor by just connecting one of the terminals.
If I were demonstrate to you an experiment which would charge a capacitor, without attaching any source to either terminal, and doesn't involve stupid things like firing ions at the terminals would that settle it for you?
To make things really clear, how about we have the capacitor as a 10cm^2 sheets of metal, separated by a suitable insulator?
The only other thing that is required is a removable switch, to discharge the capacitor when desired - say a screwdriver tip to short the two metal plates together, and of course some agreed way of measuring if the capacitor is charged or not. I suggest a standard DMM, that is used after the charging procedure?
Now two 10cm x 10cm metal sheets don't have much capacitance, so you will not get mA out of it. So don't expect wonders.. (heck this might be possible with just baking foil separated by tissues, I might give it a try!)
Are you game enough?
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(heck this might be possible with just baking foil separated by tissues, I might give it a try!)
That is how I made capacitors when I was a child.
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Yes. This will work. Make a four-layer capacitor with foil and paper. Charge the outer two layers. You will then be able to extract energy from the capacitor formed by the inner two layers.
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So current is going into, but it's not? And it's going through, but it's not? And energy is going into, but it's also not?
It seems you can't produce a coherent description of any of the particulars in these matters, in addition to use of the above words... at least not in terms that anyone else can make any meaning of. If it's so frustrating and annoying, then... why bother?
You know the phrase, "better to be thought a fool..."?
Tim
You have a 3V supply and at some particular point in time 1.5V drop across the capacitor and 1.5V drop across the resistor and say the current is 1A
Then you have 1.5V * 1 A = 1.5W going in to capacitor (being stored) with no resulting heat so no work done and you have same 1.5W on the resistor all ending up as heat so not stored.
So from the supplied 3W only 1.5W across the resistor is doing work so energy is being used and 1.5W is stored no work is done with that.
Similarly say power supply is set to output constant current 1A and you connect that to a capacitor any value and while there is current flow there is no work done as current flows in to capacitor and is being stored
Then second case you have a 1.5Ohm resistor connected to the same constant current 1A output power supply and all this 1.5W will go through resistor resulting in to heat so work is done.
If you understand the above example let me know how you explain in a better way energy flowing in/out of capacitor vs energy flowing through a resistor as there is a big difference between the two.
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Yes. This will work. Make a four-layer capacitor with foil and paper. Charge the outer two layers. You will then be able to extract energy from the capacitor formed by the inner two layers.
That is exactly like having 3 capacitors in series that you charge and then remove the middle one. It only proves what I'm saying.
If you push 3Ws worth of energy "in to" those 3 series capacitors (distance equal between plates so they have equal capacity) then each of them will contain 1Ws and none of the energy did any work it was just stored so no 3Ws worth of heat or visible photos or anything like that just stored for later use.
If my definition of going in/out or trough seems unclear let me know how will you word the difference between doing work or just storing energy ?
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You don't call wires, capacitors, coils and antennae "Maxwell stuff" do you?
Why not? Will they get offended?
Clearly you're not doing this so why don't you answer your own question?
Well you say energy flows through vacuum just to mock engineers, why can't I say it flows through wires?
You don't get it. I AM an engineer. If you want to contradict the experimental data, knock yourself out.
I don't, I'm just saying energy flows through wires, which contradicts no experimental data whatsoever.
If there were experimental data supporting the energy part of the slogan "Wires transfer charge. Space transfers energy." we would have known about them and they would be in books. ::)
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I don't, I'm just saying energy flows through wires, which contradicts no experimental data whatsoever.
If there were experimental data supporting the energy part of the slogan "Wires transfer charge. Space transfers energy." we would have known about them and they would be in books. ::)
You've not read Hayt then?
The question of where the energy is stored in an electric field has not yet been answered. Potential energy can never be pinned down precisely in terms of physical location. Someone lifts a pencil, and the pencil acquires potential energy. Is the energy stored in the molecules of the pencil, in the gravitational field between the pencil and the earth, or in some obscure place? Is the energy in a capacitor stored in the charges themselves, in the field, or where? No one can offer any proof for his or her own private opinion, and the matter of deciding may be left to the philosophers. Electromagnetic field theory makes it easy to believe that the energy of an electric field or a charge distribution is stored in the field itself...
Engineering Electromagnetics p.106 (8th Edition)
He goes on to make an argument for why the field interpretation is preferred.
And when he gets to waveguides, there is no question what Hayt thinks:
Stated more generally, all fields in a good conductor such as copper are essentially zero at distances greater than a few skin depths from the surface. Any current density or electric field intensity established at the surface of a good conductor decays rapidly as we progress into the conductor. Electromagnetic energy is not transmitted in the interior of a conductor; it travels in the region surrounding the conductor, while the conductor merely guides the waves. We will consider guided propagation in more detail in Chapter 13.
Engineering Electromagnetics p.407 (8th Edition)
I'm fine with folks digging in to "charges store and transfer energy" (even if I think their interpretation is clunky and inelegant for explaining how energy can cross an air gap so this is not what I teach in my motor laboratories) but let's not pretend there aren't reputable textbooks that firmly take the 'energy is in the fields' interpretation of Maxwell and the experiments done on electromagnetic radiation.
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One of you mentioned a Wimshurst machine (converts mechanical energy in to electrical energy) and I found a video correctly explains how it works so if you are able to understand his high quality video and more importantly correct explanation then you will understand why energy travels through wires.
There are a few places on that machine where energy travels outside metal conductors but energy is still carried by electrons and not fields as electrons travel through a small space between wires. Hes animation shows that effect very nicely.
I think that is a fantastic video and unless you do not agree that is what happens there it debunks Derek's claim that energy travels outside the wire in his experiment.
https://www.youtube.com/watch?v=nA4aCd5qFWs (https://www.youtube.com/watch?v=nA4aCd5qFWs)
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Clearly you're not doing this so why don't you answer your own question?
Cause that's your assumption.
I don't, I'm just saying energy flows through wires, which contradicts no experimental data whatsoever.
It contradicts all the experimental data since the times of the discovery of electromagnetism in the 19th century until this very day. But if you want to ignore that, be my guest.
If there were experimental data supporting the energy part of the slogan "Wires transfer charge. Space transfers energy." we would have known about them and they would be in books. ::)
They are. I even showed you where this is in the very first book on the matter. You probably haven't read many books on the subject.
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I think that is a fantastic video and unless you do not agree that is what happens there it debunks Derek's claim that energy travels outside the wire in his experiment.
https://www.youtube.com/watch?v=nA4aCd5qFWs (https://www.youtube.com/watch?v=nA4aCd5qFWs)
Absolutely nothing in this video debunks Derek's experiment. Charges are conducted by conductors, and the energy is transferred by the fields through the various dielectric materials (air, plastic, etc.)
So ¯\_(ツ)_/¯
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Absolutely nothing in this video debunks Derek's experiment. Charges are conducted by conductors, and the energy is transferred by the fields through the various dielectric materials (air, plastic, etc.)
So ¯\_(ツ)_/¯
Please pay more attention to the video.
There is no transfer of energy between the rotating disks and laden jars (large capacitors) until voltage is high enough to have electrons travel the gap
Similarly when you have a spark between the spark gap energy flows from capacitors trough conductors and finally through the arc (plasma).
There is only 20V in Derek's video and about 1m of gap thus no electrons will flow at that low of a voltage and such a large gap. The energy flow you see in those first 65ns are due to transmission line (capacitor) being charged and all energy travels through wires as it is traveling in this generator where the arc/plasma is also a conductor.
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I don't, I'm just saying energy flows through wires, which contradicts no experimental data whatsoever.
It contradicts all the experimental data since the times of the discovery of electromagnetism in the 19th century until this very day. But if you want to ignore that, be my guest.
If there were experimental data supporting the energy part of the slogan "Wires transfer charge. Space transfers energy." we would have known about them and they would be in books. ::)
They are. I even showed you where this is in the very first book on the matter. You probably haven't read many books on the subject.
What book? What experiment? What was predicted by "energy flows through wires" and not found in experiments?
Why can't I find them in all modern books: Jackson, Griffiths, Haus Melcher and so on?
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Please pay more attention to the video.
There is no transfer of energy between the rotating disks and laden jars (large capacitors) until voltage is high enough to have electrons travel the gap
Conversely, there is no wire between the collectors and the sectors. So your claim that wires carry the energy is false.
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Why can't I find them in all modern books: Jackson, Griffiths, Haus Melcher and so on?
Maybe you're functionally illiterate, who knows?
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Why can't I find them in all modern books: Jackson, Griffiths, Haus Melcher and so on?
Maybe you're functionally illiterate, who knows?
Worth a read:
https://royalsocietypublishing.org/doi/10.1098/rsta.2017.0457#RSTA20170457C39
One problem that Heaviside used to illustrate the energy current concept was that of a battery connected by a simple circuit to a resistor load in his 1886–1887 work ‘The transfer of energy and its application to wires. Energy current’ [33]—effectively an analysis of a ‘twin wire transmission line’. The standard approach is to assume that electrical energy flows inside metal wires (confinement), but Heaviside's energy current approach dictates otherwise. Poynting had first published on this arrangement [31], which was criticized by Heaviside due to Poynting's misconception of the nature of the external electric field surrounding the wires [32,34]. Poynting considered only a tangential electric field in the axis of the wire, combined with the circumferential magnetic field, resulting in an inwardly radial component of the energy flux density, W, only. Thus, given no energy flux in the direction of the electric current—how does energy get from the battery to the load? This is not answered by conventional circuit theory. Heaviside argued that this ‘Poynting component’ is simply the heat lost due to Joule heating in the conductor; however, a more prominent component exists outside the wire due to a radial electric field—the surface charges on the conductors that set up the field and maintain the electric current are responsible for the energy transfer external to the conductors. This complete field picture then presents a ‘map’ of the energy flow. Heaviside showed for the first time that a radial electric field and a circumferential magnetic field produce an ‘energy current’, a flow of electromagnetic energy in the space surrounding the electric conductors, directed from the battery along the axis of the conductor towards and entering the load.
This remarkable result is one that is rarely presented, but that Heaviside gives in great detail in his Electrical papers, vol. II, which have recently been the subject of a rigorous modern mathematical treatment [39], confirming Heaviside's results. It is readily shown that the energy current approach is compatible with the circuit theory approach by applying the integral formulation of the Poynting theorem to the problem, determining the power dissipated in the load resistor as VI, as dictated by conventional ‘confined’ circuit theory. This ‘energy current approach’ gives physical insight, but requires detailed knowledge of the electric and magnetic fields surrounding the analysed circuits. It complements his work on electromagnetic wave propagation by analysing the energy associated with the wave and also his work on electromagnetic diffusion in which he found that the current in the wire penetrates from the outside surface inwards.
Earlier in the article, the controversy is addressed head on. Even more than 140+ years later, Heaviside is still stirring the pot:
It is well known that this approach is useful in antenna theory and microwave circuits. However, Heaviside extended the use of the theorem to DC electric circuits. In doing so, he reversed the contemporary view of electric current, proposing that the electric and magnetic fields due to the current are the primitives, rather than being a result of the motion of the electronic charge in the conductor. This is a controversial viewpoint and, in his Electrical papers, the phrase ‘we reverse this’ [36], referring to the ‘current in the wire being set up by the energy transmitted through the medium around it’, reverberates even to this day. This view is supported by his work on electromagnetic diffusion (previous section) and the nature of the electromagnetic field and current density penetration of an electrical conductor subjected to a step current. This is discussed in a modern context by Feynman [37], who showed that the electromagnetic momentum is ‘required’ in order to conserve angular mechanical momentum associated with the energy flux vector W, and a detailed historical discussion is presented by Nahin [12]. The ‘uniqueness’ of the vector W and the physical existence of mysterious and counterintuitive circulating ‘energy currents’ emerging from static fields (e.g. a point electric charge with a superimposed magnetic dipole) has been the subject of some debate among many scientists over a long period of time. Heaviside was the first to consider these issues in 1893 [38].
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EEVBlog drew attention to this in his original response video to Veritasium but it's worth bringing up again, especially since Feynman gets a shout-out from the Royal Society:
https://www.feynmanlectures.caltech.edu/II_27.html (https://www.feynmanlectures.caltech.edu/II_27.html)
But it [eq.27.20] tells us a peculiar thing: that when we are charging a capacitor, the energy is not coming down the wires; it is coming in through the edges of the gap. That’s what this theory says!
And of course Feynman has this to say (which Dave also drew attention to):
You no doubt begin to get the impression that the Poynting theory at least partially violates your intuition as to where energy is located in an electromagnetic field. You might believe that you must revamp all your intuitions, and, therefore have a lot of things to study here. But it seems really not necessary. You don’t need to feel that you will be in great trouble if you forget once in a while that the energy in a wire is flowing into the wire from the outside, rather than along the wire. It seems to be only rarely of value, when using the idea of energy conservation, to notice in detail what path the energy is taking. The circulation of energy around a magnet and a charge seems, in most circumstances, to be quite unimportant. It is not a vital detail, but it is clear that our ordinary intuitions are quite wrong.
Perhaps its possible to ignore all this business and retreat back to the hydraulic arguments about electrons being like water in pipes. Hayt concedes it might just be a philosophical problem - but yet he takes a firm position on which interpretation he prefers. So did Heaviside. So did Kraus. And even Feynman to an extent. He was lecturing to a room of freshmen/sophomore physics students in the 1960s. If he were talking to a room of engineers designing waveguides, the emphasis would be very different.
Our intuition is wrong - and these properties of fields are important if we think Maxwellian Theory means anything.
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I think that is a fantastic video and unless you do not agree that is what happens there it debunks Derek's claim that energy travels outside the wire in his experiment.
Remind me, who was it that said:
Internet is also fool of java type animations and explanations of capacitors and most of them are incorrectly made as the author did not understand the physics and made wrong assumptions.
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EEVBlog drew attention to this in his original response video to Veritasium but it's worth bringing up again, especially since Feynman gets a shout-out from the Royal Society:
https://www.feynmanlectures.caltech.edu/II_27.html (https://www.feynmanlectures.caltech.edu/II_27.html)
But it [eq.27.20] tells us a peculiar thing: that when we are charging a capacitor, the energy is not coming down the wires; it is coming in through the edges of the gap. That’s what this theory says!
And of course Feynman has this to say (which Dave also drew attention to):
You no doubt begin to get the impression that the Poynting theory at least partially violates your intuition as to where energy is located in an electromagnetic field. You might believe that you must revamp all your intuitions, and, therefore have a lot of things to study here. But it seems really not necessary. You don’t need to feel that you will be in great trouble if you forget once in a while that the energy in a wire is flowing into the wire from the outside, rather than along the wire. It seems to be only rarely of value, when using the idea of energy conservation, to notice in detail what path the energy is taking. The circulation of energy around a magnet and a charge seems, in most circumstances, to be quite unimportant. It is not a vital detail, but it is clear that our ordinary intuitions are quite wrong.
Perhaps its possible to ignore all this business and retreat back to the hydraulic arguments about electrons being like water in pipes. Hayt concedes it might just be a philosophical problem - but yet he takes a firm position on which interpretation he prefers. So did Heaviside. So did Kraus. And even Feynman to an extent. He was lecturing to a room of freshmen/sophomore physics students in the 1960s. If he were talking to a room of engineers designing waveguides, the emphasis would be very different.
Our intuition is wrong - and these properties of fields are important if we think Maxwellian Theory means anything.
You found some nice quotes like:
But it [eq.27.20] tells us a peculiar thing: that when we are charging a capacitor, the energy is not coming down the wires; it is coming in through the edges of the gap. That’s what this theory says!
The standard approach is to assume that electrical energy flows inside metal wires (confinement), but Heaviside's energy current approach dictates otherwise.
which explains clearly that it is a theory dictating where energy flows, not experiments.
Much like bsfeechannel, they don't point to experiments, because they cannot.
And here:
You no doubt begin to get the impression that the Poynting theory at least partially violates your intuition as to where energy is located in an electromagnetic field. You might believe that you must revamp all your intuitions, and, therefore have a lot of things to study here. But it seems really not necessary. You don’t need to feel that you will be in great trouble if you forget once in a while that the energy in a wire is flowing into the wire from the outside, rather than along the wire. It seems to be only rarely of value, when using the idea of energy conservation, to notice in detail what path the energy is taking. The circulation of energy around a magnet and a charge seems, in most circumstances, to be quite unimportant. It is not a vital detail, but it is clear that our ordinary intuitions are quite wrong.
You can see that intuitions being "wrong" is not justified by experiments, it's just that it just intuition is not what Poynting's theory tells:
But it [eq.27.20] tells us a peculiar thing: that when we are charging a capacitor, the energy is not coming down the wires; it is coming in through the edges of the gap. That’s what this theory says!
Much like "siemens" is not the wrong version of "mho" (even when Thomson himself used mho), "energy flows through wires" is not the wrong version of "energy flows through vacuum".
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And so we come back to a central conceit of the (OP) problem: does energy "flow" matter? Does it mean anything at all?
Not really.
You can store energy, and release energy, at will. Nothing wrong with that. You can say it goes through cables, pipes, whatever. Also nothing wrong with that.
But it's not particularly important where and how it goes.
The likes of Feynman would be perfectly comfortable with such a view. Feynman was a huge advocate of Lagrangian mechanics, where the position and path don't matter on an instantaneous basis, instead you solve for them using an overall energy argument. (Which leads perfectly into QM where, not only is the path unimportant, but indeed the probability of every possible path, including classically-intuitively-absurd paths, has some effect and the total must be taken to find the correct result.) In classical mechanics of course, we don't have to deal with a superposition of possible paths, but only the one unique path that gives "least action" (minimizes energy). Then the path, the complete history for all time, of the particle or system or whatever, simply drops out of the equations and we can take its value at any particular point in space or time to find position, velocity, altitude, energy, whatever.
Applied to circuitry, I don't give a rat's ass what energy is flowing in my traces or cables or whatever, just that they handle enough voltage and current to do the job. That capacity requirement will ultimately be some function of the energy flowing, sure, but that's not what's important, and there are easier ways to calculate it than from the energy or power.
And so, it is even less important whether that energy is "carried in" or "around" the wires. Dude, it just gets there when it does!
(Even for antenna purposes, I don't know that it's very useful or meaningful, as anything near the active elements is necessarily near-field, and what's going on there, need not be representative of the radiated field. You don't optimize on near fields, you optimize on radiated fields, you want to know gain and radiation pattern -- who cares what's going on near.)
I think most people here already know this, which is why the bulk of this thread has been occupied with a... different, related curiosity.
Tim
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which explains clearly that it is a theory dictating where energy flows, not experiments.
Much like bsfeechannel, they don't point to experiments, because they cannot.
LoL. Here's two experiments that show where energy flows.
Experiment 1:
Attached a transformer to 220V AC, and get (say) 9V AC @ 0.5A out the other side. The wires never touch (due to insulation). How does that energy pass through the transformer?
The answer for some here seem to be : It can't. Because the energy flows in the wires. Those Watts coming out the transformer are not from the energy coming in.
Experiment 2:
Store energy into the middle of three capacitors in series. Energy is transferred even though no charges can pass around the entire loop. How is that possible?
The answer for some here seem to be : It isn't possible, because the energy flows in the wires. The energy in that middle capacitor isn't stored electrical energy, it's something else.
Question still not answered:
Why do electrons in a wire drift proportionally to the current (the flow of charge), and not the energy being transferred?
For some the answer is : I'm not sure that they even drift
For others the answer for some here seem to be : Electrons carry not only charge they also carry the energy. They all have little backpacks they carry their electrical energy around it, and when a battery is connect or disconnect a battery they all quickly fill or empty their backpacks as required, sometimes over great distances, exactly at the time the connection is made or broken.
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which explains clearly that it is a theory dictating where energy flows, not experiments.
Much like bsfeechannel, they don't point to experiments, because they cannot.
LoL. Here's two experiments that show where energy flows.
Experiment 1:
Attached a transformer to 220V AC, and get (say) 9V AC @ 0.5A out the other side. The wires never touch (due to insulation). How does that energy pass through the transformer?
The answer for some here seem to be : It can't. Because the energy flows in the wires. Those Watts coming out the transformer are not from the energy coming in.
Perhaps it's what they say.
What I say is that you get 4.5W consumed in the input, 4.5W produced in the output wires.
Potential is converted into momentum in the primary, this increases the momentum in the secondary which is then converted into potential (9V).
Experiment 2:
Store energy into the middle of three capacitors in series. Energy is transferred even though no charges can pass around the entire loop. How is that possible?
The answer for some here seem to be : It isn't possible, because the energy flows in the wires. The energy in that middle capacitor isn't stored electrical energy, it's something else.
I can't answer for some. But when you charge the exterior capacitors, you create a current and potential difference in the middle one, which stores electrical energy at a rate of VI.
Question still not answered:
Why do electrons in a wire drift proportionally to the current (the flow of charge), and not the energy being transferred?
For some the answer is : I'm not sure that they even drift
For others the answer for some here seem to be : Electrons carry not only charge they also carry the energy. They all have little backpacks they carry their electrical energy around it, and when a battery is connect or disconnect a battery they all quickly fill or empty their backpacks as required, sometimes over great distances, exactly at the time the connection is made or broken.
Mmh I'm not quite sure why you renamed potential energy into 'backpack', but if I were to explain electricity to children, I would consider it. Is potential a forbidden word now?
I'm curious how you see energy moving in vacuum. Is vacuum filling/emptying backpacks too? And giving to electrons/protons?
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which explains clearly that it is a theory dictating where energy flows, not experiments.
Much like bsfeechannel, they don't point to experiments, because they cannot.
LoL. Here's two experiments that show where energy flows.
Experiment 1:
Attached a transformer to 220V AC, and get (say) 9V AC @ 0.5A out the other side. The wires never touch (due to insulation). How does that energy pass through the transformer?
The answer for some here seem to be : It can't. Because the energy flows in the wires. Those Watts coming out the transformer are not from the energy coming in.
Experiment 2:
Store energy into the middle of three capacitors in series. Energy is transferred even though no charges can pass around the entire loop. How is that possible?
The answer for some here seem to be : It isn't possible, because the energy flows in the wires. The energy in that middle capacitor isn't stored electrical energy, it's something else.
Question still not answered:
Why do electrons in a wire drift proportionally to the current (the flow of charge), and not the energy being transferred?
For some the answer is : I'm not sure that they even drift
For others the answer for some here seem to be : Electrons carry not only charge they also carry the energy. They all have little backpacks they carry their electrical energy around it, and when a battery is connect or disconnect a battery they all quickly fill or empty their backpacks as required, sometimes over great distances, exactly at the time the connection is made or broken.
Experiment 1:
You have two energy storage devices with magnetic coupling. Primary creates a magnetic field and the energy to create that magnetic field will either be returned to source if nothing is connected to secondary or it can be used by the secondary as it has access to the same stored magnetic field.
All energy flows through wires so if you have 9V * 0.5A = 4.5W power output then input power will be the same assuming ideal transformer so 4.5W / 220Vac = 0.02A
Energy is being charged and discharged multiple times per second.
But let me ask you this. Why if you apply DC to primary you can not get anything on the secondary ? With DC you still have energy stored in the magnetic field when you connect the DC source and the magnetic field is still there so how come you can not transfer energy from primary to secondary ?
There will be a lot of energy wasted as heat in the primary with DC so energy travels through wire but you can not get any of that on the secondary as energy can only travel through wire (at least at this low potential).
Experiment 2:
Any number of capacitors in series will be seen as a single capacitor and energy will be divided between all those capacitors.
Energy flows only as long as it is needed to charge the capacitors and once the capacitors are fully charged no energy is flowing. None of that energy did any work so no heat or any other sort of radiation for ideal case and some amount of heat with resistance just for as long as the capacitors are not fully charged.
As for question still not answered. They were answered many times you just do not understand what energy is.
Not understanding energy, energy storage and likely much more makes you unable to predict what happens without running an experiment.
If the transfer of energy will have been possible efficiently without wires the there will be no expensive transmission lines.
You can transfer energy outside wires but that requires electrons or photons.
The youtube video I linked today about the electrostatic generator shows energy flowing both through wires and trough air. It is a special case due to very high voltages where air becomes a conductor but energy is not transferred by the electric field but by electrons that jump through dielectric in this case air and is excelentry presented in that video.
There are thousands of volts and small gaps that allows some energy transfer outside the wires in that applications but that space becomes a conductor at those voltages. Nothing like that is present in Derek's 20V and 1m gap experiment and all energy there travels through wires both during initial transient and in steady state DC.
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which explains clearly that it is a theory dictating where energy flows, not experiments.
Much like bsfeechannel, they don't point to experiments, because they cannot.
LoL. Here's two experiments that show where energy flows.
Experiment 1:
Attached a transformer to 220V AC, and get (say) 9V AC @ 0.5A out the other side. The wires never touch (due to insulation). How does that energy pass through the transformer?
The answer for some here seem to be : It can't. Because the energy flows in the wires. Those Watts coming out the transformer are not from the energy coming in.
Perhaps it's what they say.
What I say is that you get 4.5W consumed in the input, 4.5W produced in the output wires.
Potential is converted into momentum in the primary, this increases the momentum in the secondary which is then converted into potential (9V).
How is that momentum transferred? When a marble rolls past a stationary one, the stationary one doesn't just start moving. Do the charges have little sticks that the prod each other with?
Experiment 2:
Store energy into the middle of three capacitors in series. Energy is transferred even though no charges can pass around the entire loop. How is that possible?
The answer for some here seem to be : It isn't possible, because the energy flows in the wires. The energy in that middle capacitor isn't stored electrical energy, it's something else.
I can't answer for some. But when you charge the exterior capacitors, you create a current and potential difference in the middle one, which stores electrical energy at a rate of VI.
How does that current and potential get into the middle capacitor? And what what is current times potential, if it isn't energy?
Question still not answered:
Why do electrons in a wire drift proportionally to the current (the flow of charge), and not the energy being transferred?
For some the answer is : I'm not sure that they even drift
For others the answer for some here seem to be : Electrons carry not only charge they also carry the energy. They all have little backpacks they carry their electrical energy around it, and when a battery is connect or disconnect a battery they all quickly fill or empty their backpacks as required, sometimes over great distances, exactly at the time the connection is made or broken.
Mmh I'm not quite sure why you renamed potential energy into 'backpack', but if I were to explain electricity to children, I would consider it. Is potential a forbidden word now?
I'm curious how you see energy moving in vacuum. Is vacuum filling/emptying backpacks too? And giving to electrons/protons?
Here are my incoherent ramblings...
For electrostatics, the electric field fill the universe (just like gravity does). And the location of charges in that field define the electric field, just like how the location of masses define the gravitational field.
In a wire, where charges can move freely, changes drift to where they see the local field leading them, like marbles rolling down into a valley under gravity. They don't need to 'know' that there is a battery that is 10cm away to know which way to go, they just mindlessly follow the slope of the local electric field. Exactly like how water finds it's way to the outlet of a lake or dam. And as they move, their location also contributes to the electric field. Because charges are able to freely move within the wire, and their location defines the electric field, the electric field quickly becomes flat inside conductors when modest currents are flowing. The charges are not dissipating much energy, they are "doing minimal work" in the physics sense (force x distance).
At the edges and outside of the wires, where the charges can't freely move is where all the tension in the electric field occurs - that is where the fields have the most 'slope'. It is on that slope where you can extract energy from the fields. If you release a charge on such a slope it will know which way want to go - a negative charge will head in the "most positive" direction, and a positive charge will head in the "most negative" direction. If you were able to put an extra charge into a wire not much will happen - it will just drift along on the current.
You attach one end of a resistor to a just the positive wire, but leave the other end free. A small amount of charge will flow into it, but very quickly the whole resistor will have an flat electric field, just as flat as the wire. Anywhere you measure with a voltmeter on the either resistor or the wire will measure 0V. The resistor isn't releasing any of the field's energy, just moving where the field's energy is in space.
But when you attach one end of the resistor to a positive wire. and the other end to the negative wire, then you can extract energy from the field. All the semi-mobile charges in the resistor will see the "so many volts per meter" slope of the field and start moving in that direction. Those charges don't need to know how the electric field gets there, just that the field is there, and it has to follow it. This converts electrical energy into momentum of the charge.
Because the slope in the resistor is so high compared to that inside the wire, they really want to move fast. This gives the thermal heating (or light from the light bulb). That energy isn't coming from the electrons moving in the wire, but the slope in the electric field that is through the resistor, that is what accelerates the charge.
The wire supplies a steady supply of low-energy electrons to be accelerated in the resistor, and removes the low energy electrons that appear at the resistor's other end. The charge is accelerated using the energy supplied by the field, not the wire.
All this time, (assuming resistance of the wires is low compared to the resistor the wires) the electric field on the inside of the wires is flat, and the charges in the wire only transfers minimal energy. The wires set the shape of the electric field, and supplies charge, but the energy flows in the fields.
Batteries also change the shape of the electric field. They generate (and try to maintain) an electric field between their terminals. Batteries are sold as X volt batteries. The first thing you care about for a battery is the strength of the electric field it generates between its terminals. The total energy it can supply is a secondary consideration (along with size or cost). When you connect a battery to a wire, because the wire's charges are mobile the electric field around that wire changes to match that of the battery's terminal. And sure, some charge movement is required for this, but if you could look at how much charge moved how far to build up the electric field it is minimal. (That is unless somebody has put a large capacitor in there somewhere...)
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How is that momentum transferred? When a marble rolls past a stationary one, the stationary one doesn't just start moving. Do the charges have little sticks that the prod each other with?
They (electrons) have a charge thus they repel each other and so they do not collide. This electron wave moves very close to the speed of light no matter the voltage or current.
So when an electron gets out of a source (battery, charged capacitor, any sort of generator) it will travel the loop (not that electron but the wave) at around the speed of light and so another electron will enter the source on the other side.
It seems physicists think in terms of mathematical photons emitted and absorbed by the electrons but no matter the real mechanism for the electron wave it will travel in wire (metals have free electrons) unless there is sufficient energy (high voltage) to go through a dielectric (not the case with Derek's example).
How does that current and potential get into the middle capacitor? And what what is current times potential, if it isn't energy?
As mentioned above electrons will repel other electrons (force will drop with distance) So this repulsion force works across a fairly large gap like the dielectric in a capacitor or even 1m in air.
So electrons enters the left side of a capacitor that repels electrons from the right side plate of the same capacitor and the those electrons will travel (the electron wave it will not be the same exact electron) will get through wire to the plate of the second capacitor with will create the same effect on that second capacitor and then the third and so an electron will get back into the source.
This energy is stored in all 3 capacitors (energy that got out of the source ends split between the 3 capacitors with some energy lost due to wire resistance).
There is a big difference between energy being stored and energy being used to do work.
So you can say that one single electron left the source and you end up with 3 electrons one in each capacitor. But the electron that left the source had much higher energy (source had higher voltage than the voltage across the 3 series capacitors combined).
So knowing how many electrons you have is not a measure of how much energy you have. Energy as shown by the formula depends on the capacity in Farad (dependent on distance and plate area plus the dielectric) and the voltage squared.
The higher the source voltage vs the voltage across the capacitors the higher the energy loss on the wire with alone should be enough evidence to show that energy travels through wire.
A switch is a capacitor and yet you need to close the switch in order to transfer energy. If energy could flow through the capacitor witch that is what a switch is then you will never need to close the switch.
Everything that is known point at the fact that energy travels through wires else switch could not stop the energy flow when open and there will be no loss as heat inside the wires exactly matching what will be expected based on amount of energy transferred and wire resistance to current flow.
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which explains clearly that it is a theory dictating where energy flows, not experiments.
Much like bsfeechannel, they don't point to experiments, because they cannot.
It's theory explaining experiments. That's what a theory is for. I told you several pages ago that the idea that energy was transmitted in wires was already debunked in the 19th century based on experimental data. I even described what experiments they were.
Mmh I'm not quite sure why you renamed potential energy into 'backpack', but if I were to explain electricity to children, I would consider it. Is potential a forbidden word now?
I'm curious how you see energy moving in vacuum. Is vacuum filling/emptying backpacks too? And giving to electrons/protons?
What is remarkable of Derek's experiment is that it tosses in the bin all the alternatives to the Poynting theorem--a kind of discussion that only entertain physicists--especially S=VJ, so cherished by the hydraulic analogy lovers.
And when it comes to intuition, it seems that the Poynting theorem makes more sense now in the 21st century than it did way back when, as it can be stated by this underrated comment to Derek's video:
(https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/?action=dlattach;attach=1482844;image)
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As mentioned above electrons will repel other electrons (force will drop with distance) So this repulsion force works across a fairly large gap like the dielectric in a capacitor or even 1m in air.
If that is so, then that's a big problem.
A circuit with a few very large capacitors there might be a Coulomb of charge 'in play'. For an given elementary charge, it has to be interacting with 6.24x10^18 other elementary charges in the circuit to find the overall force on it. That whole F = k Q1 * Q2 / d^2", summed up over all the charges. This is very unlike marbles rolling around on a flat surface, where marbles only exchange energy when in contact. But it gets worse. I might have a big bag of protons on the shelf, so they need to factor into the equation. Dave might have a really big bag of electrons under his desk in Australia, so that needs to be included.
And then there are the rovers on Mars, waggling electrons around in their antenna. So that needs to be factored in too... the circuit may in fact be the receiver for the signals from Mars rover, so you can't say that those interactions don't matter and don't occur.
If the true reality is based on charges interacting at a distance, then for any charge to know what is expected of it needs to interact with every elementary charge in the whole universe, and that means every charge knows all the others on a personal, first name basis - they are all sending each other little messages all the time. That would be truly amazing!
Then an atom of Carbon 14 in a carbon film resistor resistor decays, that's two new elementary charges that every other elementary charge in your circuit and the while universe needs to interact with. It just never stops!
If you are happy with that situation being how reality is, then by all means see the world as the sums of interactions of bulk quantity of elementary charges. It will be near enough in most cases. A few billion electrons here in this charge, another few billions over there. That is the lumped element model. So wires running next to each other are transmission lines are inductors and capacitors all the way along them.
In that case, if electricity was gravity, you will be safe knowing that the apple will always fall towards the ground, and you can attribute it to the uncountable number of atoms in the apple are each individually being tugged on by the even more uncountable number of atoms in planet Earth, based on an inverse square law everybody learns at school - F = G m1 m2 / d^2 - which very similar to the equation for interacting charges.
The presence of the electric field resolves all this false complexity. The charge interacts with the electric field locally. The field interacts with itself (like waves on a pond), combining and distributing information. The charge experiences all the interactions with other charges in the universe though their combined effect on the electric field exactly where it is. A charge doesn't need to communicate with Mars - that information is already there in the field. Those elementary charges can now stop writing their messages to all the other charges in the universe, and just go with the flow, directed by the electric field exactly where it is.
And if something dramatic changes, like somebody connects a battery to a wire? That energy and information spreads out in the Electric field, causing charges in the wire to move until the field inside the wire is once again nearly flat (this is the "electron wave"), and all the 'potential difference' between the wires is the energy in the field between wires - there because of the battery.
(once again this all ignore the magnetic field, which is actually needed to receive signals from Mars...)
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But it's not particularly important where and how it goes.
In principle I kind of agree with you, but I think it does matter in some cases. For instance, the recent discussion of PCB stackups is a bit of a nonsense until you take fields into account.
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As mentioned above electrons will repel other electrons (force will drop with distance) So this repulsion force works across a fairly large gap like the dielectric in a capacitor or even 1m in air.
If that is so, then that's a big problem.
...
(once again this all ignore the magnetic field, which is actually needed to receive signals from Mars...)
The lumped element circuit model for the transmission line works because it is what happens in real world. If calculations based on that will not provide accurate results then it will not be used.
You maybe saw my Spice simulation for a 10m transmission line and the result is the same as what Derek got testing it in real world.
When switch is closed energy will first need to charge the line capacitance and this is the reason you see a small current through the lamp as the lamp is a wire connecting the two capacitors in series.
The question is not even as complicated as you make it to be and is just
Does energy travels through wires in Derek's low voltage experiment ?
The answer for me is a clear yes and nothing points to other conclusions as all measurements you make point to the fact that energy travels through wires.
Energy is a generic term for all forms and you can transfer energy without wires but it will be in a different form like infrared photons that will be emitted by a transmission wire if resistance of that wire is not zero.
An incandescent lamp filament is a wire with high resistance and it emits a lot of photons mostly in the infrared spectrum but as it gets hotter some in visible spectrum. And yes that is energy transferred outside the wire (those photons) but that is undesired for a transmission line so it is minimized so that most of the energy is transferred through the wire with as little loss as it will be economical.
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A photon is the elementary particle dual of an electromagnetic wave. Light is an electromagnetic wave.
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How is that momentum transferred? When a marble rolls past a stationary one, the stationary one doesn't just start moving. Do the charges have little sticks that the prod each other with?
This is, again, a non-physical question.
If you want to imagine that charges water the magnetic field, and magnetic wheat pushes on far-away charges, it works.
If you want to say that God propagates potential and momentum at the speed of light, it works too.
What really matter is: it is, and does so according to Maxwell equations.
How does that current and potential get into the middle capacitor? And what what is current times potential, if it isn't energy?
So you agree that energy is flowing in wires in the middle capacitor?
For how the potential travels: see above.
For electrostatics, the electric field fill the universe (just like gravity does). And the location of charges in that field define the electric field, just like how the location of masses define the gravitational field.
In a wire, where charges can move freely, changes drift to where they see the local field leading them, like marbles rolling down into a valley under gravity. They don't need to 'know' that there is a battery that is 10cm away to know which way to go, they just mindlessly follow the slope of the local electric field. Exactly like how water finds it's way to the outlet of a lake or dam. And as they move, their location also contributes to the electric field. Because charges are able to freely move within the wire, and their location defines the electric field, the electric field quickly becomes flat inside conductors when modest currents are flowing. The charges are not dissipating much energy, they are "doing minimal work" in the physics sense (force x distance).
At the edges and outside of the wires, where the charges can't freely move is where all the tension in the electric field occurs - that is where the fields have the most 'slope'. It is on that slope where you can extract energy from the fields. If you release a charge on such a slope it will know which way want to go - a negative charge will head in the "most positive" direction, and a positive charge will head in the "most negative" direction. If you were able to put an extra charge into a wire not much will happen - it will just drift along on the current.
You attach one end of a resistor to a just the positive wire, but leave the other end free. A small amount of charge will flow into it, but very quickly the whole resistor will have an flat electric field, just as flat as the wire. Anywhere you measure with a voltmeter on the either resistor or the wire will measure 0V. The resistor isn't releasing any of the field's energy, just moving where the field's energy is in space.
But when you attach one end of the resistor to a positive wire. and the other end to the negative wire, then you can extract energy from the field. All the semi-mobile charges in the resistor will see the "so many volts per meter" slope of the field and start moving in that direction. Those charges don't need to know how the electric field gets there, just that the field is there, and it has to follow it. This converts electrical energy into momentum of the charge.
Because the slope in the resistor is so high compared to that inside the wire, they really want to move fast. This gives the thermal heating (or light from the light bulb). That energy isn't coming from the electrons moving in the wire, but the slope in the electric field that is through the resistor, that is what accelerates the charge.
The wire supplies a steady supply of low-energy electrons to be accelerated in the resistor, and removes the low energy electrons that appear at the resistor's other end. The charge is accelerated using the energy supplied by the field, not the wire.
All this time, (assuming resistance of the wires is low compared to the resistor the wires) the electric field on the inside of the wires is flat, and the charges in the wire only transfers minimal energy. The wires set the shape of the electric field, and supplies charge, but the energy flows in the fields.
Batteries also change the shape of the electric field. They generate (and try to maintain) an electric field between their terminals. Batteries are sold as X volt batteries. The first thing you care about for a battery is the strength of the electric field it generates between its terminals. The total energy it can supply is a secondary consideration (along with size or cost). When you connect a battery to a wire, because the wire's charges are mobile the electric field around that wire changes to match that of the battery's terminal. And sure, some charge movement is required for this, but if you could look at how much charge moved how far to build up the electric field it is minimal. (That is unless somebody has put a large capacitor in there somewhere...)
Sounds mostly correct to me. But I don't see how electrons are shoved by the electric field, with a stick perhaps? And how do they 'see' it, with tiny eyes?
And if you care about the strength of the electric field, why are they sold as X volt, instead of Y volt/meter? I think you care more about the potential difference than the electric field strength.
Ah and of course, you say that 'energy is in electron' is false, but with no experimental evidence; so it's your opinion, not a fact.
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which explains clearly that it is a theory dictating where energy flows, not experiments.
Much like bsfeechannel, they don't point to experiments, because they cannot.
It's theory explaining experiments. That's what a theory is for. I told you several pages ago that the idea that energy was transmitted in wires was already debunked in the 19th century based on experimental data. I even described what experiments they were.
I must have missed it. Please explain what experiment was done, what was predicted by the S=JV folks, and what was found. (And why Poynting-Heaviside-etc. do not talk about it.)
Also, theories *predict* the result of experiments. You can see this as an 'explanation' if you like but the (main) goal is to have a discrepancy as small as possible between the theoretical result and the practical one.
If you ask questions like *how can charges push on each other?* you will be disappointed.
Mmh I'm not quite sure why you renamed potential energy into 'backpack', but if I were to explain electricity to children, I would consider it. Is potential a forbidden word now?
I'm curious how you see energy moving in vacuum. Is vacuum filling/emptying backpacks too? And giving to electrons/protons?
What is remarkable of Derek's experiment is that it tosses in the bin all the alternatives to the Poynting theorem--a kind of discussion that only entertain physicists--especially S=VJ, so cherished by the hydraulic analogy lovers.
I guess you're correct, something true is now in the bin for millions of people. How remarkable.
I also wonder how exactly all physicists proposing alternatives to the Poynting theorem never saw Derek's antennae coming in the whole 20th century. They must feel very silly now (no).
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"Sounds mostly correct to me. But I don't see how electrons are shoved by the electric field, with a stick perhaps? And how do they 'see' it, with tiny eyes?"
As with other fields discussed in physics, the E-field (in V/m) is defined in terms of the physical force exerted by the field on a physical charge.
(The gravitational field, similarly, is defined in terms of the force exerted by the field on a mass.)
F = qE, where the force F and the field E are vectors, and the charge q is a scalar.
The field from electron #1 at a separation (vector) r is Kq1 r1/r2,
where r1 is the unit vector along the separation vector r, and r is the magnitude (length) of the vector r.
The constant K, in SI (mks) units is 9 x 109 m/F, as can be found in any elementary textbook that uses "rationalized mks units".
Therefore, since the electronic charge qe = 1.6 x 10-19 C for both particles, and each has mass 9.1x10-31 kg, you can calculate the acceleration, remembering Newton's Third Law since if both electrons are free, the mutual repulsion sends both of them in opposite directions in the "lab frame".
With equal charges, one gets the familiar "inverse square" dependence on distance and q1q2 dependence on the charges.
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The real issue with this thread and others like it, is that the wrong language is being used to discuss the topic under consideration--a natural language, like English.
Such language is too imprecise, and subject to too much misinterpretation.
To be successful, the language used needs instead to be mathematical, such as shown in this video:
(The video even comes with an accidental mistake as a bonus!)
https://www.youtube.com/watch?v=XAKAlNH9dDw (https://www.youtube.com/watch?v=XAKAlNH9dDw)
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The real issue with this thread and others like it, is that the wrong language is being used to discuss the topic under consideration--a natural language, like English.
Such language is too imprecise, and subject to too much misinterpretation.
To be successful, the language used needs instead to be mathematical, such as shown in this video:
(The video even comes with an accidental mistake as a bonus!)
What was the mistake in the video ? I have not noticed any mistake.
Maxwell's equations do not say that energy flows outside or inside the wires.
The question is very simple and so is the answer for anyone that understand what energy is. And is not a limitation of the english language is a limitation of people not understanding what the word energy means in any context.
I noticed people always confusing Power and Energy and I always wanted to do a video explaining them but I now realize the confusing is much more than just between power and energy.
I always get questions (related to my business) from people asking how they can charge the Lithium battery in their RV or Boat from the vehicle alternator to compensate for times when there is less solar.
My answer is always why no add more solar since cost of solar generation is about 50x less expensive than generating un using gasoline. Theyr replay is that they need to drive anyway so why not take advantage of the "free energy" from the alternator.
So they think that their fuel consumption will be the same if they drive at constant speed say on highway and take or not energy from the alternator.
Every kWh taken from an alternator will cost about a liter of fuel so right now here gasoline is $1.5/liter thus about $1.5/kWh vs solar panels that have a cost amortisation of just around $0.02/kWh that is 75x less (the 50x was from the time gasoline was $1 not that long ago).
I use math to prove energy travels through wires and it seems it did not help convincing almost anyone here.
There is absolutely no experimental proof to show energy transfer in Derek's experiment travels outside the wires.
Derek just did not understood that two parallel wires form a capacitor and that is the reason he seen a current through the lamp well before electron wave had the time to travel through the wire.
He did not check to see that power provided by battery is higher than power used by lamp by significant margins and that power is not wasted but stored in the capacitor formed by the wires. Then latter power provided by battery will be lower than power output at the lamp so stored energy is being discharged.
[attach=1]
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A photon is the elementary particle dual of an electromagnetic wave. Light is an electromagnetic wave.
A photon is the fundamental (quasi) particle.
Photons emit em radiation, from the main central helical body of the photon.
Radio waves are em radiation.
EM radiation is a slab of E by H energy current. When i say slab, i mean there is no rolling E to H to E etc. Hertz was wrong.
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A photon is the elementary particle dual of an electromagnetic wave. Light is an electromagnetic wave.
A photon is the fundamental (quasi) particle.
Photons emit em radiation, from the main central helical body of the photon.
Radio waves are em radiation.
EM radiation is a slab of E by H energy current. When i say slab, i mean there is no rolling E to H to E etc. Hertz was wrong.
Radio waves, as first demonstrated by Hertz, are macroscopic phenomena.
Photons are microscopic.
The standard derivations of E and H in the "far field" (radiation region) have been demonstrated in RF engineering over and over, to the point that they are no longer discussed. It works.
In passing from the microscopic quantum domain to the macroscopic classical domain (such as Maxwell), the macroscopic quantities are the "expectation values" (q.v.) of the quantum mechanical description.
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A photon is the elementary particle dual of an electromagnetic wave. Light is an electromagnetic wave.
A photon is the fundamental (quasi) particle.
Photons emit em radiation, from the main central helical body of the photon.
Radio waves are em radiation.
EM radiation is a slab of E by H energy current. When i say slab, i mean there is no rolling E to H to E etc. Hertz was wrong.
Radio waves, as first demonstrated by Hertz, are macroscopic phenomena.
Photons are microscopic.
The standard derivations of E and H in the "far field" (radiation region) have been demonstrated in RF engineering over and over, to the point that they are no longer discussed. It works.
In passing from the microscopic quantum domain to the macroscopic classical domain (such as Maxwell), the macroscopic quantities are the "expectation values" (q.v.) of the quantum mechanical description.
Natural E×H waves do not exist. Hertz was wrong. As explained by Ionel Dinu. Heaviside was correct.
But non-natural manmade E×H radio waves do exist.
The slab of E×H can be made (manmade) to vary in strength in a sinusoidal way.
In the oldendays the manmade wave was always sinusoidal, koz of the sinusoidal mechanics used to make the wave, & hence this was easily confused with some non-existent apparition called a natural Hertzian E×H wave. And the confusion has carried over into the modern era (which will one day be called the Dark Age of Electricity)(ended by the coming of a Messiah), even tho nowadays there is no excuse, koz nowadays we can make square pulses, & yet the foolishness continues.
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Fourier analysis shows that any periodic waveform can be represented as a series of sinusoidal waveforms, or an integral over a continuous frequency range of sinusoids.
When building a transmitter, you will find that a sinusoidal carrier is a useful and efficient generator of radio waves.
Even Marconi's evanescent waves from ratty spark waveforms contained a sinusoidal carrier, due to the resonance of the antenna system.
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Fourier analysis shows that any periodic waveform can be represented as a series of sinusoidal waveforms, or an integral over a continuous frequency range of sinusoids.
When building a transmitter, you will find that a sinusoidal carrier is a useful and efficient generator of radio waves.
Even Marconi's evanescent waves from ratty spark waveforms contained a sinusoidal carrier, due to the resonance of the antenna system.
Yes i suppose that sinusoidal is best.
I am not sure how Hertz got a sinusoidal shape from his ordinary spark. I think that Ionel Dinu says that Hertz did not have a sinusoidal signal, ie it was faux-sinusoidal (see other thread)(i haven’t read Ionel's papers lately). If Hertz had a sinusoidal signal due to say resonance then Ionel Dinu would be wrong i think.
The Gasser-X was i think a version of Hertz-X, but mainly to do with speed. U might remember that Hertz had problems with speed, ie he got an infinite speed at one stage. Here is some of Gasser's stuff, i haven’t read it lately, but it might have some interesting stuff. If not then i apologise.
Superluminal information transfer confirmed by simple experiment.
Wolfgang G. Gasser (May, 2016).
Abstract:
A simple experiment has been performed in order to measure propagation speed of the electric field. The results show that the Coulomb interaction propagates substantially faster than at speed of light c.
Fig. 1: Schematic of the experiment. [not shown]
The experiment uses a spark gap between two conducting spheres acting as capacitors of opposite electric charge. After spark-formation, this rapidly collapsing dipole field is measured by an oscilloscope connected via probes to conducting detector-spheres. Whereas the mutual distance between the detector spheres connected to the oscilloscope remains at Δx = 1.65 m (from left probe tip to right probe tip), different distances from the spark-gap have been measured.
Table. 1 [not shown]
The measured propagation speeds v = Δx/Δt from the left to the right detector sphere, with Δt averaged over each five measurements, range from around 1.4 c to 5 c, and show a dependence on the distance from the spark gap.
The by far simplest explanation of the experiment is the hypothesis that the Coulomb interaction conforms to Coulomb, who assumed instantaneous interaction at a distance. The dependence of the measured propagation speed on the distance of the measurement setup from the spark gap is explained by dissipative losses and "image charge" complication, leading to electric currents in the ground and the walls.
https://www.electronicspoint.com/forums/threads/experimental-evidence-for-v-c-in-case-of-coulomb-interaction.168813/ (https://www.electronicspoint.com/forums/threads/experimental-evidence-for-v-c-in-case-of-coulomb-interaction.168813/)
http://www.pandualism.com/c/coulomb_experiment.html (http://www.pandualism.com/c/coulomb_experiment.html)
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Sounds mostly correct to me. But I don't see how electrons are shoved by the electric field, with a stick perhaps? And how do they 'see' it, with tiny eyes?
I know how it is done in the math. Make a very small bounding box, add up the surface integrals to get net force. As the size of the box reduces to zero, the integral remains.
The physical reality? I do not know. But I can make something up.
Just perhaps the position of a charge is delocalized - it continuously randomly jumps around at very small scale (after all, charges to tunnel through potential energy barriers). Rather than being a point it is a small fuzzy thing (some magic quantum uncertainty handwaving).
When the electric field has a slope to it, that makes small jump with the gradient easier than a jump in the other which is against the gradient of the field. This bias allows energy to transfer from the field to the charge, and allows the charge to accelerate.
This is most likely not the process, but it is a nice easy (maybe even natural) process that could underly such a thing, avoiding the need for charges to have little eyes, little (or very long) sticks to push each other, backpacks to hold energy, and direct connections to all other charges in the universe. Instead they are all just bouncing on vanishingly small pogo sitcks :-DD
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"I am not sure how Hertz got a sinusoidal shape from his ordinary spark."
This is the same method Marconi used, exciting a resonant circuit (in Hertz' case, a "Hertzian dipole" resonant antenna) to generate an evanescent waveform: a sine wave multiplied by a decaying exponential (as the circuit loss dissipates energy from the resonant circuit). I believe the first practical transmitter to generate a true "continuous wave" was the Poulsen arc, later improved into the "Federal Arc", where a carbon arc's negative resistance placed across a resonant circuit. Later, Alexandersson built a 200 kHz high-power alternator for GE. Both were made obsolete by the vacuum tube.
https://farside.ph.utexas.edu/teaching/em/lectures/node94.html for the math behind the Hertzian dipole.
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"I am not sure how Hertz got a sinusoidal shape from his ordinary spark."
This is the same method Marconi used, exciting a resonant circuit (in Hertz' case, a "Hertzian dipole" resonant antenna) to generate an evanescent waveform: a sine wave multiplied by a decaying exponential (as the circuit loss dissipates energy from the resonant circuit). I believe the first practical transmitter to generate a true "continuous wave" was the Poulsen arc, later improved into the "Federal Arc", where a carbon arc's negative resistance placed across a resonant circuit. Later, Alexandersson built a 200 kHz high-power alternator for GE. Both were made obsolete by the vacuum tube.
https://farside.ph.utexas.edu/teaching/em/lectures/node94.html for the math behind the Hertzian dipole.
Interesting.
Anyhow, if Hertz intentionally or accidentally made a sinusoidal radio wave, then what reasoning was used by everyone to say that a radio wave is a photon(s).
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See my reply no. 362 above.
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See my reply no. 362 above.
Interesting.
Anyhow, if Hertz intentionally or accidentally made a sinusoidal radio wave, then what reasoning was used by everyone to say that a radio wave is a photon(s).
Radio waves, as first demonstrated by Hertz, are macroscopic phenomena.
Photons are microscopic.
The standard derivations of E and H in the "far field" (radiation region) have been demonstrated in RF engineering over and over, to the point that they are no longer discussed. It works.
In passing from the microscopic quantum domain to the macroscopic classical domain (such as Maxwell), the macroscopic quantities are the "expectation values" (q.v.) of the quantum mechanical description.
How kum a microscopic photon can be mistaken for a macroscopic em something.
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See my reply no. 362 above.
Interesting.
Anyhow, if Hertz intentionally or accidentally made a sinusoidal radio wave, then what reasoning was used by everyone to say that a radio wave is a photon(s).
Radio waves, as first demonstrated by Hertz, are macroscopic phenomena.
Photons are microscopic.
The standard derivations of E and H in the "far field" (radiation region) have been demonstrated in RF engineering over and over, to the point that they are no longer discussed. It works.
In passing from the microscopic quantum domain to the macroscopic classical domain (such as Maxwell), the macroscopic quantities are the "expectation values" (q.v.) of the quantum mechanical description.
How kum a microscopic photon can be mistaken for a macroscopic em something.
https://en.wikipedia.org/wiki/Statistical_mechanics and so on...
In physics, statistical mechanics is a mathematical framework that applies statistical methods and probability theory to large assemblies of microscopic entities. It does not assume or postulate any natural laws, but explains the macroscopic behavior of nature from the behavior of such ensembles.
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A whole bunch of them make a macroscopic wave. Many a pickle makes a muckle. In formal language, an ensemble.
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Interesting.
Anyhow, if Hertz intentionally or accidentally made a sinusoidal radio wave, then what reasoning was used by everyone to say that a radio wave is a photon(s).
See my reply no. 362 above.
Radio waves, as first demonstrated by Hertz, are macroscopic phenomena.
Photons are microscopic.
The standard derivations of E and H in the "far field" (radiation region) have been demonstrated in RF engineering over and over, to the point that they are no longer discussed. It works.
In passing from the microscopic quantum domain to the macroscopic classical domain (such as Maxwell), the macroscopic quantities are the "expectation values" (q.v.) of the quantum mechanical description.
How kum a microscopic photon can be mistaken for a macroscopic em something.
A whole bunch of them make a macroscopic wave. Many a pickle makes a muckle. In formal language, an ensemble.
Yes a bunch of photons can be made to form a wave. Somehow that appears to be a quality of photons, they like to make formations, ie they are kind of sticky (eg lasers).
But i can't see any sensible physical scientific link tween formations of photons & the em field! What historic discovery has ever even hinted at such a link?
It is my photaenos that combine to form the em field. Photaenos are emitted by the main helical body of every photon.
A photaeno is the simplest fundamental unit of microscopic em radiation. Lots of photaenos combine to give the macroscopic em field.
Standard science is saying that lots of apples (photons) can make a bin of apples (photons), & that lots of bins of apples (photons) make an orange (an em field).
Apparently someone somewhere decided that a bin of photons becomes a bin of em radiation, & that many such bins automatically arrange themselves into formations, called waves, eg radio waves.
Whereas my photaeno idea simply says that lots of microscopic em radiations combine to make a macroscopic em field (shocking)(who could have guessed?).
How is it that bins of photons (oranges) can be lasers. And bins of photons (oranges) can be an em field.
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Two words: quantum mechanics.
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Two words: quantum mechanics.
So, free photons, emitted by atoms of gas excited by electricity in a laser, act like light etc.
And free photons emitted by a very hot glowing wire act like light etc.
And, free photons emitted by electricity in or on a wire act like charge and/or like magnets.
So, free photons can have say 3 states (light charge magnetism). And QM explains.
Question. Why cant free photons have these 3 states at the same time, ie a mixture?
Or light & charge at the same time.
Or light & magnetism at the same time.
Or charge & magnetism at the same time.
Thats a total of 7 combinations.
Dont tell me, let me guess, the probability of some of these is zero all of the time.
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Ah and of course, you say that 'energy is in electron' is false, but with no experimental evidence; so it's your opinion, not a fact.
I charge two capacitors in series, to 10V (so 5V in each). While the power is still on I remove them, and put them on the bench.
They still have charge in them.
How can you tell which one was connected to 10V, and which was connected to 0V
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Here even more proof energy travels through wire's and not outside the wire in Derek's example. No that there is any evidence to the contrary but maybe one of them will make you understand this is the reality in this universe.
Same LTspice transmission line simulation was used but I have used two examples.
First one is with Switched turned ON and staying ON and the second with switched turned ON for just 30ns so it was switched off before electrons got to the lamp/resistor.
The graphs show power leaving the source and the power on the resistor/lamp. The resistor is 1KOhm and the transmission line has 20Ohm resistance so there are some losses as energy travels through the wire.
[attach=1]
This is for switch staying ON. The energy is the area under the power graph and over the 400ns 155.05nJ where delivered by the source while at the lamp/resistor 142.1nJ in the same period
[attach=2]
This is for the switch being turned OFF (open circuit) after 30ns. In this case just 9.16nJ left the source and 7.68nJ where delivered to the lamp/resistor most of witch after the switch was OFF so open circuit.
[attach=3]
I think at least two things are very clear from this graphs.
There is energy storage involved and energy travel through wire with the missing energy being dissipated as heat on the transmission wire.
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:-DD So you're telling me wires are made of shittons of inductors and capacitors, through which the power flows, and that those inductors and capacitors -- you know, the in-circuit manifestations of fields in space -- are somehow hidden away, packed up inside a tiny hunk of metal?
Tim
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Here even more proof energy travels through wire's and not outside the wire in Derek's example. No that there is any evidence to the contrary but maybe one of them will make you understand this is the reality in this universe.
How does sunlight get to the earth? Do you see wires connected between the sun and the earth?
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Does electricity flow in air or wires?
The correct answer is BOTH..
In the video a transient energy lights the bulb quickly because of fields. But the wires take over, otherwise there could be no light output from a DC source.
It seems this whole confusion is cause by the attempt to make it one or the other.
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I agree.
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In the video a transient energy lights the bulb quickly because of fields. But the wires take over, otherwise there could be no light output from a DC source.
It seems this whole confusion is cause by the attempt to make it one or the other.
I think that's missing an important thing. No doubt we are all mostly agreed that there is some fields stuff going on before the wires are connected, but what it's really about is after that, when there is a solid wired connection. Does the energy flow in the wire, on the wire (skin) or is the wire merely a guide and the energy actually flows still in the field? As I see it, and it's sometimes tricky to remember what the argument is about, it's that last option which is the crux of the video and this discussion.
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Does electricity flow in air or wires?
The correct answer is BOTH..
In the video a transient energy lights the bulb quickly because of fields. But the wires take over, otherwise there could be no light output from a DC source.
It seems this whole confusion is cause by the attempt to make it one or the other.
The math works out for both DC and AC, and the claim is the same, the energy flow is outside/on the surface of the wire, even at DC. The entiriety of classical electrodynamics physics in built upon this.
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I think that's missing an important thing. No doubt we are all mostly agreed that there is some fields stuff going on before the wires are connected, but what it's really about is after that, when there is a solid wired connection. Does the energy flow in the wire, on the wire (skin) or is the wire merely a guide and the energy actually flows still in the field? As I see it, and it's sometimes tricky to remember what the argument is about, it's that last option which is the crux of the video and this discussion.
For me the question is entirely about DC and energy inside vs outside the wire. Nothing to do with switches, transmission lines, capacitors, inductors, transformer theory, antenna theory etc etc.
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I think that's missing an important thing. No doubt we are all mostly agreed that there is some fields stuff going on before the wires are connected, but what it's really about is after that, when there is a solid wired connection. Does the energy flow in the wire, on the wire (skin) or is the wire merely a guide and the energy actually flows still in the field? As I see it, and it's sometimes tricky to remember what the argument is about, it's that last option which is the crux of the video and this discussion.
For me the question is entirely about DC and energy inside vs outside the wire. Nothing to do with switches, transmission lines, capacitors, inductors, transformer theory, antenna theory etc etc.
Exactly!
I think that, under static conditions, all the energy is transferred by the electrons. They gain electrostatic potential energy as they move through the battery and loose electrostatic potential energy as they move through the lightbulb. The electron's electrostatic potential energy should not be confused with their kinetic energy.
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"The math works out for both DC and AC, and the claim is the same, the energy flow is outside/on the surface of the wire, even at DC. The entiriety of classical electrodynamics physics in built upon this."
But if you substituted the wires for tubes of the same diameter with copper outer shell of infinitesimal thickness the light bulb would not stay lit. Some inner copper is required beyond just a shell covering.
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Hontas Farmer is back still saying the Derek is both right and wrong acording to QFT/QED
QED (as all other QFTs) is a perturbation theory that extends Classical Electrodynamics to a quantum scale. QED is by no means a more fundamental theory than Classical Electrodynamics, and as being the extension, QED cannot contradict Classical Electrodynamics at macroscopic level. https://arxiv.org/pdf/1201.5536.pdf (https://arxiv.org/pdf/1201.5536.pdf)
Because QED is the perturbation theory, its approximate mathematical framework quickly falls apart when trying to describe classical systems (systems with billions of photons, electrons), as discussed in this paper: http://old.cft.edu.pl/~birula/publ/ClassLimit.pdf (http://old.cft.edu.pl/~birula/publ/ClassLimit.pdf)
In other words, QED has it own domain, and Classical Electrodynamics has its own domain. These domains do not overlap. Within their respective domains each theory has been validated experimentally to a very high degree.
In other words, to describe energy transfer in macroscopic systems proposed in Veritasium experiments, the only scientific tool Physics has is Classical Electrodynamics, and it is accurate at given scales.
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For me the question is entirely about DC and energy inside vs outside the wire. Nothing to do with switches, transmission lines, capacitors, inductors, transformer theory, antenna theory etc etc.
The only EM energy that flows into the wires is the energy that cause wire heating. The rest of the EM energy flows outside the wires.
Here is a paper, where authors compute energy flows around cylindrical wire: http://sharif.edu/~aborji/25733/files/Energy%20flow%20from%20a%20battery%20to%20other%20circuit%20elements.pdf (http://sharif.edu/~aborji/25733/files/Energy%20flow%20from%20a%20battery%20to%20other%20circuit%20elements.pdf)
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Ah and of course, you say that 'energy is in electron' is false, but with no experimental evidence; so it's your opinion, not a fact.
I charge two capacitors in series, to 10V (so 5V in each). While the power is still on I remove them, and put them on the bench.
They still have charge in them.
How can you tell which one was connected to 10V, and which was connected to 0V
Here is a funny exercise (not related to anything I said):
I charge 100 capacitors in series, to 1 MV.
While the power is still on I remove them, and put them on the bench.
They still have charge in them.
How can you tell which one was connected to 1MV, and which was connected to 0V?
If you're trying to say "all electrons are identical" then I agree.
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:-DD So you're telling me wires are made of shittons of inductors and capacitors, through which the power flows, and that those inductors and capacitors -- you know, the in-circuit manifestations of fields in space -- are somehow hidden away, packed up inside a tiny hunk of metal?
Tim
That is exactly right. Also it happens that the simulation using this matches experimental results perfectly.
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For me the question is entirely about DC and energy inside vs outside the wire. Nothing to do with switches, transmission lines, capacitors, inductors, transformer theory, antenna theory etc etc.
Energy flows inside the wire both for AC and DC. The difference is that with DC the energy flow is uniform inside the wire meaning the entire section of the wire is used while with AC the higher the frequency and line capacitance the more charges will flow closer to the surface forming the capacitor.
So for a coaxial cable with AC the energy in the shielded conductor flows closer to the outside surface and for the shield it flows more on the surface facing the shielded conductor.
In the case of Derek's experiment during the transient energy that is delivered by moving electrons travels in conductor but a bit more electrons are on the side facing the other conductor. But that transient is just a fraction of a second and after that the energy will flow uniformly inside the conductor.
It is fairly easy to test by measuring where energy loss in the form of heat is and for DC you will see that the conductor heats uniformly on the entire conductor cross section.
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How does sunlight get to the earth? Do you see wires connected between the sun and the earth?
Sun provides energy to earth through photons with most of the energy delivered in infrared but also what we call visible light and a bit of UV
With the photoelectric effect so photons will transfer their energy to electrons.
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Does electricity flow in air or wires?
The correct answer is BOTH..
In the video a transient energy lights the bulb quickly because of fields. But the wires take over, otherwise there could be no light output from a DC source.
It seems this whole confusion is cause by the attempt to make it one or the other.
Electric flow is due to electron flow and that happens in wire both for DC and AC/transient.
Unless you have electrons flying through the air from one conductor to another electrical energy will not travel through air.
And yes there are case like in that electrostatic generator video I posted but there there are very short distances in air and thousand of volts so that electrons are physically traveling through air delivering energy but that is not the case in Derek's experiment with 20V and 1m between conductors.
Derek's claim written in the thumbnail of his first video was "energy doesn't flow through wires" one of the most absurd claim I ever heard showing that he has no understanding of what energy means.
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you will see that the conductor heats uniformly on the entire conductor cross section
Easy for you to say. How about you demonstrate that?
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you will see that the conductor heats uniformly on the entire conductor cross section
Easy for you to say. How about you demonstrate that?
People tested that well before I was born it just seems that internet instead of being this great learning tool it becomes a space for misinformation.
What about you just measure the electrical resistance of a copper pipe with thin walls and the electrical resistance of a copper bar with same diameter then let me know if there is a difference.
If there is a difference in resistance that means electrons from your multimeter traveled through the middle of the copper bar not just close to the surface.
You will see that resistance will be proportional with the copper section area so electron wave will travel uniformly through the material.
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Hontas Farmer is back still saying the Derek is both right and wrong acording to QFT/QED
QED (as all other QFTs) is a perturbation theory that extends Classical Electrodynamics to a quantum scale. QED is by no means a more fundamental theory than Classical Electrodynamics, and as being the extension, QED cannot contradict Classical Electrodynamics at macroscopic level. https://arxiv.org/pdf/1201.5536.pdf (https://arxiv.org/pdf/1201.5536.pdf)
Because QED is the perturbation theory, its approximate mathematical framework quickly falls apart when trying to describe classical systems (systems with billions of photons, electrons), as discussed in this paper: http://old.cft.edu.pl/~birula/publ/ClassLimit.pdf (http://old.cft.edu.pl/~birula/publ/ClassLimit.pdf)
In other words, QED has it own domain, and Classical Electrodynamics has its own domain. These domains do not overlap. Within their respective domains each theory has been validated experimentally to a very high degree.
In other words, to describe energy transfer in macroscopic systems proposed in Veritasium experiments, the only scientific tool Physics has is Classical Electrodynamics, and it is accurate at given scales.
It is sometimes interesting to compare two historical processes:
[1] After Constantine adopted Christianity for the Roman Empire, he wanted it codified and there were a long series of Councils, including Nicaea and Chalcedon, to render the new religion in axiomatic form.
[2] Many centuries later, when the new quantum theories arose, there was a time between the wars when leading scientists, including Bohr, Einstein, Schrödinger, Heisenberg and others, debated theory and experiment to found what became known as Quantum Mechanics. One guiding concept was the "Correspondence Principle" that related quantum results at microscopic scale to classical results at macroscopic scale, initially disregarding special relativity. That "principle" is now less firm than before (depending on the physical situation under discussion), but a good example of a (non-handwaving) quantitative expression of this relationship is the Ehrenfest Theorem (published in 1927).
Those conversant in mathematics can find a description thereof in the Wikipedia article: https://en.wikipedia.org/wiki/Ehrenfest_theorem
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Energy flows inside the wire both for AC and DC
Not necessarily. Take a superconductive wire at DC Inside the superconductor, electric field is zero everywhere (electric field potential is the same at any point inside the superconductor).
In other words: E = 0 inside such wire.
Electromagnetic energy flow is described by Poynting vector:
S = E x H = 0 (if E is zero vector, then its cross product is also zero vector no matter H field strength).
So no, at DC electromagnetic energy does not flow inside a superconductive wire.
If you replace superconductor with a wire that has resistance, then there will be an electric field inside the wire along its axis (equal to wire’s voltage drop divided by the length of the wire).
There is also a magnetic filed inside the wire when it conducts DC current. It can be quantified by Maxwell’s 4th equation. Such magnetic field circulates around wire’s axis. It’s strength is zero at the axis, and increases to a maximum value at wire surface.
Therefore EM energy flow inside such wire will be non zero. However the direction of the flow inside the wire will be strictly perpendicular to the wire (remember direction of E, and the fact that cross product of two vectors is perpendicular to the vectors being multiplied). The total flow at surface (surface integral S over surface of the wire) is equal to Joule heating of the wire.
Remaining EM energy flows outside the wire. Using Poynting vector formula and some math, it can be shown that the energy flown outside the wire + Joule heating is all the energy that flows out of battery. From conservation of energy principal, we can declare that there no other EM energy flows.
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Energy flows inside the wire both for AC and DC
Not necessarily. Take a superconductive wire at DC Inside the superconductor, electric field is zero everywhere (electric field potential is the same at any point inside the superconductor).
In other words: E = 0 inside such wire.
Electromagnetic energy flow is described by Poynting vector:
S = E x H = 0 (if E is zero vector, then its cross product is also zero vector no matter H field strength).
So no, at DC electromagnetic energy does not flow inside a superconductive wire.
If you replace superconductor with a wire that has resistance, then there will be an electric field inside the wire along its axis (equal to wire’s voltage drop divided by the length of the wire).
There is also a magnetic filed inside the wire when it conducts DC current. It can be quantified by Maxwell’s 4th equation. Such magnetic field circulates around wire’s axis. It’s strength is zero at the axis, and increases to a maximum value at wire surface.
Therefore EM energy flow inside such wire will be non zero. However the direction of the flow inside the wire will be strictly perpendicular to the wire (remember direction of E, and the fact that cross product of two vectors is perpendicular to the vectors being multiplied). The total flow at surface (surface integral S over surface of the wire) is equal to Joule heating of the wire.
Remaining EM energy flows outside the wire. Using Poynting vector formula and some math, it can be shown that the energy flown outside the wire + Joule heating is all the energy that flows out of battery. From conservation of energy principal, we can declare that there no other EM energy flows.
Wire is used as a medium to transport energy. If wire has resistance then part of the energy will be wasted in the wire so less of the energy will be available to your load. A wire with no resistance to electrical flow like a superconductor means that no energy is lost while traveling through the wire.
That energy that is delivered before electron wave had the time to travel the length of the wire is due capacitance between wires and the lamp just happened in that particular example to be in series with that cable capacitance so current charging that capacitor made by wires will also pass through lamp/resistor.
The transmission line model provides exact predictions of what happens in a real transmission line and the phenomena associated with that.
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Of course, in vacuum, electron beams flow quite nicely.
In air, they flow less nicely in the form of arcs, sparks, and lightning.
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Of course, in vacuum, electron beams flow quite nicely.
In air, they flow less nicely in the form of arcs, sparks, and lightning.
The amount of energy flow through air can be ignored in Derek's experiment due to super low potential 20V and large gap 1m between the two conductors.
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How does sunlight get to the earth? Do you see wires connected between the sun and the earth?
Sun provides energy to earth through photons with most of the energy delivered in infrared but also what we call visible light and a bit of UV
With the photoelectric effect so photons will transfer their energy to electrons.
Are you aware of the fact that light is an electromagnetic wave? Just like radio waves.
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Wire is used as a medium to transport energy. If wire has resistance then part of the energy will be wasted in the wire so less of the energy will be available to your load.
This is interesting hypothesis, but it has nothing to do with reality, unless:
1) Classical Electrodynamics is off by huge amount at non-relativistic macroscopic scale (scale of the last Veritasium experiment).
or
2) Law of conservation of energy is broken
Proving any of these experimentally would earn a Nobel Prize
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Are you aware of the fact that light is an electromagnetic wave? Just like radio waves.
Yes visible light is part of the electromagnetic spectrum. Sun is not providing electrical energy directly we convert photons into electric current flow through different mechanisms.
Maybe thermal energy is easier to understand. You can transfer that through radiation (infrared photons) in all directions so not very effective is you want to transfer that to a particular object and then you have thermal conductors (it just happens to be that electrical conductors are also good thermal conductors) so if you want to transport thermal energy you will use a thermal conductor and energy will flow through that thermal conductor not around it.
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Wire is used as a medium to transport energy. If wire has resistance then part of the energy will be wasted in the wire so less of the energy will be available to your load.
This is interesting hypothesis, but it has nothing to do with reality, unless:
1) Classical Electrodynamics is off by huge amount at non-relativistic macroscopic scale (scale of the last Veritasium experiment).
or
2) Law of conservation of energy is broken
Proving any of these experimentally would earn a Nobel Prize
Where do you see broken law of conservation of energy in any of my statements ?
And how what I said has nothing to do with reality when that is exactly what you see in reality including Derek's experiment.
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Of course, in vacuum, electron beams flow quite nicely.
In air, they flow less nicely in the form of arcs, sparks, and lightning.
The amount of energy flow through air can be ignored in Derek's experiment due to super low potential 20V and large gap 1m between the two conductors.
Can you *calculate* how much? Because I can using classical electrodynamics, and it agrees with what Derek showed and also his original thought experiment with wires going to the moon. Both versions show that an appreciable fraction of the steady-state power can be transferred before the ends of the wire are causally relevant. The exact amount depends on the load and the wire spacing. The only part I can't "calculate" is the cutoff where I stop saying that the energy is transferred through space and it starts being transferred via the wire. That is because no such cutoff actually exists, and the practice of doing so is a convenient low frequency approximation of electrodynamics that can explains some but not all circuit phenomena.
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Can you *calculate* how much? Because I can using classical electrodynamics, and it agrees with what Derek showed and also his original thought experiment with wires going to the moon. Both versions show that an appreciable fraction of the steady-state power can be transferred before the ends of the wire are causally relevant. The exact amount depends on the load and the wire spacing. The only part I can't "calculate" is the cutoff where I stop saying that the energy is transferred through space and it starts being transferred via the wire. That is because no such cutoff actually exists, and the practice of doing so is a convenient low frequency approximation of electrodynamics that can explains some but not all circuit phenomena.
If you did not already check this post I made yestarday https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/msg4168171/#msg4168171 (https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/msg4168171/#msg4168171)
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Where do you see broken law of conservation of energy in any of my statements ?
And how what I said has nothing to do with reality when that is exactly what you see in reality including Derek's experiment.
You claim that in addition to energy that is flown away from battery by electromagnetic field (that an average STEM undergrad can compute using Poynting vector and calculus at least for DC case), there is additional flow of EM energy inside wire along its axis. This additional energy can’t come from the battery, because all battery energy either flows inside the wire perpendicular to its axis causing Joule heating, or flows outside wire towards load, as can be show by STEM uni student.
So this additional EM energy that you claim flows inside wire along its axis comes from nowhere, and further discussion should move to “over unity” section of this forum.
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Where do you see broken law of conservation of energy in any of my statements ?
And how what I said has nothing to do with reality when that is exactly what you see in reality including Derek's experiment.
You claim that in addition to energy that is flown away from battery by electromagnetic field (that an average STEM undergrad can compute using Poynting vector and calculus at least for DC case), there is additional flow of EM energy inside wire along its axis. This additional energy can’t come from the battery, because all battery energy either flows inside the wire perpendicular to its axis causing Joule heating, or flows outside wire towards load, as can be show by STEM uni student.
So this additional EM energy that you claim flows inside wire along its axis comes from nowhere, and further discussion should move to “over unity” section of this forum.
Have you seen the post where I made the Spice simulation showing energy output from the source and the energy at the lamp/resistor ?
https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/msg4168171/#msg4168171 (https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/msg4168171/#msg4168171)
You have a wrong understanding of what energy is and how it is transferred from source(battery) to load (lamp/resistor).
The graphs I posted there for two cases show exactly how energy travels.
Also an incandescent lamp is nothing more than a wire with higher resistance so is the 1KOhm resistor Derek used in his experiment.
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Have you seen the post where I made the Spice simulation showing energy output from the source and the energy at the lamp/resistor ?
https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/msg4168171/#msg4168171 (https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/msg4168171/#msg4168171)
You have a wrong understanding of what energy is and how it is transferred from source(battery) to load (lamp/resistor).
The graphs I posted there for two cases show exactly how energy travels.
Also an incandescent lamp is nothing more than a wire with higher resistance so is the 1KOhm resistor Derek used in his experiment.
You are using wrong tools. Spice by no means can be used to simulate electromagnetic field in matter and space.
Also, at DC, you did not have to go that far by drawing what looks like a transmission line. 3 resistors (one for load and two for each wire), a battery and a ground symbol are sufficient for DC analysis in Spice.
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You have a wrong understanding of what energy is and how it is transferred from source(battery) to load (lamp/resistor).
If on the one side we have electrodacus, who is right, and on the other side we have the rest of the world, who are wrong, then really all rational people would wish to be with the rest of the world and remain wrong. Apparently, in this scenario, being wrong is the right place to be.
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You are using wrong tools. Spice by no means can be used to simulate electromagnetic field in matter and space.
Also, at DC, you did not have to go that far by drawing what looks like a transmission line. 3 resistors (one for load and two for each wire), a battery and a ground symbol are sufficient for DC analysis in Spice.
That transient when you are closing the switch (with is what Derek concentrated on) is not DC.
All that was needed and if you understood the graphs is exactly what Derek got as a result just wrongly explained the reason for that result.
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If on the one side we have electrodacus, who is right, and on the other side we have the rest of the world, who are wrong, then really all rational people would wish to be with the rest of the world and remain wrong. Apparently, in this scenario, being wrong is the right place to be.
It is very clearly most engineers understand as well as I do how things work so are most Physicists.
There is no evidence against what I'm saying and no evidence for the claim that "energy doesn't flow in wires"
It is not the first time when Derek shows his inability to understand energy and energy conservation.
Way waste resources to do the transmission line experiment when we know (some of us) how to simulate that.
I got the exact same result from stimulating a transmission line as it is much easier to take measurements there including integrating power to get energy at different points.
So is not as you think me against the world it is me doing my best to educate you.
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I think that's missing an important thing. No doubt we are all mostly agreed that there is some fields stuff going on before the wires are connected, but what it's really about is after that, when there is a solid wired connection. Does the energy flow in the wire, on the wire (skin) or is the wire merely a guide and the energy actually flows still in the field? As I see it, and it's sometimes tricky to remember what the argument is about, it's that last option which is the crux of the video and this discussion.
For me the question is entirely about DC and energy inside vs outside the wire. Nothing to do with switches, transmission lines, capacitors, inductors, transformer theory, antenna theory etc etc.
For steady currents you can calculate energy density (ignoring prefactors and constants of nature) as qV + I*Phi [Phi == magnetic flux]. or E^2+B^2 and you will get the same answer. The former describes the electric energy in terms of charges, the latter in terms of fields. You can't really get away from describing the magnetic component in terms of some field in free space because there is no scalar magnetic potential, so I have picked a form where the current plays a role, but I don't need to refer to equivalent circuit elements like L.
Taking only the electric component qV, that is zero on the interior of a conductor because the net charge density is zero. There is a small electric field inside the wire to overcome the wire resistance but the net charge density is zero. The only place with a net charge density is the surface of the wire, and in the charge model that is where all the electrostatic energy is stored. Even though the current is uniformly distributed across the wire cross section there is no (electrostatic) energy density there.
The magnetic component is harder to nail down. For the field centric approach it's no problem: B is unambiguously defined everywhere, so we can just integrate up B^2. But the flux * Phi representation is sort of inherently non-local: its is the current around a loop times the magnetic flux through the loop, so a product of quantities measured at two different locations.
So at DC, you can consistently define the electric component of the energy density to the wire *surface* as an alternative to the fields. When you include the magnetic component or deal with AC or transient behavior you pretty much have to fall back to a field based approach to energy density. There is no reasonable way to quantitatively define the energy to be stored in the volume of the wires.
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That transient when you are closing the switch (with is what Derek concentrated on) is not DC.
All that was needed and if you understood the graphs is exactly what Derek got as a result just wrongly explained the reason for that result.
I thought you were referring to DC when you wrote this: “ The difference is that with DC the energy flow is uniform inside the wire meaning the entire section of the wire is used while with AC the higher the frequency and line capacitance the more charges will flow closer to the surface forming the capacitor.”
No matter DC, AC or transient analysis - you are using wrong tools.
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I'm starting to come around to the idea that the electrons carry the energy, and it's not in the fields around.
All it would take to convince me would be if somebody could design a circuit that could extract 1% of the energy travelling in the wires in the diagram attached.
If you could build it inside the inner 'box' of wires that would be perfect, as I suspect an electric field exists between the inner and outer wires... You could even use a GND reference in the box if you want (but note that none of these wires are attached to GND)
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I think that's missing an important thing. No doubt we are all mostly agreed that there is some fields stuff going on before the wires are connected, but what it's really about is after that, when there is a solid wired connection. Does the energy flow in the wire, on the wire (skin) or is the wire merely a guide and the energy actually flows still in the field? As I see it, and it's sometimes tricky to remember what the argument is about, it's that last option which is the crux of the video and this discussion.
For me the question is entirely about DC and energy inside vs outside the wire. Nothing to do with switches, transmission lines, capacitors, inductors, transformer theory, antenna theory etc etc.
For steady currents you can calculate energy density (ignoring prefactors and constants of nature) as qV + I*Phi [Phi == magnetic flux]. or E^2+B^2 and you will get the same answer. The former describes the electric energy in terms of charges, the latter in terms of fields. You can't really get away from describing the magnetic component in terms of some field in free space because there is no scalar magnetic potential, so I have picked a form where the current plays a role, but I don't need to refer to equivalent circuit elements like L.
[...]
The magnetic component is harder to nail down. For the field centric approach it's no problem: B is unambiguously defined everywhere, so we can just integrate up B^2. But the flux * Phi representation is sort of inherently non-local: its is the current around a loop times the magnetic flux through the loop, so a product of quantities measured at two different locations.
So at DC, you can consistently define the electric component of the energy density to the wire *surface* as an alternative to the fields. When you include the magnetic component or deal with AC or transient behavior you pretty much have to fall back to a field based approach to energy density. There is no reasonable way to quantitatively define the energy to be stored in the volume of the wires.
Yes there is, you replace whatever phi is by A the vector potential, and you get the potential energy.
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Actually there is a scalar magnetic potential, but it is not relevant to general field calculations.
It is used in electromagnet design, including air gaps, and is analogous to scalar electric potential in a circuit, replacing resistance (electrical) by reluctance (magnetic).
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I'm starting to come around to the idea that the electrons carry the energy, and it's not in the fields around.
All it would take to convince me would be if somebody could design a circuit that could extract 1% of the energy travelling in the wires in the diagram attached.
If you could build it inside the inner 'box' of wires that would be perfect, as I suspect an electric field exists between the inner and outer wires... You could even use a GND reference in the box if you want (but note that none of these wires are attached to GND)
First show your solution for a circuit which extract 1% of the energy travelling in the vacuum in the diagram.
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So at DC, you can consistently define the electric component of the energy density to the wire *surface* as an alternative to the fields. When you include the magnetic component or deal with AC or transient behavior you pretty much have to fall back to a field based approach to energy density. There is no reasonable way to quantitatively define the energy to be stored in the volume of the wires.
There are devices that can store energy within conductive media. They call them lasers and masers, including RF masers.
However copper tubes in Veritasium experiment hardly qualify for a maser :)
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First show your solution for a circuit which extract 1% of the energy travelling in the vacuum in the diagram.
Create vacuum in the box and heat the outer wire (cathode) high enough so sufficient number of electrons will flow towards the inner wire (anode) to tap the 1% of energy.
They call this a diode vacuum tube, don’t they?
Now is your turn.
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First show your solution for a circuit which extract 1% of the energy travelling in the vacuum in the diagram.
Create vacuum in the box and heat the outer wire (cathode) high enough so sufficient number of electrons will flow towards the inner wire (anode) to tap the 1% of energy.
They call this a diode vacuum tube, don’t they?
Now is your turn.
That was along the lines of what I was thinking...
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you will see that the conductor heats uniformly on the entire conductor cross section
Easy for you to say. How about you demonstrate that?
People tested that well before I was born it just seems that internet instead of being this great learning tool it becomes a space for misinformation.
What about you just measure the electrical resistance of a copper pipe with thin walls and the electrical resistance of a copper bar with same diameter then let me know if there is a difference.
If there is a difference in resistance that means electrons from your multimeter traveled through the middle of the copper bar not just close to the surface.
You will see that resistance will be proportional with the copper section area so electron wave will travel uniformly through the material.
Woah! No-one's talking resistance here (except you, as a diversion). You said "you will see that the conductor heats uniformly on the entire conductor cross section" - that is heat, thermal. I am wondering just how you can measure the internal temperature of a conductor, and you Internet isn't any help there.
So, just how do you see that? If you make a hole and place a probe you're affecting the conductor integrity, and even with a thermal imager you're only going to see the outside.
Or was this just another 'fact' or 'law' you made up on the spot?
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That transient when you are closing the switch (with is what Derek concentrated on) is not DC.
All that was needed and if you understood the graphs is exactly what Derek got as a result just wrongly explained the reason for that result.
I thought you were referring to DC when you wrote this: “ The difference is that with DC the energy flow is uniform inside the wire meaning the entire section of the wire is used while with AC the higher the frequency and line capacitance the more charges will flow closer to the surface forming the capacitor.”
No matter DC, AC or transient analysis - you are using wrong tools.
That is exactly what I said in reply to Dave.
I do not think I use the wrong tools as I do get the correct result and I want you to show me an alternative way to do it.
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Create vacuum in the box and heat the outer wire (cathode) high enough so sufficient number of electrons will flow towards the inner wire (anode) to tap the 1% of energy.
They call this a diode vacuum tube, don’t they?
Now is your turn.
So you understand that energy transfer is done by the wave of electrons. How many electrons flow from the wire/copper pipe in Derek's experiment ?
Should be very substantial and directional if you want to account for the energy transferred in the first 65ns. What about the rest ?
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Woah! No-one's talking resistance here (except you, as a diversion). You said "you will see that the conductor heats uniformly on the entire conductor cross section" - that is heat, thermal. I am wondering just how you can measure the internal temperature of a conductor, and you Internet isn't any help there.
So, just how do you see that? If you make a hole and place a probe you're affecting the conductor integrity, and even with a thermal imager you're only going to see the outside.
Or was this just another 'fact' or 'law' you made up on the spot?
I guess you will need to learn about another type of energy storage and that will be thermal storage.
If the electrons travel closer to the outside surface of the wire like in AC then resistive losses will show that so there will be no need to even measure the temperature.
In DC current flow the wire inductance is not relevant as magnetic field after it was created will remain constant thus no charging and discharging.
Just look at the case where I had the switch on for just 30ns vs when switch stayed ON and see the difference in energy transmission losses. All data is there for you to compare.
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So you understand that energy transfer is done by the wave of electrons. How many electrons flow from the wire/copper pipe in Derek's experiment ?
Should be very substantial and directional if you want to account for the energy transferred in the first 65ns. What about the rest ?
No. The energy is transferred from EM field to electrons.
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I hope nobody argues here that oscillating EM field does carry energy. Examples are gamma rays, X-rays, UV light, visible light, microwave radiation, radio waves from terahertz to sub-hertz bands all transfer energy. Why is it so hard to realize that a constant EM field also carries energy?
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No. The energy is transferred from EM field to electrons.
Can you be more specific ?
Before the switch placed next to battery is closed there is no magnetic field just electric field that is limited to battery and will not extend out plus it is a constant not a variable field.
The switch is a small capacitor so there is also a constant electric field between the contacts of the switch. So how come you need to close the switch to transfer energy if you already have an electric field that you say can transfer energy.
When you start moving the switch contacts you are changing the capacity of the switch so you now have redistribution of charges that results in variation in electric field (not the other way around).
Say you just moved the switch contacts but you did not closed them then depending on the direction you moved them (away from each other or closer to each other) you will have a current flowing in or out of the battery and out or in of the capacitor/switch (redistributing charges).
This charge redistribution will be completed only after the time needed for the electron wave created to travel the entire loop of wire plus the time for the wave interactions to settle down and so some time after you moved the switch a small current (the one needed to redistribute charge) will flow through the lamp.
So during this charge redistribution some of the energy was lost in the wire and the lamp/resistor.
Is there something that you do not agree with?
If there is something you disagree with please provide details explanations and the equations to back up what you are saying.
You all seems to make ambiguous statements without providing any proof.
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I hope nobody argues here that oscillating EM field does carry energy. Examples are gamma rays, X-rays, UV light, visible light, microwave radiation, radio waves from terahertz to sub-hertz bands all transfer energy. Why is it so hard to realize that a constant EM field also carries energy?
Magnets have a constant magnetic field so do they transfer any energy if sitting next to a copper wire ?
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I hope nobody argues here that oscillating EM field does carry energy. Examples are gamma rays, X-rays, UV light, visible light, microwave radiation, radio waves from terahertz to sub-hertz bands all transfer energy. Why is it so hard to realize that a constant EM field also carries energy?
Magnets have a constant magnetic field so do they transfer any energy if sitting next to a copper wire ?
You need both electric and magnetic field: S = E x H
S is energy flow in any given point. Energy flow outside a region of space is a surface integral of S over surface of the region. I’m just writing this to forestall desire to add constant E field to your example (e.g. fix the magnet between electrodes of charged capacitor). Surface integral of S would be zero over any arbitrary surface. There would be no energy transfer.
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First show your solution for a circuit which extract 1% of the energy travelling in the vacuum in the diagram.
Create vacuum in the box and heat the outer wire (cathode) high enough so sufficient number of electrons will flow towards the inner wire (anode) to tap the 1% of energy.
They call this a diode vacuum tube, don’t they?
Now is your turn.
Careful: an unevenly heated box exhibits free electron drift. A tube diode is a heat engine, albeit a very terrible one.
(https://www.seventransistorlabs.com/Images/6AL5_Spice.png)
Consider this SPICE model fitted to 1/2 6AL5: it has a negative voltage in it, corresponding to the tail of the thermal energy curve. This tail is exponential, interestingly enough (or, not at all), hence the ideal diode, and the nonlinear dependent source expresses the Child-Langmuir law at higher currents (evidently the cylindrical geometry gives an exponent somewhat lower than 3/2 (inverted as it's solving for voltage, hence 0.775).
So, for a 6AL5 that spends about a watt of heater power, you can put its two diodes in parallel, and get a whopping couple of microwatts out.
Whereas in an evenly heated box, the space charge will have some drift velocity, much as the electrons in the bulk conductor; it will be hard to extract any meaningful power from it though, as the maximum difference in energy is merely the difference in voltage, of course.
Tim
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Can you be more specific ?
Sure. Charged particle (an electron) gets accelerated in electric field.
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You have a wrong understanding of what energy is and how it is transferred from source(battery) to load (lamp/resistor).
If on the one side we have electrodacus, who is right, and on the other side we have the rest of the world, who are wrong, then really all rational people would wish to be with the rest of the world and remain wrong. Apparently, in this scenario, being wrong is the right place to be.
The other big thread used to be filled with the Maxwell/Poynting Bro's absolutely shooting down anyone who dared even hinted at suggesting anything other than a 100% Poynting explanation, and heaven forbid if you got even even the slightest direction of Poynting vector wrong, it's was flaming pitchforks.
Now all the Poynting bros have vanished and both threads are now completely dominated by the Energy In The Wire (EIT) absolutists.
LOL :-DD :popcorn:
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The truth is out there.
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The other big thread used to be filled with the Maxwell/Poynting Bro's absolutely shooting down anyone who dared even hinted at suggesting anything other than a 100% Poynting explanation, and heaven forbid if you got even even the slightest direction of Poynting vector wrong, it's was flaming pitchforks.
Now all the Poynting bros have vanished and both threads are now completely dominated by the Energy In The Wire (EIT) absolutists.
LOL :-DD :popcorn:
:) This is not a political or philosophical discussion (or at least it should not be as it is not for me).
As you showed in your own video the transmission line model where you use finite RLC elements works perfect in predicting what you see in real tests.
Claim Derek made is that energy transfer is done outside the wire. Then the most pertinent question is why you even need the wires? They are quite expensive so it will be great if you could transfer electrical energy efficiency without needing them.
Also how come that from experimental result and transmission line calculations you get the same thing and that is energy output from the source equals energy delivered to lamp/load plus the losses in wires. If wires were not delivering the energy then why will all the losses (except some ultra small leakage) are found as heat in the wires. The lamp or resistor is also a wire with higher resistance.
I'm willing to admit that I'm wrong if a pertinent alternative explanation is available with the correct math to back that up.
The simulation I have done in LTspice perfectly matches the experimental result Derek got in his last video and there you can look at all aspects of the model to see where energy is at any point in time.
To have electrical power (energy is the integral of that over time) you need both a voltage potential and a current flow. Unless voltage is so high that current flows through air the current will follow the path of least resistance and that is the wire.
Derek's only so called evidence is that he sees some energy used by the lamp way before the electron wave had time to travel through wire and that is very simply explained by the two capacitors formed by the transmission line on each side of the lamp and showed to be the case form the spice simulation of a transmission line.
That current is needed to charge the capacitors that are just an energy storage device and current cannot flow through them (ignoring again ultra small leakage) but flows in or out of them.
So everything will reduce to the question of current flows through capacitors or in/out of capacitors.
Charging a capacitor requires adding electrons on one side while removing the same amount from the other side. And the electric field that the extra electron produces to push an electron out of the other side will be there only after you have that electron imbalance. Is not the electric field that magically appears between the plates of a capacitors (they are an electric shield) but the field is the result of electron imbalance.
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The other big thread used to be filled with the Maxwell/Poynting Bro's absolutely shooting down anyone who dared even hinted at suggesting anything other than a 100% Poynting explanation, and heaven forbid if you got even even the slightest direction of Poynting vector wrong, it's was flaming pitchforks.
Now all the Poynting bros have vanished and both threads are now completely dominated by the Energy In The Wire (EIT) absolutists.
LOL :-DD :popcorn:
It seems that Mr Poynting got around a bit. I am familiar with the name from my field, chemical engineering:
https://en.wikipedia.org/wiki/Poynting_effect
(Nothing whatsoever to do with electricity.)
But it seems this is the same John Henry Poynting who is associated with the Poynting vector:
https://en.wikipedia.org/wiki/Poynting_vector
In the old days, it appears people were much more able to study everything:
https://en.wikipedia.org/wiki/John_Henry_Poynting
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It seems that Mr Poynting got around a bit. I am familiar with the name from my field, chemical engineering:
https://en.wikipedia.org/wiki/Poynting_effect
(Nothing whatsoever to do with electricity.)
But it seems this is the same John Henry Poynting who is associated with the Poynting vector:
https://en.wikipedia.org/wiki/Poynting_vector
In the old days, it appears people were much more able to study everything:
https://en.wikipedia.org/wiki/John_Henry_Poynting
That guy was dead by the time people found out what atoms are made out of. I sure not heard about him at university (Electrical engineering in some east european country).
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Yeah, the guys who figured out how atoms are constructed are now dead, too.
Funny how real progress in science often builds on previous results.
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This is not a political or philosophical discussion (or at least it should not be as it is not for me).
It seems to be 100% a philosophical discussion for you. You seem to be arguing about nothing.
As you showed in your own video the transmission line model where you use finite RLC elements works perfect in predicting what you see in real tests.
So that's good right? What is there to argue about?
Claim Derek made is that energy transfer is done outside the wire.
The claim Derek made is that enough power would arrive to light a lamp after a certain amount of time. The experiment showed this to be true. The calculations gave the same answer.
Then the most pertinent question is why you even need the wires? They are quite expensive so it will be great if you could transfer electrical energy efficiency without needing them.
Because the wires are part of the experiment. Derek did not suggest the wires are not needed. Quite the contrary, in fact. So why would you suggest the wires are not needed? Nobody even hinted at such a thing.
Also how come that from experimental result and transmission line calculations you get the same thing and that is energy output from the source equals energy delivered to lamp/load plus the losses in wires.
Because this is how things are supposed to work? Why is it strange to you that the calculations and experiment give the same result?
At this point, it is really not clear what you disagree with.
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The other big thread used to be filled with the Maxwell/Poynting Bro's absolutely shooting down anyone who dared even hinted at suggesting anything other than a 100% Poynting explanation, and heaven forbid if you got even even the slightest direction of Poynting vector wrong, it's was flaming pitchforks.
Now all the Poynting bros have vanished and both threads are now completely dominated by the Energy In The Wire (EIT) absolutists.
LOL :-DD :popcorn:
The Poynting guys are away having a party while the wirists are here licking their own wounds.
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At this point, it is really not clear what you disagree with.
I disagree with the main statement Derek made (it is in the video thumbnail) and that is "energy doesn't flow in wires"
There is absolutely no evidence for that. If you think you know what that evidence is then let me know.
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At this point, it is really not clear what you disagree with.
I disagree with the main statement Derek made (it is in the video thumbnail) and that is "energy doesn't flow in wires"
There is absolutely no evidence for that. If you think you know what that evidence is then let me know.
And yet without picking an arbitrary external reference you are unable to point to show one measurable difference between a wire that is carrying 0.1V @ 1mA and the same type of wire carrying 100V @ 1mA.
They even have the same (minimal) internal resistive heating.
They have the same magnetic field around them.
They have the same voltage drop from end to end.
And that is with an energy difference of three orders of magnitude... that is pretty strong evidence.
If you have to use an external reference, that is a pretty strong indicator that is is where it is in space that matters - that where it is in the electric field defines the energy.
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And yet without picking an arbitrary external reference you are unable to point to show one measurable difference between a wire that is carrying 0.1V @ 1mA and the same type of wire carrying 100V @ 1mA.
They even have the same (minimal) internal resistive heating.
They have the same magnetic field around them.
They have the same voltage drop from end to end.
And that is with an energy difference of three orders of magnitude... that is pretty strong evidence.
If you have to use an external reference, that is a pretty strong indicator that is is where it is in space that matters - that where it is in the electric field defines the energy.
You will have a different circuit at least a different lamp if you want your circuit to use 1mA.
At 0.1V for the voltage source the lamp resistance including the likely negligible wire resistance will need to be 100Ohm so that current is 1mA
At 100V the lamp will need to be 100kOhm to get that same 1mA.
So circuit will be very different but wires have resistance to current flow the voltage is irrelevant other than wire isolation or separation in air between wires.
So the difference is the lamp that is also a wire if we are talking about an incandescent lamp.
First lamp will be 0.1mW while second lamp will be 0.1W. If you keep the same wires the voltage drop across the wires will be the same since current is the same thus energy loss on the wire will be the same.
So I have no idea what you want to prove with this example.
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I must have missed it. Please explain what experiment was done,
Very funny. The only thing I did in this thread was to indicate all the experimental data that support the energy flowing in space.
what was predicted by the S=JV folks, and what was found. (And why Poynting-Heaviside-etc. do not talk about it.)
What they accomplished no one knows. What they didn't, we know. They didn't manage to come up with an alternative that doesn't break causality, locality, gauge invariance or a combination of these.
That's why, despite objections regarding its counterintuitive nature, Poynting is the most probable explanation for the flow of energy.
I guess you're correct, something true is now in the bin for millions of people. How remarkable.
Truth? Science is about knowledge. If you are the truth bearer, you're not doing science.
I also wonder how exactly all physicists proposing alternatives to the Poynting theorem never saw Derek's antennae coming in the whole 20th century. They must feel very silly now (no).
They didn't. But their peers did and criticized their proposals. That's why Poynting still stands.
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And yet without picking an arbitrary external reference you are unable to point to show one measurable difference between a wire that is carrying 0.1V @ 1mA and the same type of wire carrying 100V @ 1mA.
They even have the same (minimal) internal resistive heating.
They have the same magnetic field around them.
They have the same voltage drop from end to end.
And that is with an energy difference of three orders of magnitude... that is pretty strong evidence.
If you have to use an external reference, that is a pretty strong indicator that is is where it is in space that matters - that where it is in the electric field defines the energy.
You will have a different circuit at least a different lamp if you want your circuit to use 1mA.
At 0.1V for the voltage source the lamp resistance including the likely negligible wire resistance will need to be 100Ohm so that current is 1mA
At 100V the lamp will need to be 100kOhm to get that same 1mA.
So circuit will be very different but wires have resistance to current flow the voltage is irrelevant other than wire isolation or separation in air between wires.
So the difference is the lamp that is also a wire if we are talking about an incandescent lamp.
First lamp will be 0.1mW while second lamp will be 0.1W. If you keep the same wires the voltage drop across the wires will be the same since current is the same thus energy loss on the wire will be the same.
So I have no idea what you want to prove with this example.
And you are not bothered that the energy lost in the wire is identical, even though it carries vastly different levels of power (1000x)?
The energy loss in the wire changes with the current being transferred, not the energy being transferred.
To me that strongly suggests that the wires are transferring current, not energy.
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Perhaps its possible to ignore all this business and retreat back to the hydraulic arguments about electrons being like water in pipes. Hayt concedes it might just be a philosophical problem - but yet he takes a firm position on which interpretation he prefers. So did Heaviside. So did Kraus. And even Feynman to an extent. He was lecturing to a room of freshmen/sophomore physics students in the 1960s. If he were talking to a room of engineers designing waveguides, the emphasis would be very different.
Our intuition is wrong - and these properties of fields are important if we think Maxwellian Theory means anything.
Science is counterintuitive in essence. Galileo had trouble explaining that the earth spins on its axis. If you ride a fast horse you experience wind against your face. If the earth spins so fast, why don't we see a huge gale sweeping the earth? Einstein's theories, although very successful are the target of active attempts at proving them wrong.
https://www.youtube.com/watch?v=Bo4al7sNPkE (https://www.youtube.com/watch?v=Bo4al7sNPkE)
The job of science is to provide the most probable explanations due to the inductive (in the logical sense) nature of its reasoning. But some people expect the truth from science. Since theories present the most probable explanation, not the "correct", they give the opportunity for trying to find alternatives. Those who expect truth form science, not knowledge, are led to think that the theories are unsound and that there must be some kind of conspiracy to hold them in high regard.
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And you are not bothered that the energy lost in the wire is identical, even though it carries vastly different levels of power (1000x)?
The energy loss in the wire changes with the current being transferred, not the energy being transferred.
To me that strongly suggests that the wires are transferring current, not energy.
Why will I be bothered by that ?
Seems very normal to me.
Current * resistance gets you the voltage drop. If you are using the same wires in both examples and current is the same then voltage drop across the wire is the same and thus power loss on the wire is the current multiplied by the voltage drop across the wire.
The energy travels through wire and transportation losses are the same. The energy will be delivered to a different lamp so you do not have the same lamp in the two examples (your setup). The lamp is also a wire and on that wire voltage drop will be very different between the two examples. So that wire (the lamp filament) will have much more energy converted to infrared and visible light in the second example than the first.
There is nothing in your examples to show energy is not traveling through wires.
You have the two low resistance wires on each side of the bulb having the same loss in both cases to transfer the energy to the third wire with is the lamp filament. The difference is in the lamp filament between the two cases and that is where the electrical energy will do the most work.
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You have a wrong understanding of what energy is and how it is transferred from source(battery) to load (lamp/resistor).
If on the one side we have electrodacus, who is right, and on the other side we have the rest of the world, who are wrong, then really all rational people would wish to be with the rest of the world and remain wrong. Apparently, in this scenario, being wrong is the right place to be.
The other big thread used to be filled with the Maxwell/Poynting Bro's absolutely shooting down anyone who dared even hinted at suggesting anything other than a 100% Poynting explanation, and heaven forbid if you got even even the slightest direction of Poynting vector wrong, it's was flaming pitchforks.
Now all the Poynting bros have vanished and both threads are now completely dominated by the Energy In The Wire (EIT) absolutists.
LOL :-DD :popcorn:
People should focus on what Poynting's theorem says and not read extra meaning into it. You have to integrate the Poynting vector over the surface off a volume enclosing a region in space. Other than that, it doesn't have physical meaning. Don't let the pretty pictures distract you.
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Why will I be bothered by that ?
And it doesn't bother you at all that there is no circuit you can put in region A in the diagram below that can extract energy from the wires that surround it? Even a DC/DC convertor? Even if you can connect it to a GND? But extracting energy from regions B or C is a piece of cake?
And I guess that it doesn't bother you if an isolated charge is placed in region A and it stays where it is put, but if the same charge is placed in B or C it will accelerate, acquiring energy ultimately suppled from the battery, without being connected to it?
And it doesn't both you if you charge a capacitor between +110V and +100V is has exactly the same stored energy as one charged between 0V and -10V? Even though one has been charged at a higher energy? And the other has been charged at a completely different polarity?
And it doesn't bother you that a transformer can get 95%+ transfer of energy from one wire to the other, even though the wires don't touch, and no charges from the input wire get transferred to the output wire?
And it doesn't bother you that for your version of electrostatics (sum of force between charges), every charge needs to be in consistent communication with every other charge in the universe, to work out how far away they are, and at what direction?
And it doesn't bother you that a transmission line is a series of inductors and capacitors, however on inspection those capacitors and inductors can neither be identified or isolated?
And it doesn't bother you that commercial radio transmitters can get kilowatts of energy to disappear into literally thin air?
You must be a firm believer in the Lumped Element Model. It seems to work, so to you it must reflect the mechanics of reality, rather than an useful abstraction and approximation that allows you to get stuff done.
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Woah! No-one's talking resistance here (except you, as a diversion). You said "you will see that the conductor heats uniformly on the entire conductor cross section" - that is heat, thermal. I am wondering just how you can measure the internal temperature of a conductor, and you Internet isn't any help there.
So, just how do you see that? If you make a hole and place a probe you're affecting the conductor integrity, and even with a thermal imager you're only going to see the outside.
Or was this just another 'fact' or 'law' you made up on the spot?
I guess you will need to learn about another type of energy storage and that will be thermal storage.
If the electrons travel closer to the outside surface of the wire like in AC then resistive losses will show that so there will be no need to even measure the temperature.
More diversion. You said we would see, as a pillar of your argument. So we want to see it, to recognise what you said was so. Now, when asked how to do that, you say we don't need to, or that it's an obvious effect of something else.
You made it up, didn't you? You can't prove it or show it and all you can do is circular arguments hoping that's not the one you end with when the music stops.
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For consideration:
(https://www.seventransistorlabs.com/Images/Induction1401.jpg)
That was something like 3kW for a minute, at 10kHz or so, on a very rusty 1/4" thick steel plate.
The heating pattern is indicative of skin effect around the outer edge of the workpiece, though the glowing areas are much wider than the current paths due to the long heating duration. Nonetheless it's more than adequate to see the superficial current flow path, preference for long sides, and avoidance of corners.
Tim
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I sure not heard about him at university (Electrical engineering in some east european country).
The fact that you haven't heard about something doesn't mean that is wrong or unimportant. It just means that your education was bad. You also didn't know about preservation of charge which is a fundamental concept. I suggest that you work through a good book on Electromagnetics. You are clearly out of your depth.
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Why will I be bothered by that ?
And it doesn't bother you at all that there is no circuit you can put in region A in the diagram below that can extract energy from the wires that surround it? Even a DC/DC convertor? Even if you can connect it to a GND? But extracting energy from regions B or C is a piece of cake?
And I guess that it doesn't bother you if an isolated charge is placed in region A and it stays where it is put, but if the same charge is placed in B or C it will accelerate, acquiring energy ultimately suppled from the battery, without being connected to it?
And it doesn't both you if you charge a capacitor between +110V and +100V is has exactly the same stored energy as one charged between 0V and -10V? Even though one has been charged at a higher energy? And the other has been charged at a completely different polarity?
And it doesn't bother you that a transformer can get 95%+ transfer of energy from one wire to the other, even though the wires don't touch, and no charges from the input wire get transferred to the output wire?
And it doesn't bother you that for your version of electrostatics (sum of force between charges), every charge needs to be in consistent communication with every other charge in the universe, to work out how far away they are, and at what direction?
And it doesn't bother you that a transmission line is a series of inductors and capacitors, however on inspection those capacitors and inductors can neither be identified or isolated?
And it doesn't bother you that commercial radio transmitters can get kilowatts of energy to disappear into literally thin air?
You must be a firm believer in the Lumped Element Model. It seems to work, so to you it must reflect the mechanics of reality, rather than an useful abstraction and approximation that allows you to get stuff done.
Not sure where to start from but you have a lot of misconception about how things work.
Yes regions B and C have an electric field but you can still do not extract energy from that. Same with magnetic field that is also constant basically like a permanent magnet. Let me know how you extract energy from a permanent magnet and we can clarify from where you actually extracted the energy because it was sure not from the magnet.
Yes if you place a charged particle in an electric field it will be accelerated but note the action you take to put a charged particle there.
The potential difference for the capacitor is 10V in both cases and as far as energy is concerned it makes absolutely no difference. The zero/ground is something arbitrary that we chose.
If a capacitor is placed on the positive of two power supplies that have common negative tied together and we consider that the zero point then we connect a capacitor with one plate to positive of the 100V supply and the other plate to the positive of the 110V supply you just have a 10V potential and as far as capacitor is concerned that is all he will see thus it will store the exact same amount of energy as the one connected to what you call a 0 to -10V supply that can also be seen as a 0V to +10V supply is all a matter of definition or how it is connected to other things where you may already have defined a "ground"/zero point/reference point.
A single inductor is an energy storage device same as the capacitor is an energy storage device and so with transformer you can store energy by creating a magnetic field while supplying the primary and then retrieve that stored energy with the secondary or the other way around or with the same.
So by running a current through any of the two coils you are creating a magnetic field that remains there as long as there is no change in current flow.
If you suddenly stop the current flow by disconnecting the source voltage on the both coils (primary and secondary) will increase and so you can take that stored energy out through any of them.
But if you run a DC current through primary then you can not take energy out from secondary. You still have a strong magnetic field but it is constant so energy storage is maintained full.
Of course a transmission line is a series of inductors, capacitors and resistors and of course you can see them if you can see the wires of the transmission line and understand what a capacitor, inductor and resistor are.
The energy of a radio transmitter will not disappear. As a simplification is a capacitor with one plate as the transmitter the ground as one of the conductors and the other plate is the receiver again with ground as the common conductor.
The capacitor is charged and discharged multiple times per second depending on radio frequency and of course there will e a lot of power loss due to resistance in this circuit.
The lumped model works because it is a representation of reality. And yes it will be an approximation as in simulation you may use just like me hundreds of this groups for a 20m transmission line to keep the calculations manageable and the results more than close enough for what we need them.
If you think you have a better model that allows to make accurate predictions about what happens on a transmission line then please share as I will be curious to hear. I will like to see the equations not just some story.
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More diversion. You said we would see, as a pillar of your argument. So we want to see it, to recognise what you said was so. Now, when asked how to do that, you say we don't need to, or that it's an obvious effect of something else.
You made it up, didn't you? You can't prove it or show it and all you can do is circular arguments hoping that's not the one you end with when the music stops.
I already did but you may have missed that post or it went over your head.
Take a multimeter set it on resistance and measure the resistance of a copper pipe and then of a copper bar. Let me know if you will measure the same resistance.
If you measure a lower resistance for the copper bar and it just happens to be proportional with the sectional area of the copper that means electrons are free to travel through the entire section of the wire not just the outside portion for a DC current.
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For consideration:
That was something like 3kW for a minute, at 10kHz or so, on a very rusty 1/4" thick steel plate.
The heating pattern is indicative of skin effect around the outer edge of the workpiece, though the glowing areas are much wider than the current paths due to the long heating duration. Nonetheless it's more than adequate to see the superficial current flow path, preference for long sides, and avoidance of corners.
Tim
You do not understand what happens there and how an induction heater works. Also the skin effect has nothing to do with Derek's DC experiment nor it has anything to do with energy traveling through or outside the wire question.
The steel piece that you are heating is magnetic and those losses are due to resistance to magnetic field not electric current.
Nobody denies the skin effect exist for AC and at high frequency is very significant it just has nothing to do with the discussion about electrical energy flowing through wires or not.
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More diversion. You said we would see, as a pillar of your argument. So we want to see it, to recognise what you said was so. Now, when asked how to do that, you say we don't need to, or that it's an obvious effect of something else.
You made it up, didn't you? You can't prove it or show it and all you can do is circular arguments hoping that's not the one you end with when the music stops.
I already did
No, you did not.
but you may have missed that post
I thought perhaps I had, but apparently I saw what you think passes for it.
or it went over your head.
Ha ha! Just solve everything with the ad homs, eh.
Take a multimeter set it on resistance and measure the resistance of a copper pipe and then of a copper bar. Let me know if you will measure the same resistance.
If you measure a lower resistance for the copper bar and it just happens to be proportional with the sectional area of the copper that means electrons are free to travel through the entire section of the wire not just the outside portion for a DC current.
Do you normally have problems comprehending a simple question? (Answer: yes, this web forum is full of examples)
Resistance has got nothing whatever to do with this. Once again, you said "you will see that the conductor heats uniformly on the entire conductor cross section" (my emphasis because you apparently can't recognise your own words).
So once again, how does one "see that the conductor heats uniformly on the entire conductor cross section"? We don't want it implied by measuring resistance, we want to be able to see it, as you told us we could and would.
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I sure not heard about him at university (Electrical engineering in some east european country).
The fact that you haven't heard about something doesn't mean that is wrong or unimportant. It just means that your education was bad. You also didn't know about preservation of charge which is a fundamental concept. I suggest that you work through a good book on Electromagnetics. You are clearly out of your depth.
I did not studied history of electricity but applied electrical engineering based on latest understanding of the subject.
You do not understand what conservation of charge means and how it applies to an isolated system.
What you care about in the two or three parallel capacitor problem is conservation of energy.
You are unable to make predictions about a system because you do not understand how to apply the known equations.
That is why people do not think paralleling two ideal and identical capacitors (no resistance in circuit) 3V for charged one will result in 2.121V
And you can get very close to that also by using a DC-DC converter to transfer energy from one capacitor to the other.
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Yes if you place a charged particle in an electric field it will be accelerated but note the action you take to put a charged particle there.
And placing the charge in the central region takes more 'action', and it stubbornly refuses to accelerate. It just sits there.
The potential difference for the capacitor is 10V in both cases and as far as energy is concerned it makes absolutely no difference. The zero/ground is something arbitrary that we chose.
And yet you say that an charge that you say a charge in a wire is carrying high potential energy is somehow 'different' to one that has zero potential energy? You talk of "high energy electrons" moving slowly in the wires.
A single inductor is an energy storage device same as the capacitor is an energy storage device and so with transformer you can store energy by creating a magnetic field while supplying the primary and then retrieve that stored energy with the secondary or the other way around or with the same.
I can't resolve your statements with that all the energy is in the wires, when now you are talking about storing it (and even transferring it) in a magnetic field
Of course a transmission line is a series of inductors, capacitors and resistors and of course you can see them if you can see the wires of the transmission line and understand what a capacitor, inductor and resistor are.
Show me a picture of an inductor in Derick's video, or even just a capacitor
The energy of a radio transmitter will not disappear. As a simplification is a capacitor with one plate as the transmitter the ground as one of the conductors and the other plate is the receiver again with ground as the common conductor.
A capacitor with the physical dimensions of "two thin wires separated by kilometers" is so close to zero that no meaningful power transfer can occur. Strange how my phone's radio still work in my pocket, and radios even in space where there is no common ground...
If you think you have a better model that allows to make accurate predictions about what happens on a transmission line then please share as I will be curious to hear. I will like to see the equations not just some story.
Sure. Wikipedia does a better job of documenting them than I ever could.
Gauss's law - https://en.wikipedia.org/wiki/Gauss%27s_law
Gauss's law for magnetism - https://en.wikipedia.org/wiki/Gauss%27s_law_for_magnetism
Faraday's law of induction - https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction
The Ampère–Maxwell equation - https://en.wikipedia.org/wiki/Amp%C3%A8re%27s_circuital_law#Extending_the_original_law:_the_Amp%C3%A8re%E2%80%93Maxwell_equation
Between them this system of equations can explain all the little wrinkles that you are unable to.
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If you think you have a better model that allows to make accurate predictions about what happens on a transmission line then please share as I will be curious to hear. I will like to see the equations not just some story.
Sure. Wikipedia does a better job of documenting them than I ever could.
But you really shouldn't try. Electrodacus has a habit of saying things that are clearly wrong, muddled or confusing, and then inviting people to argue about them. It's no surprise that threads like this one tend to go on and one without end. That's the goal, to play people like a fiddle and goad them in to making more and more responses. The best solution is to simply ignore the posts and decline to respond. Electrodacus will eventually get bored and go away.
"Don't feed the trolls"
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And you can get very close to that also by using a DC-DC converter to transfer energy from one capacitor to the other.
You keep saying this, and demanding "proof" from others, but you have yet to demonstrate this, even though you assert it over and over. Please demonstrate that this is at all possible.
I think you will find that because the capacitor is a low impedance load (it will store all the energy you can give it) the transfer from the DC-DC convertor will be limited by the convertor's input and output impedance. When the voltages in the capacitors are equal you will end up with less than 0.5 Vinitial in both.
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That's the goal, to play people like a fiddle and goad them in to making more and more responses.
I think he genuinely believes what he says (that he is correct and everyone else "doesn't understand properly"). And much of the arguments are merely ways to not have to accept that he is wrong.
The best solution is to simply ignore the posts and decline to respond. Electrodacus will eventually get bored and go away.
If other topics are any guide, that would only be temporary until he starts up a new thread to demonstrate that, actually, he was right all along. Looks like a troll but I am not convinced.
What did turn off the lights on the faster than the wind thing was making him see he hadn't got a leg to stand on. That shut him up about it, until now (when it is far enough back that he can pretend he missed the end and insist energy storage is the answer to everything, as any fule would kno).
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And you can get very close to that also by using a DC-DC converter to transfer energy from one capacitor to the other.
You keep saying this, and demanding "proof" from others, but you have yet to demonstrate this, even though you assert it over and over. Please demonstrate that this is at all possible.
I think you will find that because the capacitor is a low impedance load (it will store all the energy you can give it) the transfer from the DC-DC convertor will be limited by the convertor's input and output impedance. When the voltages in the capacitors are equal you will end up with less than 0.5 Vinitial in both.
I will not even bother to answer your comments as it will be a never ending story. You clearly lack basic understanding.
What is your prediction if you are using say a 80% efficient DC-DC converter with constant current limiting with current limited so that wires and capacitor internal resistance is basically insignificant in therms of extra losses or say the 80% efficiency includes those losses to make the calculation simpler.
What will be the end voltage if you transfer from a charge 1F capacitor 3V to a identical 1F capacitor that is fully discharged so 0V
Unless you can offer a prediction you can not claim you understand how any of this works. And feel free to use any tool you want to come up with the answer.
Also when I mentioned that end voltage will be higher than 1.5V some of you (not sure if you included) had concerned about the fact that a DC-DC converter uses active components. This will again be a lack of understanding if you or the ones that made this comment think extra energy will somehow magically appear in this isolated circuit because of the use of a DC-DC converter that is only powered by the charged capacitor.
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But you really shouldn't try. Electrodacus has a habit of saying things that are clearly wrong, muddled or confusing, and then inviting people to argue about them. It's no surprise that threads like this one tend to go on and one without end. That's the goal, to play people like a fiddle and goad them in to making more and more responses. The best solution is to simply ignore the posts and decline to respond. Electrodacus will eventually get bored and go away.
"Don't feed the trolls"
You are welcome in providing the answer to the question I posted to hamster_nz. In fact anyone is welcome.
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And you can get very close to that also by using a DC-DC converter to transfer energy from one capacitor to the other.
You keep saying this, and demanding "proof" from others, but you have yet to demonstrate this, even though you assert it over and over. Please demonstrate that this is at all possible.
I think you will find that because the capacitor is a low impedance load (it will store all the energy you can give it) the transfer from the DC-DC convertor will be limited by the convertor's input and output impedance. When the voltages in the capacitors are equal you will end up with less than 0.5 Vinitial in both.
I will not even bother to answer your comments as it will be a never ending story. You clearly lack basic understanding.
What is your prediction if you are using say a 80% efficient DC-DC converter with constant current limiting with current limited so that wires and capacitor internal resistance is basically insignificant in therms of extra losses or say the 80% efficiency includes those losses to make the calculation simpler.
What will be the end voltage if you transfer from a charge 1F capacitor 3V to a identical 1F capacitor that is fully discharged so 0V
Unless you can offer a prediction you can not claim you understand how any of this works.
What part of "When the voltages in the capacitors are equal you will end up with less than 0.5 Vinitial in both." is not "offering a prediction"?
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What part of "When the voltages in the capacitors are equal you will end up with less than 0.5 Vinitial in both." is not "offering a prediction"?
Show me the math you use to get to that conclusion.
The question I asked involves a DC-DC converter with constant current control between the two capacitors and total efficiency of 80%.
Say you start at 3V for the charged capacitor and you stop the discharge when the charged capacitor voltage drops to 2.121V so half the initial energy.
What will be the voltage on the other capacitor the one that started empty at 0V.
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The question I asked involves a DC-DC converter with constant current control between the two capacitors and total efficiency of 80%.
Say you start at 3V for the charged capacitor and you stop the discharge when the charged capacitor voltage drops to 2.121V so half the initial energy.
What will be the voltage on the other capacitor the one that started empty at 0V.
What you are saying is "If I had a Unicorn, then I could open a Unicorn Zoo - prove me wrong!".
No, you prove me wrong. Show me your unicorn!
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I must have missed it. Please explain what experiment was done,
Very funny. The only thing I did in this thread was to indicate all the experimental data that support the energy flowing in space.
Really, where then?
Please explain what experiment was done, what S=JV predicted, what Poynting predicted and what was found.
There's none.
what was predicted by the S=JV folks, and what was found. (And why Poynting-Heaviside-etc. do not talk about it.)
What they accomplished no one knows. What they didn't, we know. They didn't manage to come up with an alternative that doesn't break causality, locality, gauge invariance or a combination of these.
That's why, despite objections regarding its counterintuitive nature, Poynting is the most probable explanation for the flow of energy.
You gave no example of an alternative breaking any of this ;D . Coincidence?
I also wonder how exactly all physicists proposing alternatives to the Poynting theorem never saw Derek's antennae coming in the whole 20th century. They must feel very silly now (no).
They didn't. But their peers did and criticized their proposals. That's why Poynting still stands.
Sure so in the 20th century and 21st physicists never heard of antennae. ::)
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What you are saying is "If I had a Unicorn, then I could open a Unicorn Zoo - prove me wrong!".
No, you prove me wrong. Show me your unicorn!
You will think that it is a unicorn because you do not understand the subject. It is in fact just a simple horse :)
But here is the proof
[attachimg=1]
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Why will I be bothered by that ?
And it doesn't bother you at all that there is no circuit you can put in region A in the diagram below that can extract energy from the wires that surround it? Even a DC/DC convertor? Even if you can connect it to a GND? But extracting energy from regions B or C is a piece of cake?
And it doesn't bother you at all that energy is supposedly flowing through the region A yet there is no circuit you can put in region A in the diagram that can extract energy from it?
(Hopefully the answer is no)
And it doesn't bother you that for your version of electrostatics (sum of force between charges), every charge needs to be in consistent communication with every other charge in the universe, to work out how far away they are, and at what direction?
Are you bothered???
Do you think that if every electron could communicate with far away electrons, humans could make machines hacking this property to communicate messages?
And it doesn't bother you that a transmission line is a series of inductors and capacitors, however on inspection those capacitors and inductors can neither be identified or isolated?
A wire is an inductor. With 2 you have an inductor and half a capacitor. And it works even for infinitesimal wires. 8)
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Yes. A piece of wire has inductance. And conductive hollow cylinders (tubes) are in fact legit transmission lines. AFAIK.
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Why will I be bothered by that ?
And it doesn't bother you at all that there is no circuit you can put in region A in the diagram below that can extract energy from the wires that surround it? Even a DC/DC convertor? Even if you can connect it to a GND? But extracting energy from regions B or C is a piece of cake?
And it doesn't bother you at all that energy is supposedly flowing through the region A yet there is no circuit you can put in region A in the diagram that can extract energy from it?
(Hopefully the answer is no)
No, with no voltage potential over region A, there is no way you can extract energy from the field in just that region.
And it doesn't bother you that for your version of electrostatics (sum of force between charges), every charge needs to be in consistent communication with every other charge in the universe, to work out how far away they are, and at what direction?
Are you bothered???
Do you think that if every electron could communicate with far away electrons, humans could make machines hacking this property to communicate messages?
It did bother me just a little - Every charge being constantly aware of every other charge in the universe does not have the feel of being fundamental to the universe.
An electric field can still be used to communicate messages. "Ripples in the fields" is preferable to "very small pushes and shoves over great distances".
And it doesn't bother you that a transmission line is a series of inductors and capacitors, however on inspection those capacitors and inductors can neither be identified or isolated?
A wire is an inductor. With 2 you have an inductor and half a capacitor. And it works even for infinitesimal wires. 8)
Fair call. But I still feel that a lumped model of a continuous thing is (very useful) approximation.
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What you are saying is "If I had a Unicorn, then I could open a Unicorn Zoo - prove me wrong!".
No, you prove me wrong. Show me your unicorn!
You will think that it is a unicorn because you do not understand the subject. It is in fact just a simple horse :)
But here is the proof
(Attachment Link)
You are a hard person to goad something out of, but eventually it works. :)
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You are a hard person to goad something out of, but eventually it works. :)
Is that your replay ? How about you understand now that if you transfer energy more efficiently from one capacitor to the other that is identical you get close to 0.707 the voltage of the charged capacitor in both.
It should not be me that is doing the work.
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Really, where then?
Now you're trolling.
You gave no example of an alternative breaking any of this ;D . Coincidence?
If you couldn't recognize it, this means you didn't go very far in your understanding of the problem.
Sure so in the 20th century and 21st physicists never heard of antennae. ::)
Who knows? What is important to understand is that the alternatives never prospered.
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Don't let the pretty pictures distract you.
Like this one by Poynting himself (https://royalsocietypublishing.org/doi/epdf/10.1098/rstl.1884.0016), for instance, showing the energy going from a battery to a resistive wire loop through the empty space.
(https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/?action=dlattach;attach=1485253;image)
Don't look at that. It'll shatter your faith in the wires.
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You are a hard person to goad something out of, but eventually it works. :)
Is that your replay ? How about you understand now that if you transfer energy more efficiently from one capacitor to the other that is identical you get close to 0.707 the voltage of the charged capacitor in both.
It should not be me that is doing the work.
You keep on bringing up DC-DC converters
Reply 118:
Just test with a DC-DC converter and you will see very close to ideal is possible. If the DC-DC converter was 100% efficient (so ideal) then you get 0.707 Vi but you do not need a DC-DC converter if there is no resistance to get the same result.
Everybody (even I) agree that it is possible:
Replay 119:
(And of course if you have an ultra efficiency DC-DC convertor to make the transfer then you might get close to 71%, but we don't. We have two ideal caps, some ideal wire and an ideal switch).
But you keep on tuning it back to "oh, if I use a DC/DC converter it proves something". So I called your bluff.
So finally we have the proof that everybody else missed. That if you transfer energy to and from a magnetic field, using an inductor, (thus taking the energy "out of the wire") that energy only flows in the wires. :-//
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You keep on bringing up DC-DC converters
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So finally we have the proof that everybody else missed. That if you transfer energy to and from a magnetic field, using an inductor, (thus taking the energy "out of the wire") that energy only flows in the wires. :-//
At first you and others mentioned that you can not have 0.707 * Vi because that will violate the conservation of charge witch likely you confused with energy.
Then after pointing out the formula for energy stored in a capacitor it was still argued that 0.707 * Vi is not possible in an circuit made with superconductors so zero resistance.
Then I mentioned something that is way easier to test (a DC-DC converter) and claim remained the same that voltage at the end of the experiment can not be higher than 0.5 Vi.
The energy seen through the bulb in the first 65ns in Derek's experiment is related to line capacity so no energy travels through air.
Any transmission line has both inductance and capacitance (both are energy storage devices) so the transmission line model that any electrical engineer uses is a realistic and accurate way to predict what happens.
I simplified the two capacitors spice model even more and below is the schematic and the graph
In the schematic you have a 3V supply to charge one of the 1000uF capacitors then the supply is disconnected by the switch S1 after 20ms and then starting at 40ms the S2 switch is closed for just 6ms then open.
The charged capacitor C1 will discharged from 3V down to around 2V while the C2 that started fully discharged is charged using that energy at 2V so you start with C1 at 3V and you end up with both identical capacitors charged at 2V using only energy available in C1
The only other two components are the 47mH inductor and a diode. The inductor is an intermediary energy storage helping reduce the losses due to circuit resistance. There is no outside energy coming in to the circuit and some is still lost as heat but much less than just paralleling the two capacitors directly.
The red graph is the control for the S2 switch showing the period switch was closed. The blue is the voltage across C1 and green voltage across C2.
[attachimg=1]
[attachimg=2]
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Here's another one:
Let's connect and ideal DC voltage source V_d to a fully discharged capacitor C at t=0s:
1. How much energy is delivered by the source?
2. How much energy is stored in the capacitor at t=1s?
3. How much charge is delivered by the source?
4. How much charge ends up in the capacitor?
You can try the same with a current source and an inductor.
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You keep on bringing up DC-DC converters
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So finally we have the proof that everybody else missed. That if you transfer energy to and from a magnetic field, using an inductor, (thus taking the energy "out of the wire") that energy only flows in the wires. :-//
At first you and others mentioned that you can not have 0.707 * Vi because that will violate the conservation of charge witch likely you confused with energy.
Then after pointing out the formula for energy stored in a capacitor it was still argued that 0.707 * Vi is not possible in an circuit made with superconductors so zero resistance.
Then I mentioned something that is way easier to test (a DC-DC converter) and claim remained the same that voltage at the end of the experiment can not be higher than 0.5 Vi.
The energy seen through the bulb in the first 65ns in Derek's experiment is related to line capacity so no energy travels through air.
Any transmission line has both inductance and capacitance (both are energy storage devices) so the transmission line model that any electrical engineer uses is a realistic and accurate way to predict what happens.
I simplified the two capacitors spice model even more and below is the schematic and the graph
In the schematic you have a 3V supply to charge one of the 1000uF capacitors then the supply is disconnected by the switch S1 after 20ms and then starting at 40ms the S2 switch is closed for just 6ms then open.
The charged capacitor C1 will discharged from 3V down to around 2V while the C2 that started fully discharged is charged using that energy at 2V so you start with C1 at 3V and you end up with both identical capacitors charged at 2V using only energy available in C1
The only other two components are the 47mH inductor and a diode. The inductor is an intermediary energy storage helping reduce the losses due to circuit resistance. There is no outside energy coming in to the circuit and some is still lost as heat but much less than just paralleling the two capacitors directly.
The red graph is the control for the S2 switch showing the period switch was closed. The blue is the voltage across C1 and green voltage across C2.
(Attachment Link)
(Attachment Link)
You cannot add inductors and diodes or DC-to-DC converters. The problem is about connecting two capacitors in parallel without any other components.
How do you think energy is transferred through a capacitor? Let's take a parallel plate capacitor with vacuum between the plates.
What about the airgap of a transformer?
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You cannot add inductors and diodes or DC-to-DC converters. The problem is about connecting two capacitors in parallel without any other components.
How do you think energy is transferred through a capacitor? Let's take a parallel plate capacitor with vacuum between the plates.
What about the airgap of a transformer?
Electrodacus already knows that, from message #169 when a DC-DC convertor was originally suggested:
It is also 'cheating'. I too could get any answer I want if I am free to add to the system. If I said "let me put an inductor in there, a diode, a switch and a trained imp that can push the switch really quickly" would you not agree that that is not the same problem any more?
Even knowing that Electrodacus went there, and put the diode, an inductor, a switch, and even the trained imp to push it. :-//
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The problem is that circuit theory doesn't take the physical dimensions and the electrodynamic behavior of the system into account. That is why we get a contradiction.
Let's look at the more realistic example of using two lossless coaxial cables as "capacitors".
Let's see if Electrodacus' modern outlook on electronics can derive the equations that govern this example.
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Is it possible to have an electric field without an associated particle? Is it possible to have an electric field without a charged particle?
I'm sure there are situations where it is much easier to calculate considering only the fields. I guess it is important to remember that numbers can be used to roughly describe reality and roughly predict things but they are not reality itself. It would be preposterous to think otherwise, absolutely posterior backwards. However, it does seem more common than not for people to think that words make reality rather than reality dictating the words (the little angels, little to no connection to reality).
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Is it possible to have an electric field without an associated particle? Is it possible to have an electric field without a charged particle?
Absolutely. Changing magnetic field creates electric field (Faraday’s law).
And you don’t need charged particles to have a magnetic field. For example, neutron has magnetic dipole moment. Magnetars are neutron stars that create strongest magnetic fields in the Universe.
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The pathological nature of the Two Capacitor Paradox problem leads to the observation that only the Law of Conservation of Charge is satisfied, but the Law of Conservation of Energy is violated.
https://en.m.wikipedia.org/wiki/Charge_conservation
According to the Wikipedia entry, half of the energy is 'lost' in the initially-charged capacitor after the switch is closed as shown in the mathematical expression below:
\$W_i = \frac{1}{2}CV_i\$
\$W_f = \frac{1}{2}CV_f^2 + \frac{1}{2}CV_f^2 = CV_f^2 = C(\frac{V_i}{2})^2 = \frac{1}{4}CV_i^2 = \frac{1}{2}W_i\$
Source: en.m.wikipedia.org/wiki/Two_capacitor_paradox
In order to satisfy Kirchhoff's Voltage Law after the switch is closed, both capacitors must have equal voltage. In this case, it is half the initial voltage of the initially-charged capacitor. Also, when the initially-discharged capacitor is connected after the switch is closed, this has the effect of doubling the total capacitance. If Q = CV and total capacitance doubles, then the voltage must be halved (V = Q/C) assuming charge is conserved. It then follows that if voltage is energy per unit charge, and charge is conserved after the switch is closed, then the initial energy must be halved if voltage is halved (across both capacitors). If the Law of Conservation of Charge is satisfied after the switch is closed, half of the initial energy must be lost to satisfy Kirchhoff's Voltage Law.
https://en.m.wikipedia.org/wiki/Voltage
Using the swimming pool analogy, assume that a swimming pool is filled with water and the floor of the swimming pool has a surface area of A (metres squared) and a height of h (metres). The weight and the height of the water is analogous to the charge and the voltage of the capacitor. If weight of the water can be expressed as F = mg then the gravitational potential energy of the water can be expressed as E = mgh.
https://en.m.wikipedia.org/wiki/Gravitational_energy
If the area of the floor of the swimming pool is increased from A to 2A (analogous to adding the initially-discharged capacitor) then the height of the water (analogous to the voltage across the capacitors) must be halved from h to h/2. The gravitational potential energy of the water (analogous to the electrical potential energy of the capacitors) must be halved, too. The weight of the water (analogous to the charge on the capacitors) does not change when the floor area (analogous to the total capacitance) is doubled.
Elaborating further with the swimming pool analogy, if the surface area of the floor is increased to the point where all of the water molecules are in direct contact with the floor then the effective height of the water would be the height of a single water molecule. The effective height of the water in the swimming pool would be approaching zero metres. The gravitational potential energy would be approaching zero, too.
I found a YouTube video showing the magnetic field of a toroidal transformer. You can see that the field exists only within the windings of the electromagnet. It's as though there are discrete magnets stacked with north to south poles arranged in a loop. What if the electromagnet was replaced with batteries connected in a loop with the positive terminal of one battery connecting to the negative terminal of the next battery? Would the electric field only be within the batteries or would it also extend/exist outside of the batteries?
Toroidal Electromagnet video: https://youtu.be/0-KaSGlDQ-8
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If the area of the floor of the swimming pool is increased from A to 2A (analogous to adding the initially-discharged capacitor) then the height of the water (analogous to the voltage across the capacitors) must be halved from h to h/2. The gravitational potential energy of the water (analogous to the electrical potential energy of the capacitors) must be halved, too. The weight of the water (analogous to the charge on the capacitors) does not change when the floor area (analogous to the total capacitance) is doubled.
Indeed if we perform this experiment, using a pool with one wall able to move freely yet sealed water tight, we get a piston, which as it's moved back and forth, in the steady state, bears a force of F = ρ g l h^2 / 2 (for water height h, gravitational acceleration g, water density ρ, and wall length l). Thus we do work on it when pushing (makes pool smaller, taller), or vice versa. Going between the x and 2x cases (for pool width x, 2x), this provides precisely the energy required to make up for the apparent halving of energy while conserving charge. Energy of the system (pool AND actuator) is conserved, exactly as we should expect; but considering too limited of a subset (i.e. the pool by itself), energy isn't conserved, it's been moved in/out of that boundary somehow.
We also have the dynamic case, though an overly messy example of it (Navier-Stokes fluid equations are a bitch!). Suppose we partition a pool in half, filling only one half; then suddenly remove the partition. The water sloshes into the formerly-unoccupied side, and continues to slosh back and forth until friction has absorbed the "AC" energy. Indeed superfluids exist, so we could perform this experiment with such, and demonstrate an apparent perpetual motion machine, where the mean free surface level gives the charge-conserved and half-energy figure, while the AC component (gravity waves on the free surface) continues, in motion, with exactly half the initial energy of the system.
I think, because of the nonlinearity of gravity waves, the (potentially?) biphasic nature of superfluids, the free surface moving in gas rather than pure vacuum (the only known superfluids exist at very low temperatures and fairly low pressures, indeed being cooled by evaporation), there will be too many kinds of dispersion (i.e. one long, blocky wave breaks up into numerous higher order waves), friction (due to the gas in the chamber itself, and the liquid's normal phase component) and other effects, dissipating the wave energy in a real superfluid experiment; but even so, again we have the argument: where is the energy BEFORE it's been dissipated to heat, or equivalently, exchanged into other forms of energy besides where it was? And again the answer is clear, it's stored in both static (or average/DC) and dynamic (AC) modes.
Tim
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The pathological nature of the Two Capacitor Paradox problem leads to the observation that only the Law of Conservation of Charge is satisfied, but the Law of Conservation of Energy is violated.
Law of conservation of energy has never been broken/violated. There is no paradox related to the two parallel capacitors so that page on Wikipedia is misinformation as anyone can write a wiki page and write whatever it wants.
As I clearly mentioned a few times all energy is accounted for. Due to resistance in the two identical capacitors circuit (circuit perfectly symmetrical) after closing the switch half of the energy is wasted as heat in the conductors (that includes wires, switch and capacitor plates) the other half remains in the two capacitors quarter of the energy in each capacitor.
There is absolutely no mystery in the two parallel capacitor problems unless you do not understand what capacitors are.
Adding an inductor (another energy storage device that people do not understand) and a diode will not bring any extra energy as system is still isolated and the initial total energy at the beginning of the test is the same and contained all in the charged capacitor.
The end result with the inductor and diode added is closer to ideal case where circuit will have had no resistance at all case where no energy will be wasted as heat.
Did you even looked closely at the equations you posted from wikipedia to see how absurd and stupid they are.
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The pathological nature of the Two Capacitor Paradox problem leads to the observation that only the Law of Conservation of Charge is satisfied, but the Law of Conservation of Energy is violated.
Law of conservation of energy has never been broken/violated. There is no paradox related to the two parallel capacitors so that page on Wikipedia is misinformation as anyone can write a wiki page and write whatever it wants.
As I clearly mentioned a few times all energy is accounted for. Due to resistance in the two identical capacitors circuit (circuit perfectly symmetrical) after closing the switch half of the energy is wasted as heat in the conductors (that includes wires, switch and capacitor plates) the other half remains in the two capacitors quarter of the energy in each capacitor.
There is absolutely no mystery in the two parallel capacitor problems unless you do not understand what capacitors are.
Adding an inductor (another energy storage device that people do not understand) and a diode will not bring any extra energy as system is still isolated and the initial total energy at the beginning of the test is the same and contained all in the charged capacitor.
The end result with the inductor and diode added is closer to ideal case where circuit will have had no resistance at all case where no energy will be wasted as heat.
Did you even looked closely at the equations you posted from wikipedia to see how absurd and stupid they are.
Your point is moot. The whole point is that all the conductors are assumed to be perfect. There is no resistance.
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Your point is moot. The whole point is that all the conductors are assumed to be perfect. There is no resistance.
Where is that assumption made ? Are you referring to that article on wikipedia ?
Take this example two 1F identical capacitors one charged at 3V and one discharged 0V with no resistance anywhere in the circuit including the wires, switch and capacitor plates.
Initial energy in the system 0.5 * 1F * 3V2 = 4.5Ws
After the switch is closed half of the energy from the charged capacitor is transferred to the discharged capacitor as there are no resistive losses when transfer is done thus final energy in the system will be the same 4.5Ws just that now it is stored in 2F instead of 1F
Voltage on both capacitors after switch is closed and energy was transferred is
2.121V
0.5 * 2F * 2.1212 = 4.5Ws no energy is lost to outside as the circuit has zero resistance to current flow.
It is as simple as that not the wrong equations used in the Wikipedia page.
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Check if your solution satisfies the law of preservation of charge.
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Also, check what the charge and voltage ratios as a function of the series resistance and what the limits as R-->0 are.
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Check if your solution satisfies the law of preservation of charge.
You do not understand what conservation of charge means and how it is applied.
The clue is in the fact that adding another charged particle to a capacitor that is at 1V and same capacitor that is at 2V requires different amounts of energy.
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Also, check what the charge and voltage ratios as a function of the series resistance and what the limits as R-->0 are.
Any resistance different from zero will result in the same exact final state for two identical parallel capacitors and that will be that half of the energy will be lost as heat in conductors due to resistance to current flow.
You basically have a resistor divider. The value of the resistance will only influence the time it takes for the energy to be transferred and wasted as heat.
To test this just add a 1 Ohm resistor between the two capacitors and then add a 1KOhm resistor and you will see that same amount of energy will be wasted as heat and you will have the same half of the original voltage in the two identical capacitors.
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Exactly.
Since the final result is independent of the finite resistance in the circuit, it will also be the result in the limit as R-->0.
Although the famous conservation of energy law is true, it is not always relevant to a given situation unless you can include the exact losses due to, for example, resistor heat loss and EM radiation.
In first-year physics, we learned to distinguish between conservation of momentum (true) and conservation of kinetic energy (not always true) in simple mechanical calculations.
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Exactly.
Since the final result is independent of the finite resistance in the circuit, it will also be the result in the limit as R-->0.
Can you rephrase that ? Id not not understand what you mean.
As long as resistance is higher than absolute zero half of the energy will be lost as heat in that resistance so you end up with just half the energy in this particular systemic case of two identical capacitors.
If capacitors are not identical then percentage of loss as heat will be different than half.
All energy is accounted for in any setup. Q on the other hand that is Q = C*V will not be the same in any other experiment other than the two identical capacitor case with resistance higher than zero. This is due to symmetry so it is just a coincidence and not a rulle.
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I mean that in the context of mathematical limit calculations.
(I was once criticized here for suggesting "epsilon-delta" (q.v.) calculations of limits, but that's how it works.)
Even if the resistance is 1x10-6 ohms (unphysical), the loss of total energy is still 1/2.
Therefore, for any specified resistance value the loss is unchanged.
The factor (1/2), of course, is only valid for two equal capacitors.
In the calculation, the voltages across the two capacitors must be equal to each other after a long time, with corresponding charges (unequal if the capacitors are unequal).
This is an example of a one-sided limit, since a physical passive resistor (as opposed to an active device with external power applied, such as a tunnel diode or transitron vacuum tube) cannot have a negative value.
A physical inductor with finite resistance will give an "evanescent" (exponentially dying sinusoidal) solution, which should go to the same asymptotic state at long time as the non-inductive resistor.
DC-DC converters are not passive, and will have different behavior. For example, if you want to charge a capacitor efficiently, ramping the voltage in a controlled manner is better than throwing a knife switch through a resistance from a battery.
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I mean that in the context of mathematical limit calculations.
(I was once criticized here for suggesting "epsilon-delta" (q.v.) calculations of limits, but that's how it works.)
Even if the resistance is 1x10-6 ohms (unphysical), the loss of total energy is still 1/2.
Therefore, for any specified resistance value the loss is unchanged.
The factor (1/2), of course, is only valid for two equal capacitors.
Using an inductor to store that energy that otherwise will end up as heat and then discharging that energy stored in inductor in to the discharged capacitor can get you very close to ideal energy transfer with very little loss.
Even from my simple setup I got slightly above 2V in both capacitors at the end of the energy transfer.
In my example I used 1000uF = 1mF capacitors so
0.5 * 1mF * 32 = 4.5mWs as start energy
At the end with say rounded 2V in both we have
0.5 * 2mF * 22 = 4mWs
So 0.5mWs still lost as heat but it is way less than having to lose half and is close to 90% energy transfer efficiency.
Q = C * V at the start is 1mF * 3V = 3mC = 0.833uAh
Q at the end is 2mF * 2V = 4mC = 1.111uAh
Edit: What difference does it make if you use active or passive components when we talk about energy ?
No energy is added to the isolated system by adding non charged components in the experiment.
inductor same as capacitor is a passive component and if you want I can eliminate the diode replace that with another switch and show you the same thing.
Not that the diode can add any energy in to the system unless it is a photo diode and is hit by photons coming from outside of this system but then it will no longer be an isolated system.
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In the Spice model you posted above (reply 477), there is zero resistance in series with your finite inductor.
The switches have finite resistance, but the second one is not on for a long time.
However, the diode model (an actual Schottky power diode) included in the circuit can bleed charge from the system, assuming it includes losses.
In a resistor-only circuit with two capacitors, only resistors in parallel with the capacitors (e.g., capacitor leakage) will remove charge from the system as the capacitors slowly discharge with the power switch open.
(PS: when I distinguished active and passive circuits, I mean that an active circuit has an external power source. So a diode in the dark is passive, but it has losses.)
What happens if you leave S2 on for a long time in your simulation?
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In the Spice model you posted above (reply 477), there is zero resistance in series with your finite inductor.
The switches have finite resistance, but the second one is not on for a long time.
However, the diode model included in the circuit can bleed charge from the system, assuming it includes losses.
In a resistor-only circuit with two capacitors, only resistors in parallel with the capacitors (e.g., capacitor leakage) will remove charge from the system as the capacitors slowly discharge with the power switch open.
Sorry I did not mentioned but the inductor has a 0.3Ohm resistance is just not visible in the diagram.
Yes the reverse current of the diode will slowly discharge the capacitor but that can be disconnected after the energy was transferred by adding another switch.
And yes capacitors are not perfect and have some leakage thus energy will be lost over time but that is not the point of the experiment.
Main point of this experiment is to show that in the two parallel capacitors half of the energy is lost because energy travels through wires and wires have a resistance to current flow.
If energy was not transferred through wires then you will either have no loss associated with the wire resistance or the losses will be manifested somewhere else other than in wires.
You need to move electrons through wires from one capacitor to another in order to move energy. Unless voltage is so high that air can become a conductor and electrons can be transferred that way from one capacitor to another the energy will travel through wires.
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(PS: when I distinguished active and passive circuits, I mean that an active circuit has an external power source. So a diode in the dark is passive, but it has losses.)
What happens if you leave S2 on for a long time in your simulation?
What do you think it will happen ?
The energy will flow back and forth between the two capacitors and inductor (RLC resonant circuit) until half of the initial energy will be dissipated as heat so you end up with half the voltage.
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Really, where then?
Now you're trolling.
You gave no example of an alternative breaking any of this ;D . Coincidence?
If you couldn't recognize it, this means you didn't go very far in your understanding of the problem.
Right so there are plenty of problems with S=JV but you are forbidden to discuss them.
Sure so in the 20th century and 21st physicists never heard of antennae. ::)
Who knows? What is important to understand is that the alternatives never prospered.
Prosperity is knowledge. Or science I don't know. Or is it scientology?
And influencers like Derek are needed to keep physicists in the right direction. It's definitely how science works.
And it doesn't bother you at all that energy is supposedly flowing through the region A yet there is no circuit you can put in region A in the diagram that can extract energy from it?
(Hopefully the answer is no)
No, with no voltage potential over region A, there is no way you can extract energy from the field in just that region.
So what really matters is the potential, while Poynting's energy flow give no useful information.
Are you bothered???
Do you think that if every electron could communicate with far away electrons, humans could make machines hacking this property to communicate messages?
It did bother me just a little - Every charge being constantly aware of every other charge in the universe does not have the feel of being fundamental to the universe.
An electric field can still be used to communicate messages. "Ripples in the fields" is preferable to "very small pushes and shoves over great distances".
Every night, you can check that electrons in your eye are "aware" of what's going on with electrons in far away stars.
A wire is an inductor. With 2 you have an inductor and half a capacitor. And it works even for infinitesimal wires. 8)
Fair call. But I still feel that a lumped model of a continuous thing is (very useful) approximation.
Yes it's quite bizarre: the lumped model approximation has no reason to be a good approximation with light-speed processes, yet it is an extremely precise one with transmission lines.
I guess the main reason is they hardly behave as antennae, due to their counter-currents.
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(PS: when I distinguished active and passive circuits, I mean that an active circuit has an external power source. So a diode in the dark is passive, but it has losses.)
What happens if you leave S2 on for a long time in your simulation?
What do you think it will happen ?
The energy will flow back and forth between the two capacitors and inductor (RLC resonant circuit) until half of the initial energy will be dissipated as heat so you end up with half the voltage.
Yes, with two equal capacitors and any passive two-terminal network between them, that is what I expect. Each capacitor has 1/2 the original voltage, and therefore 1/2 the original charge and 1/4 the original total energy in both capacitors, until the leakage resistance not included in the model eventually discharges both of them. I'm busy this weekend, but I will do a very simple experiment next week with polypropylene capacitors and a 10 megohm voltmeter.
If I use two 10 uF capacitors, and then one 10 uF (original charge) and a second 20 uF (originally discharged), I will have discharge time constants of 200 sec (both caps) and 300 sec (both caps), so I will lose 1% of the charge due to the parasitic voltmeter in 2 sec or 3 sec, respectively, after I throw the switch, first disconnecting C1 from the battery, and then connecting C1 to C2 (DPDT CO, L&N huge honking switch).
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Yes, with two equal capacitors and any passive two-terminal network between them, that is what I expect. Each capacitor has 1/2 the original voltage, and therefore 1/2 the original charge and 1/4 the original total energy in both capacitors, until the leakage resistance not included in the model eventually discharges both of them. I'm busy this weekend, but I will do a very simple experiment next week with polypropylene capacitors and a 10 megohm voltmeter.
If I use two 10 uF capacitors, and then one 10 uF (original charge) and a second 20 uF (originally discharged), I will have discharge time constants of 200 sec (both caps) and 300 sec (both caps), so I will lose 1% of the charge due to the parasitic voltmeter in 2 sec or 3 sec, respectively, after I throw the switch, first disconnecting C1 from the battery, and then connecting C1 to C2 (DPDT CO, L&N huge honking switch).
Not sure I understand what you want to test.
Will you parallel two 10uF capacitors so that you get a 20uF and use that 20uF capacitor as the discharged capacitor and then use a 10uF charged capacitor from witch you will charge the 20uF one ?
If that is the case then you will see that voltage will drop down to 33% so 1V if your charged capacitor is at 3V initially.
So a lot more energy will be lost as heat compared to the two identical capacitor test.
You start with 45uWs of energy 0.5 * 10uF * 32
And you end up with 15uWs of energy 0.5 * 30uF * 12
The other 30uWs ends up as heat in the conductors.
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A simple demonstration at maybe 2% accuracy:
Connect voltmeter across C1, and discharge C2
Charge C1 = 10 uF to 9 V (battery), measure the voltage immediately after opening the battery-C1 switch (must be faster than a few seconds with the 100 sec time constant of 10 uF & 10 megohm).
Close C1 to C2 switch and measure voltage as soon as it settles, paying attention to the 200 sec discharge time constant.
Do again with C2 = 20 uF to demonstrate unequal capacitor case.
It should agree with your calculation, to show that total charge rather than total energy in the open system is conserved.
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A simple demonstration at maybe 2% accuracy:
Connect voltmeter across C1, and discharge C2
Charge C1 = 10 uF to 9 V (battery), measure the voltage immediately after opening the battery-C1 switch (must be faster than a few seconds with the 100 sec time constant of 10 uF & 10 megohm).
Close C1 to C2 switch and measure voltage as soon as it settles, paying attention to the 200 sec discharge time constant.
Do again with C2 = 20 uF to demonstrate unequal capacitor case.
It should agree with your calculation, to show that total charge rather than total energy in the open system is conserved.
Total energy is always conserved in an isolated system.
So in this case 1)
charged 10uF charging a discharged 10uF half of the energy will remain in the two combined capacitors and half of the energy will be dissipated as heat.
In case 2)
charged 10uF charging a discharged 20uF one third of the energy will remain stored in the capacitors while two thirds will be dissipated as heat so again energy is conserved.
There is no rule about charge being conserved it just happen due to symmetry as there is resistance everywhere including the capacitor plates.
If you add a discharged inductor in the circuit you are not adding either charge or energy and you are just adding another energy storage device as an intermediary.
The point is that energy is always conserved in an isolated system and using this equation is easy to see where energy ended up (in the conductors) from witch should be easy to conclude what path the energy has traveled.
If energy transfer was not done through conductors then there will be no reason to have all the energy loss (exactly all) inside the conductors.
The symmetry is broken when you add an inductor even if you are actually not bringing any charge from outside and system is still isolated. Inductor is basically a fairly long wire.
Edit:
I'm thinking of ways you can break the symmetry but unless one of the capacitors has superconductor plates I do not see how it can be done other than by adding an inductor as intermediary energy storage device.
Parallel capacitors will always have this symmetry as they have resistance (they are like a transmission line).
Adding an inductor will just break this symmetry reducing the loss as heat in wire.
I will try to think of a mechanical analog.
It will not make sense to do the experiment as you probably agree with me on what the result of that test will be.
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Yes, “isolated system” is the key point.
If resistor heat dissipation can escape from the bench top, then the system is not isolated.
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Yes, “isolated system” is the key point.
If resistor heat dissipation can escape from the bench top, then the system is not isolated.
You can drop the circuit in to an insulated box then it is isolated :) It will have some small leakage but still isolated.
I hope you do not doubt that all that missing energy is not ending up as heat due to resistance.
It is basically the equivalent of charging a 1.5V cell from a 3V supply through a linear regulator or resistor so half the energy is wasted as heat in the resistor or linear regulator.
The total charge Q will of course always the the same in this scenarios and a lot of energy will be lost as heat.
Adding an extra energy storage device that is first charged from the source and then discharged on the load can increase this efficiency significantly and in that case the Q will no longer be the same at the start and the end of the experiment.
Conservation of charge is no more even if you did not added charge from any external source you just used the available energy more efficiently.
So energy conservation is a law that can not be violated where charge conservation is more of a coincidence when resistive losses are involved and there is no mechanism to reduce those losses like adding a inductor as intermediary energy storage.
The important part of all this is that there is no mystery in the identical parallel capacitors and all energy is accounted for and another important part is that energy travels through wires as that is where all this heat loss originate.
On the long transmission line you have both inductive and capacitive storage but the capacitive storage is responsible for the initial energy flow through lamp as the lamp is in series with the two capacitors being charged so lamp becomes part of the total resistance of the circuit so part of the losses will be on the lamp (significant part when you use a 1KOhm resistor as a lamp) and thick copper pipes as transmission line.
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Yes, you can do calorimeter on the insulated box and measure the thermal portion of the system energy.
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Yes, you can do calorimeter on the insulated box and measure the thermal portion of the system energy.
But you think that is needed ? Is clear missing energy is ending as heat in the wires/conductors.
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Spell check error: “calorimetry”.
No, I don’t think it’s necessary. Just an example of how energy conservation is not the most relevant calculation method in certain situations.
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Spell check error: “calorimetry”.
No, I don’t think it’s necessary. Just an example of how energy conservation is not the most relevant calculation method in certain situations.
The relevant part is that energy conservation can not be violated.
If you were to measure energy going out of the source in Derek's experiment and energy received by the lamp/resistor you will see the situation in this graph
[attach=1]
That shows that much more power exits the source than gets to the lamp and that is explained by the fact that energy is stored in the line capacitance.
To make things even more clear the second graph shows what happens if the switch is only closed circuit for 30ns so less than half it is required for the electron wave to travel the transmission line.
[attach=2]
In both graphs the green is the power from source so the integral of that is the energy and the purple line is the lamp/resistor power.
The difference that is missing ends as heat on the line but most of the energy gets to lamp after the time needed for the electron wave to get there.
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A selection of references for the two-capacitor problem (one of which has already been given in this very thread)
The two-capacitor problem with radiation
Timothy B. Boykin, Dennis Hite, and Nagendra Singh
Department of Electrical and Computer Engineering,
The University of Alabama in Huntsville, Huntsville, Alabama 35899
~Received 12 June 2001; accepted 8 November 2001; Published Online: 11 March 2002
American Association of Physics Teachers. @DOI: 10.1119/1.1435344
available upon request to the author from this ResearchGate page:
https://www.researchgate.net/publication/243492397_The_two-capacitor_problem_with_radiation (https://www.researchgate.net/publication/243492397_The_two-capacitor_problem_with_radiation)
Capacitors can radiate: Some consequences of the two-capacitor problem with radiation
14 May 2003
T.C. CHOY
arXiv:physics/0305062v1 [physics.class-ph]
https://www.researchgate.net/publication/2168891_Capacitors_can_radiate_-_some_consequences_of_the_two-capacitor_problemwith_radiation (https://www.researchgate.net/publication/2168891_Capacitors_can_radiate_-_some_consequences_of_the_two-capacitor_problemwith_radiation)
freely available at ResearchGate (direct link)
https://www.researchgate.net/profile/Tuck-Choy/publication/2168891_Capacitors_can_radiate_-_some_consequences_of_the_two-capacitor_problemwith_radiation/links/5935528daca272fc5556a317/Capacitors-can-radiate-some-consequences-of-the-two-capacitor-problem-with-radiation.pdf (https://www.researchgate.net/profile/Tuck-Choy/publication/2168891_Capacitors_can_radiate_-_some_consequences_of_the_two-capacitor_problemwith_radiation/links/5935528daca272fc5556a317/Capacitors-can-radiate-some-consequences-of-the-two-capacitor-problem-with-radiation.pdf)
Radiative effects and the missing energy paradox in the two capacitor problem
Gilberto A Urzua, Omar Jimenez, Fernando Maass and Alvaro Restuccia
Departamento de Fisica, Universidad de Antofagasta, Casilla 170, Antofagasta, Chile
E-mail: gilberto.urzua@uantof.cl, omar.jimenez@uantof.cl, fernando.maass@uantof.cl, alvaro.restuccia@uantof.cl
Journal of Physics: Conference Series 720 (2016) 012054
doi:10.1088/1742-6596/720/1/012054
https://www.researchgate.net/publication/303980311_Radiative_effects_and_the_missing_energy_paradox_in_the_ideal_two_capacitors_problem (https://www.researchgate.net/publication/303980311_Radiative_effects_and_the_missing_energy_paradox_in_the_ideal_two_capacitors_problem)
freely available at ResearchGate (direct link)
https://www.researchgate.net/publication/303980311_Radiative_effects_and_the_missing_energy_paradox_in_the_ideal_two_capacitors_problem/fulltext/57a4bae308ae455e8539f85d/Radiative-effects-and-the-missing-energy-paradox-in-the-ideal-two-capacitors-problem.pdf (https://www.researchgate.net/publication/303980311_Radiative_effects_and_the_missing_energy_paradox_in_the_ideal_two_capacitors_problem/fulltext/57a4bae308ae455e8539f85d/Radiative-effects-and-the-missing-energy-paradox-in-the-ideal-two-capacitors-problem.pdf)
The two-capacitor problem revisited: a mechanical harmonic oscillator model approach
Keeyung Lee
Department of Physics, Inha University, Incheon, 402-751, Korea
Received 17 July 2008, in final form 4 September 2008; Published 6 November 2008
EUROPEAN JOURNAL OF PHYSICS
Eur. J. Phys. 30 (2009) 69–74 doi:10.1088/0143-0807/30/1/007
Online at stacks.iop.org/EJP/30/69
Preprint available from ArXiv: https://arxiv.org/abs/1210.4155 (https://arxiv.org/abs/1210.4155)
Entropic Considerations of the Two-Capacitor Problem
January 19, 2012
V.O.M. Lara, A. P. Lima, and A. Costa
Instituto de Fısica - Universidade Federal Fluminense
https://www.researchgate.net/publication/51990322_Entropic_considerations_on_the_Two-Capacitor_Problem (https://www.researchgate.net/publication/51990322_Entropic_considerations_on_the_Two-Capacitor_Problem)
freely available at ResearchGate (direct link)
https://www.researchgate.net/profile/Ap-Lima/publication/51990322_Entropic_considerations_on_the_Two-Capacitor_Problem/links/55e9aa6208ae21d099c302fb/Entropic-considerations-on-the-Two-Capacitor-Problem.pdf (https://www.researchgate.net/profile/Ap-Lima/publication/51990322_Entropic_considerations_on_the_Two-Capacitor_Problem/links/55e9aa6208ae21d099c302fb/Entropic-considerations-on-the-Two-Capacitor-Problem.pdf)
The Paradox of Two Charged Capacitors -- A New Perspective
August 2013
Authors: Ashok K. Singal
Indian Space Research Organization arXiv
https://www.researchgate.net/publication/256762725_The_Paradox_of_Two_Charged_Capacitors_--_A_New_Perspective (https://www.researchgate.net/publication/256762725_The_Paradox_of_Two_Charged_Capacitors_--_A_New_Perspective)
(direct link)
https://www.researchgate.net/profile/Ashok-Singal/publication/256762725_The_Paradox_of_Two_Charged_Capacitors_--_A_New_Perspective/links/559e349008ae76bed0bb6d46/The-Paradox-of-Two-Charged-Capacitors--A-New-Perspective.pdf (https://www.researchgate.net/profile/Ashok-Singal/publication/256762725_The_Paradox_of_Two_Charged_Capacitors_--_A_New_Perspective/links/559e349008ae76bed0bb6d46/The-Paradox-of-Two-Charged-Capacitors--A-New-Perspective.pdf)
The idea I have come to so far:
You start with energy E, you end up with energy E/2 split between the two caps and E/2 lost in some way (doesn't actually matter which one and in which measure). Charge is conserved, voltage is the same.
The transformation is irreversible, therefore some energy MUST be lost and entropy must increase.
Turns out that energy it can be lost in different ways, depending on the actual circuit: if there is appreciable resistance (a handful of microohms might suffice), it goes basically all into heat. If there is nearly no resistance or exactly zero resistance it goes into radiation (even without inductance in the loop) either by magnetic dipole or by electric dipole. If there is appreciable inductance it can go back and forth rapidly enough to lose energy by radiation of the LC oscillator (but it's usually the other ways that predominate).
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A selection of references for the two-capacitor problem (one of which has already been given in this very thread)
The idea I have come to so far:
You start with energy E, you end up with energy E/2 split between the two caps and E/2 lost in some way (doesn't actually matter which one and in which measure). Charge is conserved, voltage is the same.
The transformation is irreversible, therefore some energy MUST be lost and entropy must increase.
Turns out that energy it can be lost in different ways, depending on the actual circuit: if there is appreciable resistance (a handful of microohms might suffice), it goes basically all into heat. If there is nearly no resistance or exactly zero resistance it goes into radiation (even without inductance in the loop) either by magnetic dipole or by electric dipole. If there is appreciable inductance it can go back and forth rapidly enough to lose energy by radiation of the LC oscillator (but it's usually the other ways that predominate).
While I have not read any of those papers the fact that there are so many is ridiculous.
If you read any of those paper can you tell me if any of them actually did the test ?
Even if they have a challenge understanding the theory the experimental result will show them exactly what happens and that is the missing energy is all found in the conductors proportional with their resistance.
I guess none of them had the budget to do the superconductor experiment as they will have found out all energy remains stored and it is not radiated as some of the titles of those papers you link may suggest.
Even my simple simulation shows that losses in the case of two identical capacitors can be reduced from 50% to something like 12% or less by using an inductor and a diode to transfer the energy.
I can even provide you with a simple way to test that all loss is thermal loss not radiated (unless you call infrared photons radiated energy).
Experiment will be fairly simple:
Take to large capacitors either large electrolytic or super capacitors (just because is easier to measure the energy lost as heat not that is needed as it can be calculated).
I will give this example
3V 1F for the charged capacitor
identical 1F discharged capacitor.
Take this for example https://www.mouser.com/datasheet/2/40/AVX_SCC_3_0V-1128335.pdf (https://www.mouser.com/datasheet/2/40/AVX_SCC_3_0V-1128335.pdf)
SCCR12E105PRB it has a max DC ESR of 860mOhm I will rounded up to 1Ohm.
Use a 8Ohm series resistor just that total resistance in the circuit is a round 10Ohm the 1Ohm from the two capacitors than the 8Ohm you add.
Now just use a multimeter with data logging to measure the voltage drop across the 8Ohm resistor.
You know for sure that voltage drop divided by resistance will give you the power dissipated as heat on the resistor so no need to measure any temperature and you can have very accurate measurement.
Make 5 or 10 measurements per second depending on how capable your logging multimeter is as experiment will likely need to run for about one minute to properly charge the discharged capacitor from the charged one.
Then integrate that power measurement over time to get energy and divide by 0.8 to get the entire energy dissipated as heat including the heat dissipated on the capacitor's plates that 1Ohm DC ESR.
You will get exactly all the missing energy so half of 4.5Ws that you started as as heat and the other half is split between the two capacitors as stored energy.
With accurate equipment and careful measurement you will find that not even a faction of a percent of energy is unaccounted for.
But for anyone that understand physics this experiment will not even be needed.
And yes it is sad to see all those universities and research institutions wasting resources and making fools of themselves.
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While I have not read any of those papers the fact that there are so many is ridiculous.
Not necessarily: the first one made the analysis for a magnetic dipole radiator, obtaining under certain simplifications a nonlinear differential equation of third order that gives an exponentially decaying solution.
IIRC, the second one generalized the results by making the analysis more general, obtaining a fifth order differential equation and finding a new type of solution that goes to zero in a finite time ('sudden death'). The third one considers radiation by electric dipole, as well.
The others consider the problem from a more general point of view (increase in entropy and loss of energy). But get the same conclusion about the necessity to lose half the initial energy. So, they are not necessarily in contrast with one another, but rather explore different ways to see the same problem.
Incidentally, there is a video by youtuber SimplyPut that reaches the same conclusions about entropy in a rather intuitive way. The guy is a bus driver, but in my opinion he got the problem right.
If you read any of those paper can you tell me if any of them actually did the test ?
I guess they had a little problem in finding ideal capacitor and getting a grant to set up the superconducting wires. So, no, I don't think you can test the ideal problem in this world. But the first paper has a very interesting graph that gives you how much of the energy goes into heat and how much into radiation, based on the value of resistance in the wires.
Even if they have a challenge understanding the theory the experimental result will show them exactly what happens and that is the missing energy is all found in the conductors proportional with their resistance.
Did you perform the experiment with an ideal capacitor and superconducting wires?
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Did you perform the experiment with an ideal capacitor and superconducting wires?
No ideal capacitor or superconducting wires are needed. To prove that all energy that is not in capacitors at the end of the test is found as heat on the resistive elements (basically all conductors in the circuit).
There will be no energy left that is not accounted for either as energy stored in capacitor or as heat so no energy is "radiated away" other than as heat.
Hope you understand that supercapacitors have nothing to do with superconductors as I mentioned supercapacitors for the test just because they have higher capacity so discharge is slower and you can use a multimeter but you can use any other type of capacitors and an oscilloscope tho you lose the vertical resolution compared to a multimeter so accuracy may be a bit lower.
If you did not heard of supercapacitors before you can check the spec that I listed for the example.
If you understand the mechanisms you do not need to do the test to know what the results will be. This is my main hobby (energy generation and energy storage) so it will be boring to do a test for witch I know exactly the result.
If you measure a voltage drop across a resistor you know that all that energy calculated form that ended up as heat.is not like an LED where some was radiated as visible photons there will be infrared radiated from the resistor but that is not the type of radiation discussed in those papers.
I feel frustrated not being able to properly explain what looks like such a simple problem.
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You can just admit you don't know how superconductors behave, like, it's okay to not know things sometimes?? I know it's awfully late in the thread to do so [sunk cost fallacy], but late is still better than never?
The whole reason superconductors were brought up is you claim a conservation of energy, when none exists in any possible conceivable experiment that can be constructed, using such materials and no active circuitry (e.g. DC-DC).
You've had many "outs" presented to you throughout this thread:
- Energy goes into AC mode
- "Oops, I meant RMS voltage"
- Admit the trivial circuit (two capacitances and ""ideal"" wires) cannot be constructed
- Admit the trivial circuit (two capacitances and real wires, superconducting or otherwise) behaves as expected
After the first few pages, the main curiosity in this thread, I think, revolves around the peculiar psychology causing this reluctance; the technical topics are all well settled, after all.
Tim
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Check if your solution satisfies the law of preservation of charge.
You do not understand what conservation of charge means and how it is applied.
The clue is in the fact that adding another charged particle to a capacitor that is at 1V and same capacitor that is at 2V requires different amounts of energy.
:palm: What you are trying to do is not the law of conservation of charge.
Just integrate the current through the capacitors with respect to time to calculate the charge in each capacitor.
Think of it this way: You are in a sealed room with an empty box on the floor and a box full of tennis balls high op on a shelf. You move half of the balls from the full box to the empty box. In the process some of the gravitational potential energy of the balls is converted to another form (probably heat). The point is that the number of tennis balls stays the same. So we have the law of conservation of tennis balls. The law of conservation of charge is the same concept. It is about the number of charged particles, not their electric potential energy!
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You can just admit you don't know how superconductors behave, like, it's okay to not know things sometimes?? I know it's awfully late in the thread to do so [sunk cost fallacy], but late is still better than never?
The whole reason superconductors were brought up is you claim a conservation of energy, when none exists in any possible conceivable experiment that can be constructed, using such materials and no active circuitry (e.g. DC-DC).
You've had many "outs" presented to you throughout this thread:
- Energy goes into AC mode
- "Oops, I meant RMS voltage"
- Admit the trivial circuit (two capacitances and ""ideal"" wires) cannot be constructed
- Admit the trivial circuit (two capacitances and real wires, superconducting or otherwise) behaves as expected
After the first few pages, the main curiosity in this thread, I think, revolves around the peculiar psychology causing this reluctance; the technical topics are all well settled, after all.
Tim
I never did experiments with superconductors.
- The properties of superconductors are zero resistance to current flow (not super low resistance but zero).
- Conservation of energy in an isolated system is always true no matter if superconductor materials are used or not.
What are you calling an active circuit ? If the circuit is powered by the only source of energy at the start of the experiment meaning the charged capacitor in this case it is completely irrelevant what components you are using as long as the system stay isolated.
The inductor and simple diode is all that is needed as I have shown in spice simulation. Neither the diode nor the inductor have any stored energy at the beginning of the experiment.
If I can demonstrate that significantly less energy is lost by just adding the diode and inductor and sowed that conservation of charge is no longer a thing then what makes you so shure that can not be shown with superconductors where no conductor in circuit will have any resistance.
I also hope you understand the concept of an isolated system.
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:palm: What you are trying to do is not the law of conservation of charge.
Just integrate the current through the capacitors with respect to time to calculate the charge in each capacitor.
Think of it this way: You are in a sealed room with an empty box on the floor and a box full of tennis balls high op on a shelf. You move half of the balls from the full box to the empty box. In the process some of the gravitational potential energy of the balls is converted to another form (probably heat). The point is that the number of tennis balls stays the same. So we have the law of conservation of tennis balls. The law of conservation of charge is the same concept. It is about the number of charged particles, not their electric potential energy!
In a circuit with resistance you have charged capacitor in series with resistor in series with discharged capacitor.
Since all this elements are in series current will be the same through the loop.
So at the start you have 3V in charged capacitor say 3Ohm total circuit resistance an 0V on discharged capacitor.
Thus almost all energy is dissipated as heat on the resistor as you have 3V / 3Ohm = 1A * 3V = 3W dissipated as heat on the resistor.
A bit latter you have 2V on the charged capacitor and 1V on the discharged capacitor as current exiting the charged capacitor can only be equal with the one entering the discharged capacitor thus the special case of charge being conserved and half of the energy lost as heat.
Now you will have (2V - 1V ) / 3Ohms = 0.333A * 1V = 0.333W loss on the resistor as heat.
If you do the integration either as a calculation or as a measurement you will see that no energy is missing and half of initial energy is still in the two capacitors while the other half was lost as heat on the resistor. There is no energy unaccounted for so no paradox.
This is as basic of electrical knowledge as it gets. If you want to claim that there is energy lost some other way other that as heat on the circuit resistance then you are welcome to prove that either with math or with an experiment.
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Again, in that simple ideal circuit with a finite-value resistor connecting the second (initially uncharged) capacitor to the first (charged) capacitor, and no additional lossy elements (such as capacitor ESR or leakage) the total energy lost by the circuit itself (into heat) is independent of the resistance value, and my earlier statement holds, that it is the same value in the limit as R-->0. Of course, the time required to transfer the charge is directly proportional to that resistor value and the limit as R-->0 of transfer time (to a specific settling fractional value such as 1 ppm, or 14 time constants) is zero. Here, "limit" is the normal mathematical meaning of the term.
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Again, in that simple ideal circuit with a finite-value resistor connecting the second (initially uncharged) capacitor to the first (charged) capacitor, and no additional lossy elements (such as capacitor ESR or leakage) the total energy lost by the circuit itself (into heat) is independent of the resistance value, and my earlier statement holds, that it is the same value in the limit as R-->0. Of course, the time required to transfer the charge is directly proportional to that resistor value and the limit as R-->0 of transfer time (to a specific settling fractional value such as 1 ppm, or 14 time constants) is zero. Here, "limit" is the normal mathematical meaning of the term.
Capacitor DC ESR is already included in that resistance. So there is no additional loss. Energy that is not found stored in the two capacitors at the end of the test will be found as heat in the resistive elements and that is super obvious includes the DC ESR witch is the resistance of the capacitor plates which are no different from wires.
If you admit that there is no energy unaccounted for either as stored in capacitors or as heat in the conductors than you admit that there is no paradox and there is no magical energy that is not explained by this two.
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Again, in that simple ideal circuit with a finite-value resistor connecting the second (initially uncharged) capacitor to the first (charged) capacitor, and no additional lossy elements (such as capacitor ESR or leakage) the total energy lost by the circuit itself (into heat) is independent of the resistance value, and my earlier statement holds, that it is the same value in the limit as R-->0. Of course, the time required to transfer the charge is directly proportional to that resistor value and the limit as R-->0 of transfer time (to a specific settling fractional value such as 1 ppm, or 14 time constants) is zero. Here, "limit" is the normal mathematical meaning of the term.
Capacitor DC ESR is already included in that resistance. So there is no additional loss. Energy that is not found stored in the two capacitors at the end of the test will be found as heat in the resistive elements and that is super obvious includes the DC ESR witch is the resistance of the capacitor plates which are no different from wires.
If you admit that there is no energy unaccounted for either as stored in capacitors or as heat in the conductors than you admit that there is no paradox and there is no magical energy that is not explained by this two.
Yes, all the energy is accounted for when you include the "loss" away from the circuit as heat. If it were important, one could do calorimetry to track that loss. Note that Spice simulations do not handle heat or EM radiation, only voltages and currents in the discrete components of the Spice model. My comments about the limits just are to emphasize that the "lost" energy is independent of the total circuit (series) resistance, and that the electrical components themselves are not an isolated system in the thermodynamic meaning of "isolated". I never considered the two-capacitor system to be paradoxical.
At a trade show, I had an interesting conversation with an engineer from SBE, who make very low-loss polypropylene capacitors. One of their series, used in high-voltage pulse systems, has such low loss that they indeed used calorimetry to measure it.
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Yes, all the energy is accounted for when you include the "loss" away from the circuit as heat. If it were important, one could do calorimetry to track that loss. Note that Spice simulations do not handle heat or EM radiation, only voltages and currents in the discrete components of the Spice model. My comments about the limits just are to emphasize that the "lost" energy is independent of the total circuit (series) resistance, and that the electrical components themselves are not an isolated system in the thermodynamic meaning of "isolated". I never considered the two-capacitor system to be paradoxical.
At a trade show, I had an interesting conversation with an engineer from SBE, who make very low-loss polypropylene capacitors. One of their series, used in high-voltage pulse systems, has such low loss that they indeed used calorimetry to measure it.
The important part is where the loss originated in order to know what medium energy traveled through.
Since all the loss originated in wires (not outside the wires) that means energy traveled through wires.
Yes energy in the form of infrared photons will be radiated from the wires to outside the wire with circuit in a vacuum but that is after the energy was already transported.
Of course the loss is independent in this particular case of the resistance as we are talking about energy that is power integrated over time.
With a lower resistance you have higher power loss but for a shorter time duration and with a higher resistance you have lower power loss but over a longer period of time.
Glad to hear that you do not consider the two capacitor problem a paradox.
That loss is in dielectric like a parallel resistance exists with any capacitor as we do not have an ideal insulator like we have an ideal conductor.
The main point of discussion is if energy travels through wire or not.
Electrical energy is electrical power integrated over time and electrical power is electrical potential multiplied with electrical current.
Since electrical current is the rate at which the charge is moving and charge in this case is an electron and they will flow through wires unless energy is so high that air becomes a conductor (not the case in Derek's experiment).
Thus from the above the conclusion will be that energy travels through wire.
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I don't see that we disagree about energy traveling through a wire and being dissipated as it travels.
With respect to capacitor resistance, a physical capacitor (either with a dielectric, or some other insulators to hold the plates in fixed locations, such as a vacuum capacitor) will have a series resistance and a parallel resistance.
When engineers discuss "ESR" of a real capacitor, they must take into account its frequency dependence.
At a single frequency, the series and parallel capacitances can be combined into a single ESR (or into a equivalent parallel resistance), by elementary circuit theory.
A physical resistance in series with the capacitor (plates and wires) will dissipate energy as heat, but conserves the charge.
A parallel (leakage) resistance will also dissipate energy as heat, but will eventually discharge the capacitors, which does not conserve the charge.
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I don't see that we disagree about energy traveling through a wire and being dissipated as it travels.
With respect to capacitor resistance, a physical capacitor (either with a dielectric, or some other insulators to hold the plates in fixed locations, such as a vacuum capacitor) will have a series resistance and a parallel resistance.
When engineers discuss "ESR" of a real capacitor, they must take into account its frequency dependence.
At a single frequency, the series and parallel capacitances can be combined into a single ESR (or into a equivalent parallel resistance), by elementary circuit theory.
A physical resistance in series with the capacitor (plates and wires) will dissipate energy as heat, but conserves the charge.
A parallel (leakage) resistance will also dissipate energy as heat, but will eventually discharge the capacitors, which does not conserve the charge.
For this examples where a fairly large capacitor is discharged over a few seconds in to another capacitor the ESR of the capacitor will basically be the DC ESR witch is dependent on the DC resistance of the plates.
The parallel leakage resistance is so high for typical capacitors that will be within the measurement error of such a setup so it will not be relevant.
When Derek's says that "energy doesn't travel in wires" he is not referring to leakage he is implying that all energy travels outside the wires.
His proof is that energy arrives at the lamp in the time it takes light to travel 1m (distance between the source and the lamp/resistor) instead of the much longer transmission line.
That argument is completely irrelevant because he ignores the fact that the transmission line has capacitance. He completely ignored the line capacitance in the first video and while he acknowledged that in second video he ignored the effect of that even if that capacitance is what explains fully that energy through the lamp in first 65ns.
This can be seen as a charged parallel plate capacitor (open circuit) with large gap between the plates where you insert another plate of negligible thickness. This will transform that single capacitor in to two capacitors in series. That plate will see charges separate on each side of the plate so electrons move from one face of the plate to the other. When the plate is removed the charges will get back to neutral.
Now if you instead of one plate insert two plates connected together with a lamp then when the electrons move from one plate to the other they will pass through the lamp and the same in reverse will happen when you get those plates out of the electric field.
The energy in the charged capacitor will remain the same so energy was not provided by that but by the person that inserted those plates.
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The pathological nature of the Two Capacitor Paradox problem leads to the observation that only the Law of Conservation of Charge is satisfied, but the Law of Conservation of Energy is violated.
Law of conservation of energy has never been broken/violated. There is no paradox related to the two parallel capacitors so that page on Wikipedia is misinformation as anyone can write a wiki page and write whatever it wants.
Of course. And you absolutely need to stop saying that other people on this thread are claiming conservation of energy is violated. That is just an outright lie and it isn't ok. Stop it.
The point of the paradox is to show that the initial setup is non physical. Postulating zero loss and zero inductance and requiring that you reach a steady state is a non physical setup and the equivalent to dividing by zero. Any conclusions drawn from it are meaningless. Your entire premise is based on something that cannot exist and is meaningless.
With no inductance or resistance the voltage is discontinuous and the current flow is infinite. That is impossible. If you add any finite resistance or inductance -- even an attohenry everything becomes finite and does not agree with your claim. This is trivially verifiable analytically, by simulation, or by experent.
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Even "dividing by zero" can be handled analytically with proper limits.
As an example, the famous sinc function
y=sin(x)/x obviously goes to 1 at x=0,
even though both the numerator and denominator are 0 at x=0.
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Of course. And you absolutely need to stop saying that other people on this thread are claiming conservation of energy is violated. That is just an outright lie and it isn't ok. Stop it.
The point of the paradox is to show that the initial setup is non physical. Postulating zero loss and zero inductance and requiring that you reach a steady state is a non physical setup and the equivalent to dividing by zero. Any conclusions drawn from it are meaningless. Your entire premise is based on something that cannot exist and is meaningless.
With no inductance or resistance the voltage is discontinuous and the current flow is infinite. That is impossible. If you add any finite resistance or inductance -- even an attohenry everything becomes finite and does not agree with your claim. This is trivially verifiable analytically, by simulation, or by experent.
You will be spiking for yourself in regards to energy conservation as it seems many people disagree with that either knowingly or unknowingly like Derek did in the Direct downwind faster than wind video where what he basically presented there was an overunity device so getting more power out than in.
You can not get rid of inductance but you can get rid of resistance.
And you can get inductance as a intermediary energy storage in a normal circuit with resistance and reduce the effect of energy loss as I already demonstrated very close to ideal case of zero energy loss that is possible for a circuit with no resistance.
The two capacitor problem is useful if understood to debunk the main claim Derek made and that is the "energy doesn't travel in wires".
All energy including the initial transient (AC) and the DC after that will travel from the source to the load (lamp/resistor) through wires.
There is absolutely no evidence of Derek's main claim and he's supposed evidence is to show that there is energy arriving at the lamp before the electron wave had the time to travel the entire transmission line about 65ns in his experiment.
The explanation for that small amount of energy he sees before those 65ns is easy to explain if you understand that a transmission line is made of distributed capacitance and inductance elements for the entire length of the transmission line and that the capacitance being an energy storage device is the one responsible for that current seen through the lamp while energy is stored in the two capacitors each side of the lamp.
There is no current flow through a capacitors when charging but there is a current flow in to a capacitor and that just means energy travels through wires as electrical energy is the integral over time of electrical power which is the product of electrical potential and electrical current.
Since current is not possible through the dielectric of a capacitor in this case 1 meter of air electrical energy can only travel through wires.
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A question (for Tim, ejeffrey, or whoever):
How do we precisely define the circumstances in which conservation of charge is applicable when analyzing a system, and when it is not?
For example, we may have capacitor A holding a charge Q0, terminals connected to one side of a black box. A second capacitor B starts out uncharged and is connected to the other side of the black box. The black box draws power from capacitor A and uses it to charge capacitor B. At the end of the process, capacitor A has remaining charge Q1 and capacitor B has charge Q2. We can imagine that Q1 + Q2 may not always be equal to Q0.
We know that if the black box just contains a series resistor, then conservation of charge will apply. But if the black box contains a DC/DC converter, then maybe not.
What is the rigorous technical statement about when and where conservation of charge applies? Is it related to the number of closed loops in the system topology? That conservation of charge applies around each loop? But then what about branches where two loops overlap? Or is it that conservation of charge should apply within each separable island in the topology, such as on each side of a transformer?
Or is this just overcomplicating things, and one should just apply KCL around each node in the system and sum over all the nodes?
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I would love to hear electrodacus's analysis of the behavior of this circuit when the switch is closed, with Vi of 3 V and C of 3 F, with ideal capacitor, wires and switches, of course, (so no dissipative elements).
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I would love to hear electrodacus's analysis of the behavior of this circuit when the switch is closed, with Vi of 3 V and C of 3 F, with idea capacitor, wires and switches, of course, (so no dissipative elements).
It will cause a rift in the fabric of space-time and open up a wormhole to another dimension ;D
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A question (for Tim, ejeffrey, or whoever):
How do we precisely define the circumstances in which conservation of charge is applicable when analyzing a system, and when it is not?
Charge conservation, considered as a physical conservation law, implies that the change in the amount of electric charge in any volume of space is exactly equal to the amount of charge flowing into the volume minus the amount of charge flowing out of the volume.
If we agree that there is no transfer to charge through the capacitors, and there is no charge flowing in from outside of the schematic, then hopefully we can agree that that the charge in the upper half (A) and the charge in the lower half (B) will remain the same before and after the switch is closed.
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I would love to hear electrodacus's analysis of the behavior of this circuit when the switch is closed, with Vi of 3 V and C of 3 F, with idea capacitor, wires and switches, of course, (so no dissipative elements).
You will have an ideal LC circuit thus current through the inductor will increase gradually until capacitor will be fully discharged so a finite current that will create a finite magnetic field that will then collapse and charge the capacitor all they way back up to 3V.
At any point after you close the switch the energy will be there in an isolated circuit as no energy is lost through resistive loss as there is no resistance and no energy will be dissipated despite the variable magnetic field around the circuit.
But if the circuit is not isolated meaning you have some other loop in range of this and that has resistance then energy will be lost to that eventually all the initial energy will be lost in that magnetically coupled loop. But this will no longer be considered an isolated circuit.
The point is that none of the energy will be radiated away but always be there and if you open the switch at the correct time when all energy is in the capacitor then you will be in the same state as the initial state.
That is how the charging of a capacitor from another using an inductor as intermediary storage works in order to reduce the amount of lost energy during transfer.
Adding the inductor adds no energy to the circuit yet it helps getting very close to ideal where most of the energy is still stored after transfer and not lost as heat.
So no energy is "radiated away" by in this case magnetic field as the electric field exists only between the capacitor plates but that also is not "radiated away"
While in Derek's example there is of course energy stored in magnetic field around the wires it is not what transfers the energy from one wire to the other.
The current in those first 65ns is due to capacitor being charged as current cannot flow through 1m of air at 20V potential (leakage so low that it will not register). Since current flow is not present trough air no energy travels outside the wire let alone all energy as Derek claims.
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A question (for Tim, ejeffrey, or whoever):
Just noticed that your profile photo is an illustration of energy traveling through air as there electrical potential is large enough to make the air a usable conductor. Not the case with the 20V in Derek's experiment.
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But if the circuit is not isolated meaning you have some other loop in range of this and that has resistance then energy will be lost to that eventually all the initial energy will be lost in that magnetically coupled loop.
Are you saying if there was a suitable loop in range (say with a resistor in it), then all the energy would be lost from this circuit? That all the energy would be transferred through the air to this loop, and it would end up heating the resistor?
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But if the circuit is not isolated meaning you have some other loop in range of this and that has resistance then energy will be lost to that eventually all the initial energy will be lost in that magnetically coupled loop.
Are you saying if there was a suitable loop in range (say with a resistor in it), then all the energy would be lost from this circuit? That all the energy would be transferred through the air to this loop, and it would end up heating the resistor?
Yes but it may take hundreds of years if the distance of that loop is relatively far. And that will not be considered an isolated system. The discussion of isolated system can be just theoretical in real tests it will be close enough to an isolated system in order to draw some conclusions depending on what the test wants to show.
It is very similar with a superconductor ring where you induce a current at that remains there indefinitely https://www.open.edu/openlearn/science-maths-technology/engineering-technology/superconductivity/content-section-2.2 (https://www.open.edu/openlearn/science-maths-technology/engineering-technology/superconductivity/content-section-2.2)
Of course you can not have an ideal capacitor as you can not have a dielectric without losses so the LC even if made of superconductor materials it will eventually dissipate the energy as heat in the dielectric material.
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But if the circuit is not isolated meaning you have some other loop in range of this and that has resistance then energy will be lost to that eventually all the initial energy will be lost in that magnetically coupled loop.
Are you saying if there was a suitable loop in range (say with a resistor in it), then all the energy would be lost from this circuit? That all the energy would be transferred through the air to this loop, and it would end up heating the resistor?
Yes but it may take hundreds of years if the distance of that loop is relatively far. And that will not be considered an isolated system. The discussion of isolated system can be just theoretical in real tests it will be close enough to an isolated system in order to draw some conclusions depending on what the test wants to show.
It is very similar with a superconductor ring where you induce a current at that remains there indefinitely https://www.open.edu/openlearn/science-maths-technology/engineering-technology/superconductivity/content-section-2.2 (https://www.open.edu/openlearn/science-maths-technology/engineering-technology/superconductivity/content-section-2.2)
Of course you can not have an ideal capacitor as you can not have a dielectric without losses so the LC even if made of superconductor materials it will eventually dissipate the energy as heat in the dielectric material.
You seem to be saying:
"A portion of the energy is outside of the wires, and can be transferred to completely galvanically isolated circuits. The amount of energy transferred between the two depend on the physical geometry of their physical arrangement - so size, distance, orientation and so on."
Do you get dielectric losses if you use a vacuum as the dielectric in capacitors? What heats up if there is nothing there?
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like Derek did in the Direct downwind faster than wind video where what he basically presented there was an overunity device so getting more power out than in.
You were, and are, wrong about that. You wanna pop back over to that thread and answer the questions you conveniently forgot about?
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You seem to be saying:
"A portion of the energy is outside of the wires, and can be transferred to completely galvanically isolated circuits. The amount of energy transferred between the two depend on the physical geometry of their physical arrangement - so size, distance, orientation and so on."
Do you get dielectric losses if you use a vacuum as the dielectric in capacitors? What heats up if there is nothing there?
You can only transfer energy while the magnetic field strength is variable so it will not work at DC.
I try to stay on the subject and magnetic field is not involved in what happens in those first few ns when switch is closed.
The line capacity is what induces that current through the lamp/resistor.
Vacuum will also have losses as there is no such thing as real vacuum (But I'm not a physicist so I will not pretend to know all the subatomic particles that may or not pop up or out of existence at random).
Main subject is electrical energy and if it flows through wires or not. Since we are ignoring any super small leakage current through air or even vacuum the main electric current will travel through wires and you need current on top of electrical potential in order to have power with integrated over time will be energy.
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You were, and are, wrong about that. You wanna pop back over to that thread and answer the questions you conveniently forgot about?
I will go there and check but you will need to also go there and point to me the question you are referring to as I'm sure there are a lot of them.
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Again, in that simple ideal circuit with a finite-value resistor connecting the second (initially uncharged) capacitor to the first (charged) capacitor, and no additional lossy elements (such as capacitor ESR or leakage) the total energy lost by the circuit itself (into heat) is independent of the resistance value, and my earlier statement holds, that it is the same value in the limit as R-->0. Of course, the time required to transfer the charge is directly proportional to that resistor value and the limit as R-->0 of transfer time (to a specific settling fractional value such as 1 ppm, or 14 time constants) is zero. Here, "limit" is the normal mathematical meaning of the term.
My biggest gripe with the R->0 limit is that it necessarily lead to an unphysical model. If charge needs to move from cap1 to cap 2 at a finite distance, the zero transfer time implies faster than light charge motion. Radiation takes you out of this untenable position: the charges move from here to there, but they need to accelerate and decelerate - hence there will be radiation.
The two-cap problem is a model breaker. One way or the other, something's gotta give.
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My biggest gripe with the R->0 limit is that it necessarily lead to an unphysical model. If charge needs to move from cap1 to cap 2 at a finite distance, the zero transfer time implies faster than light charge motion. Radiation takes you out of this untenable position: the charges move from here to there, but they need to accelerate and decelerate - hence there will be radiation.
The two-cap problem is a model breaker. One way or the other, something's gotta give.
Not sure I understand what you want to say.
If resistance is zero you still have inductance both in connection between the capacitors and capacitor plates. So current and thus the time it will take charges to move from one capacitor to another will be limited.
I do not see the problem with the two capacitors either with resistance or without.
The first electric field in between the two capacitor plates on the discharged capacitor will only happen when the first electron gets to one of the plates and it will not be that electron that jumped first the space between the switch contacts when they got close enough but it will be caused by that forming a cascading wave traveling at the speed of light through wire.
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So no energy is "radiated away" by in this case magnetic field as the electric field exists only between the capacitor plates but that also is not "radiated away"
Maybe in the world of trivial Spice simulations electric field exists only between capacitor’s plates, but not in the real world.
A hint: add a second ideal inductor close to that ideal inductor that you mentioned and check if you can measure any voltage across terminals of the new inductor during oscillations. Then think about where does that voltage come from. The same question puzzled Michael Faraday about 191 years ago. He found the answer, and kids now learn the Faraday’s law in the 8th grade (well, at least in the country where I went to middle school half a century ago).
Another hint (since you are a Spice fan): by the words “close” and “ideal” I mean coupling parameter’s value 1 in K statement.
PS. Dave ought to add more educational videos to his channel. If people did not study electromagnetic induction in school, maybe Dave can close that gap too…
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Maybe in the world of trivial Spice simulations electric field exists only between capacitor’s plates, but not in the real world.
A hint: add a second ideal inductor close to that ideal inductor that you mentioned and check if you can measure any voltage across terminals of the new inductor during oscillations. Then think about where does that voltage come from. The same question puzzled Michael Faraday about 191 years ago. He found the answer, and kids now learn the Faraday’s law in the 8th grade (well, at least in the country where I went to middle school half a century ago).
Another hint (since you are a Spice fan): by the words “close” and “ideal” I mean coupling parameter’s value 1 in K statement.
PS. Dave ought to add more educational videos to his channel. If people did not study electromagnetic induction in school, maybe Dave can close that gap too…
There will be a voltage drop across the inductor even if it had no resistance but the inductor has series resistance as parameter in spice and for the example with 47mH inductor that was set at 0.3Ohm. Capacitors and switch also have series resistance included in the simulation.
I'm not a spice fan. Spice is just a tool that if used correctly will provide correct results.
The spice simulations I showed get the same results as the experimental tests.
There is a reason a transmission line in spice is simulated as a series of LC elements as that is the best approximation of what happens and it is confirmed by the results that are not in contradiction to experimental results including the one Derek made.
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There will be a voltage drop across the inductor even if it had no resistance but the inductor has series resistance as parameter in spice and for the example with 47mH inductor that was set at 0.3Ohm. Capacitors and switch also have series resistance included in the simulation.
I'm not a spice fan. Spice is just a tool that if used correctly will provide correct results.
The spice simulations I showed get the same results as the experimental tests.
I probably needed to express myself more clear. I asked you to create an ideal transformer by adding the second inductor.
Anyway, you are wrong. Changing magnetic field of the inductor induces changing electric field that in its turn induces changing magnetic field, and so on. And all that radiates into the Universe at a speed of light…
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Maybe in the world of trivial Spice simulations electric field exists only between capacitor’s plates, but not in the real world.
A hint: add a second ideal inductor close to that ideal inductor that you mentioned and check if you can measure any voltage across terminals of the new inductor during oscillations. Then think about where does that voltage come from. The same question puzzled Michael Faraday about 191 years ago. He found the answer, and kids now learn the Faraday’s law in the 8th grade (well, at least in the country where I went to middle school half a century ago).
Another hint (since you are a Spice fan): by the words “close” and “ideal” I mean coupling parameter’s value 1 in K statement.
PS. Dave ought to add more educational videos to his channel. If people did not study electromagnetic induction in school, maybe Dave can close that gap too…
There will be a voltage drop across the inductor even if it had no resistance but the inductor has series resistance as parameter in spice and for the example with 47mH inductor that was set at 0.3Ohm. Capacitors and switch also have series resistance included in the simulation.
I'm not a spice fan. Spice is just a tool that if used correctly will provide correct results.
The spice simulations I showed get the same results as the experimental tests.
There is a reason a transmission line in spice is simulated as a series of LC elements as that is the best approximation of what happens and it is confirmed by the results that are not in contradiction to experimental results including the one Derek made.
If you watch long enough you will see get the general pattern.
- Nobody is allowed an electric field, except in the dielectric of capacitors
- Nobody is allowed a magnetic field, except in for in inductors or transformers
- Nobody is allowed to have magnetic or electric interaction between wires, but transmission lines are allowed
... leading the result that if you deny any existence of the electric and magnetic fields and their ability to actually transfer the energy, then it is a fait accompli that energy flows only in the wires.
The argument from electrodacus all along is that Lumped Element model the "level 0" description of electrical reality, and the underlying physics (e.g. electro magnetic field simulations) are misguided, do not reflect reality, and can be ignored, because against intuition they show that the energy does not flow in the conductors.
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There is an interesting thought experiment which was hinted at earlier in the thread I think.
You are allowed to observe two wires floating in free space, which stretch out of sight in both directions. There is nothing near either wire except you and any instruments you care to have about your person.
Each wire is identical in all physical respects, and each wire is carrying exactly 1 amp DC (which you can verify by measuring the magnetic field, or by measuring the voltage drop along a short length of the wire, or by measuring the heat being radiated).
One wire is transmitting 10 W and one wire is transmitting 1000 W. You are required to determine, without interfering with the wires in any way, which one is which. Can it be done?
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I probably needed to express myself more clear. I asked you to create an ideal transformer by adding the second inductor.
Anyway, you are wrong. Changing magnetic field of the inductor induces changing electric field that in its turn induces changing magnetic field, and so on. And all that radiates into the Universe at a speed of light…
Are you trying to say magnetic energy going in to creating the magnetic field around a wire or inductor (same thing) will be energy lost ?
Because that is not according to any evidence.
That magnetic field is conservative so when you disconnect the circuit all that energy will be put back in to the circuit.
It seems you are not the only one having this incorrect view about an inductor/wire.
If you looked at any of my simulations maybe the most relevant is that simulating a transmission line you will see that all energy is accounted for.
The one where switch is turned ON for just 30ns is particularly relevant for your concern.
[attach=1]
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If you watch long enough you will see get the general pattern.
- Nobody is allowed an electric field, except in the dielectric of capacitors
- Nobody is allowed a magnetic field, except in for in inductors or transformers
- Nobody is allowed to have magnetic or electric interaction between wires, but transmission lines are allowed
... leading the result that if you deny any existence of the electric and magnetic fields and their ability to actually transfer the energy, then it is a fait accompli that energy flows only in the wires.
The argument from electrodacus all along is that Lumped Element model the "level 0" description of electrical reality, and the underlying physics (e.g. electro magnetic field simulations) are misguided, do not reflect reality, and can be ignored, because against intuition they show that the energy does not flow in the conductors.
The lumped element model will not have been used if it did not provide accurate predictions of what happens. And it is accurate because it represents reality just a reduced form of that in terms of resolution.
Where is the electric field inside a charged capacitor transferring energy ?
The switch itself is also a capacitor and there is an electric field before closing the switch so how come you need to actually close the switch to transfer energy.
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Not sure I understand what you want to say.
If resistance is zero you still have inductance...
No, I was talking of a circuit where the only dissipation mechanism is that through Joule loss. The reasoning is that the loss is the same irregardless of the value of R, so all dissipation is accounted for when R= 1 ohm, or 0.1 ohm, or 0.00001 ohm or... and then we take the limit for R->0 and say that the loss is still the same - so it is all accounted for by ohmic losses even when R->0.
The first paper in my list above (the paper that was posted earlier by someone else here) avoid this problem by computing the radiation loss contribute considering NOT and oscillating LC circuit, but simply the radiation associated with the accelerated charges. It shows that for a circuit with IIRC a diameter of 10cm but still negligible self-inductance, the radiation loss becomes relevant only when the resistance of the loop falls under a handful of microohm (I don't have it at hand now, but the numbers are in that ballpark).
So, ohmic loss till a certain value of resistance, but under that it's radiation that takes over. At first they share the losses, then radiation becomes dominant and account for nearly all losses.
The current does not oscillate: the solution is a decaying exponential - and with a more advanced model - there are also solutions where the current dies off in a finite time (paper 2 or 3 in the list above).
No more paradox of the missing energy (it is extracted from the circuit)
No more paradox of ohmic loss with R=0 (it is taken care of by radiation)
No more need for an inductance that makes the circuit oscillate at a frequency 1/Sqrt[L Ceq].
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One wire is transmitting 10 W and one wire is transmitting 1000 W. You are required to determine, without interfering with the wires in any way, which one is which. Can it be done?
You will need to clarify what you mean it transmits 10W and 1000W. To where and in what form ?
I'm assuming you mean this are two isolated systems each with its own source but then where is the return wire ?
As you mentioned wires are identical and both cary 1A the difference will be the electric field between those wires and their respective return wires.
As far as wires are concerned the loss as heat will be the same IR
There may be no load just one wire 100x longer than the other one and so a higher voltage supply in order to be able to push 1A through a higher resistance wire (higher resistance due to extra length).
This is how my house heating system is designed as I have loops of 80m of 18AWG wire connected to about 30V DC and so the wires are the heating elements embedded in the concrete floor witch acts as thermal mass storage.
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No, I was talking of a circuit where the only dissipation mechanism is that through Joule loss. The reasoning is that the loss is the same irregardless of the value of R, so all dissipation is accounted for when R= 1 ohm, or 0.1 ohm, or 0.00001 ohm or... and then we take the limit for R->0 and say that the loss is still the same - so it is all accounted for by ohmic losses even when R->0.
The first paper in my list above (the paper that was posted earlier by someone else here) avoid this problem by computing the radiation loss contribute considering NOT and oscillating LC circuit, but simply the radiation associated with the accelerated charges. It shows that for a circuit with IIRC a diameter of 10cm but still negligible self-inductance, the radiation loss becomes relevant only when the resistance of the loop falls under a handful of microohm (I don't have it at hand now, but the numbers are in that ballpark).
So, ohmic loss till a certain value of resistance, but under that it's radiation that takes over. At first they share the losses, then radiation becomes dominant and account for nearly all losses.
The current does not oscillate: the solution is a decaying exponential - and with a more advanced model - there are also solutions where the current dies off in a finite time (paper 2 or 3 in the list above).
No more paradox of the missing energy (it is extracted from the circuit)
No more paradox of ohmic loss with R=0 (it is taken care of by radiation)
No more need for an inductance that makes the circuit oscillate at a frequency 1/Sqrt[L Ceq].
I'm not aware of such an effect and it will not make sense as it will be a discontinuity for the zero ohm case (superconductors) where a current induced in a superconductor ring will flow forever with no loss so magnetic field around the ring and the current is maintained with no losses of any type.
But say that what you claim is true how will this change the fact that energy transfer is through wires ? Derek's example had a 1 or 1.1KOhm resistor as the load so total circuit resistance was fairly high so even with your claim energy will still travel through wires.
If energy did not travel through wires then wires will not have been needed or at least not a closed loop.
The loss is not the same if you introduce an inductor as an intermediary energy storage device.
The inductor is ideal for this job as when you apply a voltage across the current will start slowly to increase as energy is stored in the magnetic field created.
So if you connect a 0.1Ohm resistor across a 3V capacitor the current will be 3/0.1 = 30A * 3V = 90W of power loss as heat while an inductor with same resistance 0.1Ohm will just slowly increase current as it stores that energy and not waste it as heat then you just connect the charged inductor across a discharged capacitor and transfer all that stored energy to that capacitor.
You can get 90% efficiency with two identical capacitors and inductor vs just 50% when paralleling the capacitors directly or through an additional resistor.
When you look at the numbers for both just capacitors or capacitors and inductor all energy is accounted for nothing lost through radiation.
So I will like to see what equation have you used where radiated energy loss is present in an isolated system.
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One wire is transmitting 10 W and one wire is transmitting 1000 W. You are required to determine, without interfering with the wires in any way, which one is which. Can it be done?
You will need to clarify what you mean it transmits 10W and 1000W. To where and in what form ?
Electrical power, from a source, to a destination.
I'm assuming you mean this are two isolated systems each with its own source but then where is the return wire ?
The return path is far away and out of sight.
As you mentioned wires are identical and both cary 1A the difference will be the electric field between those wires and their respective return wires.
If you look at the electric field, this is outside the wire. But you say the all the power is carried inside the wire, so doesn't that mean the electric field outside the wire is not important? If it is not important, why do you wish to consider it?
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One wire is transmitting 10 W and one wire is transmitting 1000 W. You are required to determine, without interfering with the wires in any way, which one is which. Can it be done?
You will need to clarify what you mean it transmits 10W and 1000W. To where and in what form ?
I'm assuming you mean this are two isolated systems each with its own source but then where is the return wire ?
As you mentioned wires are identical and both cary 1A the difference will be the electric field between those wires and their respective return wires.
As far as wires are concerned the loss as heat will be the same IR
There may be no load just one wire 100x longer than the other one and so a higher voltage supply in order to be able to push 1A through a higher resistance wire (higher resistance due to extra length).
This is how my house heating system is designed as I have loops of 80m of 18AWG wire connected to about 30V DC and so the wires are the heating elements embedded in the concrete floor witch acts as thermal mass storage.
I assume IanB is talking about something like this schematic attached (of course this is one possible hypothetical arrangement, in your test cell you can only see the two wires).
If you want to pedantic about the resistive loss in the wires, at the source the voltages are adjusted until 1A is flowing, so the 10W and 1000W is being delivered to the loads.
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There is an interesting thought experiment which was hinted at earlier in the thread I think.
You are allowed to observe two wires floating in free space, which stretch out of sight in both directions. There is nothing near either wire except you and any instruments you care to have about your person.
Each wire is identical in all physical respects, and each wire is carrying exactly 1 amp DC (which you can verify by measuring the magnetic field, or by measuring the voltage drop along a short length of the wire, or by measuring the heat being radiated).
One wire is transmitting 10 W and one wire is transmitting 1000 W. You are required to determine, without interfering with the wires in any way, which one is which. Can it be done?
For they were, all of them, deceived, as another wire was forged, one to rule them all, and at infinity, bind them [read: circulate their currents].
;D
Tim
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If you look at the electric field, this is outside the wire. But you say the all the power is carried inside the wire, so doesn't that mean the electric field outside the wire is not important? If it is not important, why do you wish to consider it?
Electric field will be directional towards the return wire so if you knew where the return wire is and at what distance you will be able to make a distinction between the two.
If as you say you have no clue where the return wires are and how far you will not be able to distinguish between the two super long wires.
It is irrelevant as far as wires are concerned if they transport 10W or 1000W as both transport 1A and are identical same resistance per unit of length.
So you can have say a 10Vdc 1A trough say a 10km long wire loop and 1000Vdc 1A through 1000km long loop but all you see is maybe 1km or so of wire from each and there will be no way to distinguish them.
And yes all energy is carried inside the wire in this particular example it will be 1W per km of wire all lost as heat.
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The pathological nature of the Two Capacitor Paradox problem leads to the observation that only the Law of Conservation of Charge is satisfied, but the Law of Conservation of Energy is violated.
Law of conservation of energy has never been broken/violated. There is no paradox related to the two parallel capacitors so that page on Wikipedia is misinformation as anyone can write a wiki page and write whatever it wants.
The paradox exists if only the DC steady state is considered and not the AC transient state.
Did you even looked closely at the equations you posted from wikipedia to see how absurd and stupid they are.
Did you notice my accidental typo?
If those two capacitors are identical (same capacity) the voltage after switch is closed will be 0.707 * Vi
How did you calculate 0.707*Vi for the voltage across the capacitors? Either you were refering to the Vrms of the sinusoidal voltages due to the oscillation (AC conditions) or you were refering to the DC steady state as 0.707*Vi. Vrms is 0.707*Vpeak or 0.707*Vi in this case. If Vi was 3Vdc then the Vrms would be 2.121Vrms with a steady state 1.5Vdc (or 0.5*Vi) across both capacitors. Under these conditions, I think that both the Law of Conservation of Charge and Energy would be satisfied without any violation.
Quoting the Wikipedia article (https://en.m.wikipedia.org/wiki/Two_capacitor_paradox#Solutions):
If the wires are assumed to have inductance but no resistance, the current will not be infinite, but the circuit still does not have any energy dissipating components, so it will not settle to a steady state, as assumed in the description. It will constitute an LC circuit with no damping, so the charge will oscillate perpetually back and forth between the two capacitors; the voltage on the two capacitors and the current will vary sinusoidally. None of the initial energy will be lost, at any point the sum of the energy in the two capacitors and the energy stored in the magnetic field around the wires will equal the initial energy.
Here is a post from a few pages back in this thread:
(https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/?action=dlattach;attach=1477765;image)
From the left, the voltage steps down corresponding to the switch turning on.
Subsequently, the capacitor voltages (Ch1, Ch3) oscillate as predicted.
If those two capacitors are identical (same capacity) the voltage after switch is closed will be 0.707 * Vi
If you were refering only to the DC steady state voltage of 0.707*Vi then I think only the Law of Conservation of Charge would be satisfied.
For DC steady state only, \$E_i\$ is the initial energy (before the switch is closed) and \$E_f\$ is the final energy (after the switch is closed).
\$E_i = \frac{1}{2}CV_i^2\$
\$E_f = \frac{1}{2}CV_f^2 + \frac{1}{2}CV_f^2 = CV_f^2\$
\$E_f = E_i\$
\$CV_f^2 = \frac{1}{2}CV_i^2\$
\$V_f^2 = \frac{1}{2}V_i^2\$
\$V_f = \frac{1}{\sqrt{2}}V_i \approx 0.707V_i\$
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Are you trying to say magnetic energy going in to creating the magnetic field around a wire or inductor (same thing) will be energy lost ?
Because that is not according to any evidence.
Yes, exactly. Changing EM field carries away energy. Marconi and Popov were the first to demonstrate that in early 20th century.
In 1970s Russians had to build Chernobyl power plant just to power a single Duga radar (huge step up from Popov’s transmitter).
Today I charge my phone and watch wirelessly.
Need more examples?
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The paradox exists if only the DC steady state is considered and not the AC transient state.
What is the paradox at DC steady state ?
How did you calculate 0.707*Vi for the voltage across the capacitors? Either you were refering to the Vrms of the sinusoidal voltages due to the oscillation (AC conditions) or you were refering to the DC steady state as 0.707*Vi. Vrms is 0.707*Vpeak or 0.707*Vi in this case. If Vi was 3Vdc then the Vrms would be 2.121Vrms with a steady state 1.5Vdc (or 0.5*Vi) across both capacitors. Under these conditions, I think that both the Law of Conservation of Charge and Energy would be satisfied without any violation.
Energy since that is what we are discussing here stored in a capacitor is 0.5 * C * V2
So initial energy before closing the switch
0.5 * 1 * 32 = 4.5Ws
If none of the energy was lost as heat then you will expect the same 4.5Ws split between the two capacitors so a 2F capacitor
0.5 * 2 * 2.1212 = 4.5Ws
With just 1.5V at the end of the test half of the energy was lost as heat in the wires due to wire/conductor resistance and the capacitor plates are also wires/conductors.
0.5 * 2 * 1.52 = 2.24Ws just half of the energy you started with.
If you were refering only to the DC steady state voltage of 0.707*Vi then I think only the Law of Conservation of Charge would be satisfied.
For DC steady state only, \$E_i\$ is the initial energy (before the switch is closed) and \$E_f\$ is the final energy (after the switch is closed).
\$E_i = \frac{1}{2}CV_i^2\$
\$E_f = \frac{1}{2}CV_f^2 + \frac{1}{2}CV_f^2 = CV_f^2\$
\$E_f = E_i\$
\$CV_f^2 = \frac{1}{2}CV_i^2\$
\$V_f^2 = \frac{1}{2}V_i^2\$
\$V_f = \frac{1}{\sqrt{2}}V_i \approx 0.707V_i\$
Final energy includes the energy stored in the two capacitors in this case 2.25Ws and the energy dissipated as heat in the wire resistance in this case the remaining 2.25Ws so there is no paradox and all is accounted for.
If you look at few posts back you will see that I showed close to 90% energy transfer efficiency by just adding an inductor to the circuit as intermediary storage device so final voltage was around 2V in both capacitors and just a smaller part of the energy ended as heat in the wires.
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Yes, exactly. Changing EM field carries away energy. Marconi and Popov were the first to demonstrate that in early 20th century.
In 1970s Russians had to build Chernobyl power plant just to power a single Duga radar (huge step up from Popov’s transmitter).
Today I charge my phone and watch wirelessly.
Need more examples?
You do not understand how both of those examples work that is why you think the energy is lost.
They are not isolated systems.
The phone is an inductive charger basically a transformer so it works for very short distance a few mm and very inefficiently compared to wires directly.
If you checked the spice simulation I have done with the extra inductor and diode you will see that energy was stored in the magnetic field around the inductor and all of that was retrieved back to charge the empty capacitor and the only losses about 10 or 11% of the total where due to wire resistance and that diode and none of it lost due to that magnetic field.
You can do the real test and you will get the same results.
If half of the energy was lost as radiated energy in the two capacitor test then how come I reduced that loss from 50% to just 11% by adding the inductor and diode (they do not contain any energy at the start of the test and do not collect energy from outside).
With just the capacitors starting with 3V on the charged one and ending with 1.5V in both half the energy is lost as heat in wires (can be both measured and calculated so no mystery) then you add two extra components the inductor and diode and you end up with about 2V in each capacitor so close to 90% of the initial energy and just 10% was lost again all in the wires/conductors and the diode which is a semiconductor.
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You do not understand how both of those examples work that is why you think the energy is lost.
They are not isolated systems.
The three examples are not isolated systems - they draw power from outside (Duga from Chernobyl station, wireless charge from power grid) :)
Now, are you saying that a radio transmitter does not radiate EM energy, unless there is a receiver?
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Right so there are plenty of problems with S=JV but you are forbidden to discuss them.
Of course there's a conspiracy of mainstream scientists whose aim is to silence any dissenters.
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You do not understand how both of those examples work that is why you think the energy is lost.
They are not isolated systems.
The three examples are not isolated systems - they draw power from outside (Duga from Chernobyl station, wireless charge from power grid) :)
Now, are you saying that a radio transmitter does not radiate EM energy, unless there is a receiver?
We are discussing if energy travels through wires or outside the wires.
For example the phone charger where you have two inductors. Say you put 1Ws of energy into transmitter coil from a source then energy will travel through wire/inductor and say 10% is lost as heat due to IR and half creates a magnetic field around it (all conserved) then the receiver inductor can supply part of that stored magnetic field to a load so say 50% of that magnetic field energy is collected by this inductor and again about 10% of that is lost as heat IR in this inductor the rest is delivered to load. Then the rest of the magnetic field can be retrieved back to the sender coil and put back in to the source or just wasted as heat.
It is basically an inefficient air transformer as the magnetic coupling is not as great as if the inductors share a common high magnetic field conductor core.
In a transformer you also put some energy in with each half wave in the primary and if the secondary is open circuit all that magnetic field will be returned to primary thus the loss will only be in the wire and a bit in the iron core.
But again main question on this thread is if energy is delivered through wires to the lamp/resistor or outside the wire.
Since electrical energy is electrical power integrated over time and electrical power is the product of electrical potential and electrical current and the current can only travel through the wire it means energy is delivered through the wire.
There is absolutely zero evidence that "energy doesn't travel in wire" main claim Derek makes based only on the small current observed at the lamp before the electron wave had the time to travel through wire. That small current is just due to line capacitance being charged so the equivalent of two capacitors on each side of the lamp that are being charged by the battery so since current can not travel through 1m of air that is the dielectric of this capacitors (not at 20V) then no energy travels outside the wire. Not during transient and not after during steady state.
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But again main question on this thread is if energy is delivered through wires to the lamp/resistor or outside the wire.
Since electrical energy is electrical power integrated over time and electrical power is the product of electrical potential and electrical current and the current can only travel through the wire it means energy is delivered through the wire.
There is absolutely zero evidence that "energy doesn't travel in wire" main claim Derek makes based only on the small current observed at the lamp before the electron wave had the time to travel through wire. That small current is just due to line capacitance being charged so the equivalent of two capacitors on each side of the lamp that are being charged by the battery so since current can not travel through 1m of air that is the dielectric of this capacitors (not at 20V) then no energy travels outside the wire. Not during transient and not after during steady state.
The small current observed at the lamp before the electron wave had the time to travel through the wire was energy emitted by the lamp. That energy had to come from the battery (there is no other power source), and it didn't travel through the wire (because there wasn't time), so it traveled through the air.
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The small current observed at the lamp before the electron wave had the time to travel through the wire was energy emitted by the lamp. That energy had to come from the battery (there is no other power source), and it didn't travel through the wire (because there wasn't time), so it traveled through the air.
You are ignoring the capacitors formed by the transmission line wires one on each side of the lamp.
When you are charging two capacitors in series from a source you will not say that energy passed through capacitors but energy went in the capacitor and was stored. While energy was pushed into capacitors it traveled through the wires. You can can not have electric current through a dielectric in this case 1m of air and with no current you have no energy.
This graph I made to show power at the supply and power at the lamp if understood will explain everything.
The switch is only closed for 30ns then open and stays open
With green is power in mW delivered by the supply and with magenta is power used by the lamp.
The area under the graphs represents the energy.
Notice that energy from supply was delivered only during those 30ns that switch was closed circuit and as soon as circuit was broken no more energy is supplied by the source. The total energy delivered by the source in those initial 30ns is 9.16nJ as you can see in the grey window on the left last value.
The lamp was still getting energy well after the switch was closed. Most of the energy arrives at the lamp after the switch was closed and first large chunk exactly after the energy had the time to travel the entire distance through wire.
In the end 7.68nJ of energy where delivered to the lamp and the difference is all accounted by energy lost as heat in the wire's
There is no energy radiated away all energy is accounted for and all will eventually be radiated as photons form the lamp and infrared photons from the wire as heat is lost to environment.
[attach=1]
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You are ignoring the capacitors formed by the transmission line wires one on each side of the lamp.
While in theory there would be some pumping via line capacitance, are you really saying that the amount shown in the experiment is solely due to the capacitance of two wires 1m apart? Seriously?
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Right so there are plenty of problems with S=JV but you are forbidden to discuss them.
Of course there's a conspiracy of mainstream scientists whose aim is to silence any dissenters.
A good thing Derek is there to stop it. Youtubers shall prevail over mainstream scientists!
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You are ignoring the capacitors formed by the transmission line wires one on each side of the lamp.
While in theory there would be some pumping via line capacitance, are you really saying that the amount shown in the experiment is solely due to the capacitance of two wires 1m apart? Seriously?
Yes that is exactly what I'm saying and the simulation for that showed the exact same result. I did rough estimate of the line capacitance based on two pipes 2.5cm diameter 1 meter apart and so values I used in the simulation should be fairly equivalent to real test Derek made and there result where also very similar.
You need to remember that we are talking about 1Kohm resistor as the lamp/load and about 65ns
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Wouldn't such a small capacitance be quickly discharged, so the scope should show a downward trend right after the leading edge? The time constant would be really really small, after all.
Edit: let's say the wire is 5mm across (actually a tube) and 5m long. The capacitance would be 0.21pf and the time constant 2.10e-12 aka 2.1femtoseconds.
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Wouldn't such a small capacitance be quickly discharged, so the scope should show a downward trend right after the leading edge? The time constant would be really really small, after all.
Edit: let's say the wire is 5mm across (actually a tube) and 5m long. The capacitance would be 0.21pf and the time constant 2.10e-12 aka 2.1femtoseconds.
It is a distributed capacitance and there is inductance involved so charging of each capacitor (seens a series of lumped components) will be delayed.
You can see the downward trend when the switch is open after was just 30ns closed but the wave and associated storage both as inductance and capacitance will travel all around the transmission line until it gets to lamp after about 65ns or so.
So the transmission line is not just a single capacitance and inductance plus resistance it is a continuous set of this.
I think I used 100 RLC elements for the simulation of a 10m transmission line so I divided the total line capacity in 100 small pcs then use that as a repeating circuit.
You where probably considering the capacitance of a single wire while that is fairly small in Dereks experiment you have the capacitance between two parallel wires that are 1m apart and I assumed 25mm diameter pipes (looked like 1" copper pipe).
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[As a board newbie] there is not much I can add to the original Veritasium debate other than voice my own opinion. Which is, all Veritasium did was to prove the concept of close coupled inductance and mutual capacitance (cross talk) by building the world's longest (hypothetical) dipole antenna /slash speaker cable. Nice thought experiment, but how does this prove the hypothesis in a universe owned by the EMC demons of EMF?
I ponder what would happen if, rather than being stretched out one light second apart, the cables are laid out serpentine fashion - like differential traces on a PCB? It's the same distance for the electron wave to propogate. So is the outcome the same if the round trip time is the same?
A not so hypothetical experiment is to stand under a high tension power line whilst holding a flourescent tube. Spooky stuff. Is this an example of "vampire capacitance?"
On this theme, [thanks to Dave's debunking videos :o ] I note there are free energy harvesting devices than can run entire hospitals from a single wifi signal. So maybe we don't know how electricity really travels? So then, why does MY electricy have to flow through a meter when, I can just run out a very long wire and close couple with the ether for free? :bullshit:
ERROR :palm:
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On this theme, [thanks to Dave's debunking videos :o ] I note there are free energy harvesting devices than can run entire hospitals from a single wifi signal. So maybe we don't know how electricity really travels? So then, why does MY electricy have to flow through a meter when, I can just run out a very long wire and close couple with the ether for free? :bullshit:
ERROR :palm:
When I was a young child in the late 80's I used to listen to a radio that was made only out of a speeked (the high impedance one from an old rotary dial phone) and a diode in parallel with that.
One end will be connected to any large metallic object that was not grounded like the water gutters on the house acting as the receive antenna and the other end connected to my body (a finger was enough) with me basically being capacitively connected to ground that was the return path.
This way I was able to listen to the closest AM (amplitude modulated) radio station with no batteries using energy from the radio station antenna.
The two antennas are just the plates of a capacitor and the ground is the return path for the loop.
There were stories that people leaving very close to AM transmit antenna will steal energy by capacitively coupling to the radio station antenna.
A bit later I build a similar radio but with an inductor and variable capacitor in order to be able to tune any of the close by AM stations and also then I powered a red LED in a similar way.
But this will not be correctly named wireless energy transfer as I was in a closed loop in series with a capacitor that was charged and discharged at AM frequency of the radio station.
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A good thing Derek is there to stop it. Youtubers shall prevail over mainstream scientists!
Unless he's colluded with them.
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[As a board newbie] there is not much I can add to the original Veritasium debate other than voice my own opinion. Which is, all Veritasium did was to prove the concept of close coupled inductance and mutual capacitance (cross talk) by building the world's longest (hypothetical) dipole antenna /slash speaker cable. Nice thought experiment, but how does this prove the hypothesis in a universe owned by the EMC demons of EMF?
Inductance, capacitance and "antennance" are just aspects of the same phenomenon.
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It is a distributed capacitance and there is inductance involved so charging of each capacitor (seens a series of lumped components) will be delayed.
Doesn't matter if it's distributed - the entire thing has a very very short time constant. All you're saying is that if you imagine it comprised 100 smaller capacitors, each will be 2.1ff (i.e. 2.1pf/100).
And... if there's a 18V pulse going in, how come there's only 5V coming out the other side?
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It is a distributed capacitance and there is inductance involved so charging of each capacitor (seens a series of lumped components) will be delayed.
Doesn't matter if it's distributed - the entire thing has a very very short time constant. All you're saying is that if you imagine it comprised 100 smaller capacitors, each will be 2.1ff (i.e. 2.1pf/100).
And... if there's a 18V pulse going in, how come there's only 5V coming out the other side?
You are looking at the capacitance of a single conductor. The way that Derek's experiment is setup the capacity is between two wires with 1m distance between them and 10m long in each direction.
So total capacitance will be around 42pf for the two 10m long parallel copper pipes and I have that split in to 100 so 0.42pf for each element.
Voltage on the resistor is proportional with the current passing through the resistor while this distributed capacitance charges so it forms a divider with the line impedance. And there are two lines on each side of the resistor.
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It is a distributed capacitance and there is inductance involved so charging of each capacitor (seens a series of lumped components) will be delayed.
Doesn't matter if it's distributed - the entire thing has a very very short time constant. All you're saying is that if you imagine it comprised 100 smaller capacitors, each will be 2.1ff (i.e. 2.1pf/100).
And... if there's a 18V pulse going in, how come there's only 5V coming out the other side?
Just for fun I divided capacitors like this, and the prompts thoughts that energy is stored in the dielectric between the plates. >:D
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Is the electric field really between the plates?
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Is the electric field really between the plates?
For a capacitor yes. The plates have large surface area and distance between plates is super small in comparison thus field lines are all between the plates.
With wires separated by 1m of air it still between the wires but a bit more "loose" if this is the correct word.
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Doesn't matter if it's distributed - the entire thing has a very very short time constant. All you're saying is that if you imagine it comprised 100 smaller capacitors, each will be 2.1ff (i.e. 2.1pf/100).
Ues this calculator https://www.emisoftware.com/calculator/wire-pair-capacitance/ (https://www.emisoftware.com/calculator/wire-pair-capacitance/)
You can use 12mm wire radius 1m separation and 10m length and you will get 62pF total
I also checked with 1mm radius for fire and got about 40pf so I used 42pf in my simulation just to be conservative.
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For the sake of argument I went with 60pF. Derek's result I've attached along with my conservative (that is, relatively slow rise time) simulation. They look completely different, and I'm not convinced that splitting the 1 cap into however many your simulator can put up with will change things significantly. Specifically, the simulation shows a peak of close to the input level and then dies off very quickly, even on this much shorter timescale than the real thing.
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Doesn't matter if it's distributed - the entire thing has a very very short time constant. All you're saying is that if you imagine it comprised 100 smaller capacitors, each will be 2.1ff (i.e. 2.1pf/100).
Ues this calculator https://www.emisoftware.com/calculator/wire-pair-capacitance/ (https://www.emisoftware.com/calculator/wire-pair-capacitance/)
You can use 12mm wire radius 1m separation and 10m length and you will get 62pF total
I also checked with 1mm radius for fire and got about 40pf so I used 42pf in my simulation just to be conservative.
A lossless transmission line looks like a resistor at its input, not a capacitor. Please read Haus and Melcher or any other good book.
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For the sake of argument I went with 60pF. Derek's result I've attached along with my conservative (that is, relatively slow rise time) simulation. They look completely different, and I'm not convinced that splitting the 1 cap into however many your simulator can put up with will change things significantly. Specifically, the simulation shows a peak of close to the input level and then dies off very quickly, even on this much shorter timescale than the real thing.
It makes a big difference when you split the capacitance and also inductance.
I guess you did not see my spice simulation that I posted earlier in the thread.
So here it is again. Keep in mind I simulated only half the circuit the other side of the resistor is directly connected to source negative not through another transmission line and I did not add the influence of oscilloscope probes.
The important part is that shape is exactly the same with a slow rise and then a flat response until about 65ns
[attach=1]
[attach=2]
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A lossless transmission line looks like a resistor at its input, not a capacitor. Please read Haus and Melcher or any other good book.
This is a real line and during transient the inductance and capacitance of the line play an important role as see from both experimental result and simulation.
During DC after the transient effect is gone it will be just a resistive line.
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Please read the book.
The experimental line will be close to a lossless line. The theory for lossy lines has also been worked out. Just read and you will learn a lot!
You are openly displaying your own ignorance.
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Or look at this note I posted a few months ago.
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Please read the book.
The experimental line will be close to a lossless line. The theory for lossy lines has also been worked out. Just read and you will learn a lot!
You are openly displaying your own ignorance.
I took a looked at the pdf you linked and it seems he only considered the impedance and resistance of the circuit and not the energy storage capacity.
I do not think the ignorance is on my part. The results of my simulation using proper lumped elements perfectly matches the waveform shape that Derek obtained with his real world test.
The results show that the initial current through the lamp/resistor is due to transmission line capacitance charging so energy flowing through the lamp/resistor is due to resistor being in series with line capacitance (an energy storage device) being charged.
It is the equivalent of charging a capacitor from a voltage source through a lamp/resistor. So during the initial connection the current flows into capacitor (not through it) and that is what is seen in those first 65ns in Derek's experiment as well as in my spice simulation.
After that initial transient phase when at steady state the line will act as a purely resistive line so energy is transferred from the source to the lamp/resistor through the wires and current going in to the lamp/resistor equals current going out of the source.
Voltage across the resistor will be slightly lower due to energy loss on the transmission wires.
The main point Derek tried to make with both videos is that energy flows from source to load outside the wire is completely incorrect not only for DC but for AC/transient as well.
You can consider the lamp and transmission line one and the same so you can imagine a long filament 10m in both directions then you can understand that all energy provided by the source is dissipated in that filament in an uniform way when in DC steady state and a bit uniform during the initial transient due to capacitive and inductive characteristics of the line which is now also the lamp/resistor.
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The main point Derek tried to make with both videos is that energy flows from source to load outside the wire is completely incorrect
There is no wire between the, uh, two wires. There is energy passing from one side to the other, and you say that's due to the capacitance of the wires. So, let's go with that - the inside of the capacitor is empty, so has not the energy transferred across inside of the capacitor but outside of the wires?
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The main point Derek tried to make with both videos is that energy flows from source to load outside the wire is completely incorrect
There is no wire between the, uh, two wires. There is energy passing from one side to the other, and you say that's due to the capacitance of the wires. So, let's go with that - the inside of the capacitor is empty, so has not the energy transferred across inside of the capacitor but outside of the wires?
The energy is not passing from one side to the other. Energy is being stored in the capacitor made by the parallel wires.
In order to transfer electrical energy you need a current and that 1m of air between wires is an almost perfect electrical insulator so no current flow no power and if there is no power there is no energy transfer but there is energy storage that everyone seems to want to ignore.
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The results show that the initial current through the lamp/resistor is due to transmission line capacitance charging so energy flowing through the lamp/resistor is due to resistor being in series with line capacitance (an energy storage device) being charged.
If current flows through the lamp/resistor while the capacitance is charging, then the lamp/resistor is consuming power and radiating it as heat and light. (Power = I2R) Where did that power come from?
It is the equivalent of charging a capacitor from a voltage source through a lamp/resistor. So during the initial connection the current flows into capacitor (not through it) and that is what is seen in those first 65ns in Derek's experiment as well as in my spice simulation.
Yes, and during that initial charging period the lamp lights up and consumes power. How did the power reach the lamp?
The main point Derek tried to make with both videos is that energy flows from source to load outside the wire is completely incorrect not only for DC but for AC/transient as well.
But in the transient the load receives power and therefore converts energy to heat and light. Where did that power come from during the transient?
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You could say the potential energy is qV so that there'll be two wires at different potentials connected to the resistor-lamp. So energy will flow from the wire into the lamp.
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If current flows through the lamp/resistor while the capacitance is charging, then the lamp/resistor is consuming power and radiating it as heat and light. (Power = I2R) Where did that power come from?
Yes, and during that initial charging period the lamp lights up and consumes power. How did the power reach the lamp?
But in the transient the load receives power and therefore converts energy to heat and light. Where did that power come from during the transient?
You have a source (say battery) and you are charging a capacitor from that.
How if the energy from the source transferred to the capacitor ? Through wires.
There is some energy loss on the wires that ends up as heat in the wire and then is radiated to outside world.
Now in series on one of this wires you add a lamp. All you did was making one of those wires a higher resistance as lamp is also a wire.
You are asking how the energy is delivered from source to lamp and answer is through the wires.
All energy used by the lamp flowed through wires and when capacitor is fully charged so same voltage as the source lamp can no longer glow as no energy flows through it (basically a wire).
I guess the confusion is mostly related to what a capacitor is and the fact that energy flowing in capacitor is stored/conserved not used to do any work.
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Yes, OK. So let's put a capacitor on both sides, and put the lamp between the two capacitors. Now there is an air gap completely separating the lamp from the battery. But still the lamp lights up?
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Yes, OK. So let's put a capacitor on both sides, and put the lamp between the two capacitors. Now there is an air gap completely separating the lamp from the battery. But still the lamp lights up?
We are going back to where we were in reply #266
electrodacus will say that the lamp (or in this case LEDs) are lit, but power doesn't flow through the capacitors, only in and out at the same time :-//
https://www.youtube.com/watch?v=5qdDqOwGomY (https://www.youtube.com/watch?v=5qdDqOwGomY)
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Yes, OK. So let's put a capacitor on both sides, and put the lamp between the two capacitors. Now there is an air gap completely separating the lamp from the battery. But still the lamp lights up?
There is no difference between one capacitor on one side of the lamp or capacitors on both sides.
An electron moved on one plate of a capacitor will push of an electron from the other plate and back in to battery with a single capacitor but if there are two capacitors in series then current will flow on the wire connecting the two capacitors as that one electron pushed off from the plate of first capacitor will on up on the plate of the other capacitor traveling through wire (is not the same electron that it will show up on the plate of the second capacitor but one displaced by that) then that electron on the second capacitor will push an electron in to the battery.
Again this current flow same as with just a capacitor will flow through wires only until capacitor/capacitors are charged. That energy stored in capacitor can be used latter to do work.
Like if you remove the battery and then connect a wire instead of the battery all energy that flowed into capacitors will now flow back out.
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We are going back to where we were in reply #266
electrodacus will say that the lamp (or in this case LEDs) are lit, but power doesn't flow through the capacitors, only in and out at the same time :-//
Not "in and out" just in and when capacitors are charged no current will flow other than for the small leakage depending on the quality of those capacitors.
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Er, hate to point this out but in the simplified circuit it goes: supply -> capacitor -> resistor -> return.
Now, the cap is not charged. It is 0V both sides. Connect the supply and now there is a voltage across the resistor. Where did that come from if not through the capacitor?
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Er, hate to point this out but in the simplified circuit it goes: supply -> capacitor -> resistor -> return.
Now, the cap is not charged. It is 0V both sides. Connect the supply and now there is a voltage across the resistor. Where did that come from if not through the capacitor?
In to capacitor not through capacitor.
You can not have a current "through" the dielectric.
You understand that when the capacitor is full may be 30ns or 30s but once the capacitor is fully charged (and I mean charged so it stores energy) current flow will stop. And current flows through wires but in to capacitor not through it.
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Er, hate to point this out but in the simplified circuit it goes: supply -> capacitor -> resistor -> return.
Now, the cap is not charged. It is 0V both sides. Connect the supply and now there is a voltage across the resistor. Where did that come from if not through the capacitor?
You can not have a current "through" the dielectric.
Demonstrations such as placing an ammeter on both legs of a capacitor as it charges, showing the same current flowing both terminals, at the same time are complete fabrications...
https://www.youtube.com/watch?v=V7Rc1rCI3H8 (https://www.youtube.com/watch?v=V7Rc1rCI3H8)
Unless it has to be the same electron flowing out as flowed in - in which case long wires are out due to the slow drift velocity. On a 1m x 2mm diameter copper wire, carrying 1A DC , it will take on average about 12 hours for an electron to travel down that wire.
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Something goes through to use up energy across that resistor (or lamp, or anything else that consumes power). Where does it come from? It is energy of some form, isn't it?
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Demonstrations such as placing an ammeter on both legs of a capacitor as it charges, showing the same current flowing both terminals, at the same time are complete fabrications...
Unless it has to be the same electron flowing out as flowed in - in which case long wires are out due to the slow drift velocity. On a 1m x 2mm diameter copper wire, carrying 1A DC , it will take on average about 12 hours for an electron to travel down that wire.
It is absolutely not the same electron not even the same electron after some delay that you find living the other plate of the capacitor.
On wires is not the electron that flows into the wire that gets on the other end at the speed of light but the electron that enters the wire will create a wave that pushes out an electron on the other end after about the time light needs to travel that distance.
Electron drift speed is a different subject.
So when you charge a capacitor from a battery none of the electrons entering one plate will be found on the other side.
The electrons leaving from the other plate where already on that plate and they leave in order for that plate to become positively charged.
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Something goes through to use up energy across that resistor (or lamp, or anything else that consumes power). Where does it come from? It is energy of some form, isn't it?
No energy will not go through but in to capacitor.
Those electrons that travel through the resistor (basically a wire) where never in the battery or the wire connected on the other side of the capacitor.
The analogy I can make is a compressed air cylinder that is split in two by an rubbery elastic membrane.
Discharged cylinder means the membrane is relaxed in the middle then when you push more air molecules in one of the sides same amount of air molecules will leave from the other side but they will not be the same as that elastic membrane (dielectric) will prevent any air molecules passing from one half of the cylinder to the other.
Electrons can not pass through dielectric so no current no energy flows through.
As I mentioned before a switch is also a capacitor where close circuit it means you shorted the two plates. As long as switch is open so just a capacitor no energy will flow through it.
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Something goes through to use up energy across that resistor (or lamp, or anything else that consumes power). Where does it come from? It is energy of some form, isn't it?
No energy will not go through but in to capacitor.
Well, hang on a minute. There is 4mA going through that resistor, enough to light up an LED. Energy must be involved there, no? Where did that come from?
The analogy I can make is a compressed air cylinder that is split in two by an rubbery elastic membrane.
In that case the air on one side pushes against the membrane and that pushes against whatever is on the other side. You've transferred energy from the pushing side to the pushed side, haven't you?
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Well, hang on a minute. There is 4mA going through that resistor, enough to light up an LED. Energy must be involved there, no? Where did that come from?
In that case the air on one side pushes against the membrane and that pushes against whatever is on the other side. You've transferred energy from the pushing side to the pushed side, haven't you?
Yes those 4mA are due to electrons leaving that plate of the capacitor (the discharged capacitor has equal amount of free electrons on each plate) so when you charge a capacitor you push electrons on to one plate while the same amount of electrons leave the other plate same as in that compressed air tube analogy air molecules enters one side and others are leaving from the other side.
So a discharged compressed air cylinder with two partitions will have air in it say 500 molecules one one side and 500 molecules on the other side.
Say you charged this discharged cylinder from another one that is charged but 10x larger that has 8000 in one side and 2000 on the other side.
Now with two hoses you connect the large one to the small one to charge it
You will get almost 800 on one side and more than 200 on the other side (to lazy to do the exact calculation) then you are left with about 7700 on the large one and about 2300 on the other side. You get the idea the pressure (voltage) will equalize.
After pressure has equalized there is no flow so no energy transfer. You can add some turbine (lamp) in between the two and it will work during transfer but stop when pressure is equalized.
As any analogy you can not go to far and still be correct but for the limited purpose of explaining why energy flows in or out of capacitors but not through should be good enough.
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Something goes through to use up energy across that resistor (or lamp, or anything else that consumes power). Where does it come from? It is energy of some form, isn't it?
No energy will not go through but in to capacitor.
Well, hang on a minute. There is 4mA going through that resistor, enough to light up an LED. Energy must be involved there, no? Where did that come from?
The analogy I can make is a compressed air cylinder that is split in two by an rubbery elastic membrane.
In that case the air on one side pushes against the membrane and that pushes against whatever is on the other side. You've transferred energy from the pushing side to the pushed side, haven't you?
Careful, you may encourage them to go full Humpty Dumpty mode on you:
'When I use a word,' Humpty Dumpty said, in rather a scornful tone, 'it means just what I choose it to mean — neither more nor less.'
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Well, hang on a minute. There is 4mA going through that resistor, enough to light up an LED. Energy must be involved there, no? Where did that come from?
In that case the air on one side pushes against the membrane and that pushes against whatever is on the other side. You've transferred energy from the pushing side to the pushed side, haven't you?
Yes those 4mA are due to electrons leaving that plate of the capacitor (the discharged capacitor has equal amount of free electrons on each plate) so when you charge a capacitor you push electrons on to one plate while the same amount of electrons leave the other plate same as in that compressed air tube analogy air molecules enters one side and others are leaving from the other side.
OK, I guess we are on the same page, then. The energy going into the capacitor on the plus side is pushing energy out of the capacitor on the negative side, right? In a simplistic manner of speaking.
So, back to our capacitor with 1m of free air in the middle... there is a net energy transfer from one side to the other. The PSU is pushing some energy into the cap which is consequently pushing the same amount of energy out to the resistor which is consuming it (aka converting to heat/light). The PSU has lost it, the resistor has used it, yes?
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Please look at Equation 18. Zo is real. It is a resistance and has no imaginary part. Note that the two transmission lines and the resistor form a resistive divider. Let's assume that the line is perfectly matched, i.e. R = 2Zo. Before the first reflection arrives back, 3/4 of the energy delivered by the battery is sorted in the two transmission lines. 1/4 is dissipated in the resistor. No energy is transferred from the transmission line to the resistor.
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Not sure why I did not get notifications but likely google AI trying to protect me from wasting more time :)
In any case you all seems to completely ignore energy storage is like that does not even exist for you.
It all gets down to you thinking energy pases trough a capacitor that will mean current passing through a capacitor instead on in to a capacitor as that energy is stored energy doing no work.
The wires connecting the capacitor to the supply will have a loss and you can add resistor or lamps in series with those and have even more loss but all the energy that will flow is to charge the capacitor not going through capacitor.
That is why with DC current will only flow for a short time as much as it is needed to charge the capacitor. If there is any current after that it will be super small and is the leakage current for that capacitor that is the equivalent for a high value resistor in parallel with the capacitor and has nothing to do with the results measured by Derek as that leakage current is so small it will not register on a 8 or even 12bit oscilloscope.
The definition of an electrical current is "a stream of charged particles like electrons or ions moving through an electrical conductor".
There are no electrons let alone ions traveling through that 1m of air between the lines.
If there is no electrical current then there is no electrical power and thus no energy traveling outside the wire.
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It all gets down to you thinking energy pases trough a capacitor...
So you're agreed that energy does indeed pass through a capacitor? That 1m gap without any wires is actually allowing energy transfer through it?
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It all gets down to you thinking energy pases trough a capacitor...
So you're agreed that energy does indeed pass through a capacitor? That 1m gap without any wires is actually allowing energy transfer through it?
No electrical energy is transferred through that 1m gap.
There are no electrons flying from one wire to the other wire that is 1m apart.
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It all gets down to you thinking energy pases trough a capacitor...
So you're agreed that energy does indeed pass through a capacitor? That 1m gap without any wires is actually allowing energy transfer through it?
No electrical energy is transferred through that 1m gap.
There are no electrons flying from one wire to the other wire that is 1m apart.
Then what is the resistor burning up?
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Then what is the resistor burning up?
The resistor is in series with the capacitor so is basically a high resistance wire connecting the capacitor to the source.
There are capacitors on both sides in Derek's experiment but it is no different is just electrons moving from plate of one capacitor to plate on the other capacitor so still a high resistance wire connecting the two capacitors in series.
Do you agree that no electrons travels through the 1m space from one wire to the other?
If you do then you must understand that no electrical energy travels through that 1m gap.
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If you do then you must understand that no electrical energy travels through that 1m gap.
If no electrical energy travels through the gap, what kind of energy does travel through the gap? What name will you give it?
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Then what is the resistor burning up?
The resistor is in series with the capacitor so is basically a high resistance wire connecting the capacitor to the source.
There are capacitors on both sides in Derek's experiment but it is no different is just electrons moving from plate of one capacitor to plate on the other capacitor so still a high resistance wire connecting the two capacitors in series.
Do you agree that no electrons travels through the 1m space from one wire to the other?
If you do then you must understand that no electrical energy travels through that 1m gap.
I agree that no electrons travel through the 1m space from one wire to the other, but it does not follow that no electrical energy travels through that 1m gap.
I am sure you agree that no electrons will pass through the two capacitors either side of the LED+Resistor. However, electrical energy does pass through the capacitors, and it does light the LED. That energy is coming from the DC supply!
https://www.youtube.com/watch?v=5qdDqOwGomY (https://www.youtube.com/watch?v=5qdDqOwGomY)
And as pointed out before, I can replace the LED+R with a third capacitor in series, and charge the middle capacitor from the DC supply, remove it form the circuit and measure it. It does have charge (and therefore energy), even though no electrons have moved through any of the capacitors.
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If you do then you must understand that no electrical energy travels through that 1m gap.
If no electrical energy travels through the gap, what kind of energy does travel through the gap? What name will you give it?
If you ignore a few infrared photons then there is no energy that travels through that gap.
All energy travels through wires.
Here is a good video about capacitors. He should have mentioned that the initial positive charge plate that he starts with is charged relative to ground
https://www.youtube.com/watch?v=HQPcOCuVG1I (https://www.youtube.com/watch?v=HQPcOCuVG1I)
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I agree that no electrons travel through the 1m space from one wire to the other, but it does not follow that no electrical energy travels through that 1m gap.
I am sure you agree that no electrons will pass through the two capacitors either side of the LED+Resistor. However, electrical energy does pass through the capacitors, and it does light the LED. That energy is coming from the DC supply!
And as pointed out before, I can replace the LED+R with a third capacitor in series, and charge the middle capacitor from the DC supply, remove it form the circuit and measure it. It does have charge (and therfore energy), even though no electrons have moved through any of the capacitors.
No enenrgy pases trough capacitors.
What you see as work done on the LED is the result of electrons moving from the plate of first capacitor to a plate on the second capacitor as the two capacitors in series are charged from the supply.
If you disconnect the battery and then short the wires that were before connected to battery you are discharging the capacitors so electrons will flow back between the two plates making them neutral again.
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No enenrgy pases trough capacitors.
What you see as work done on the LED
Since it takes energy to do work, and the energy did not pass through the capacitors, how did it get from the battery to the LED?
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Since it takes energy to do work, and the energy did not pass through the capacitors, how did it get from the battery to the LED?
The LED is in series with the capacitor/capacitors so that is wasted energy during charging process.
Is the same as waste energy in the wires as they carry the energy from the source to the capacitors (energy storage device) you just add an LED and thus extra loss to the process of charging the capacitor.
Electrical energy travels through wires and travels through LED/lamp/resistor but does not travels through capacitor but in capacitor.
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There are capacitors on both sides in Derek's experiment
Let's not complicate matters unnecessarily.
There is a power source, cap, resistor, return. That's it. We've decided the cap is the two wires 1m apart, and we've gone with your values for it. We know the resistor is converting energy to heat.
The only issue is that there is the 1m gap without wires in that circuit. What, exactly, is passing from the power supply side of the capacitor to the other side that lets the resistor heat up? It must be energy, mustn't it, since that's what everything ultimately reduces to.
So, if energy is not being transferred across that gap, what IS going across? That resistor isn't using nothing to heat up, you know.
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There are capacitors on both sides in Derek's experiment
Let's not complicate matters unnecessarily.
There is a power source, cap, resistor, return. That's it. We've decided the cap is the two wires 1m apart, and we've gone with your values for it. We know the resistor is converting energy to heat.
The only issue is that there is the 1m gap without wires in that circuit. What, exactly, is passing from the power supply side of the capacitor to the other side that lets the resistor heat up? It must be energy, mustn't it, since that's what everything ultimately reduces to.
So, if energy is not being transferred across that gap, what IS going across? That resistor isn't using nothing to heat up, you know.
Electrons from the source flow through wires in to one plate of the capacitor. Same number of free electrons leave the plate on the other side of the capacitor and return to the source.
All energy flows through wires in to the plates that are also wires and no energy flows through the capacitor dielectric in this case 1m of air.
The resistor is nothing more than a wire with higher resistance so energy flows through it same as it flows through wires.
if source is disconnected this charge imbalance remains there and if you short this wires then electrons from one plate can travel to the other plate that has a deficit of electrons in order for both plates to become neutral so you are discharging the capacitor and current will flow in reverse through wires.
Electrons (charged particles) are the ones that carry the energy and they do so through wires.
While air even 1m of it can become conductive the potential (voltage) will need to be way higher than 20V so no energy was transferred outside the wires in Derek's experiment.
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So let's stop talking about energy and start talking about power.
The battery is a source of power. The LED/resistor is a consumer of power. The LED lights up, so power was transferred from the battery to the LED. The capacitor is between the battery and the LED. What kind of power went through the capacitor on its journey from the battery to the LED?
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So let's stop talking about energy and start talking about power.
The battery is a source of power. The LED/resistor is a consumer of power. The LED lights up, so power was transferred from the battery to the LED. The capacitor is between the battery and the LED. What kind of power went through the capacitor on its journey from the battery to the LED?
Why it is hard to understand the concept of energy storage.
There is nothing going "through" capacitor but in to capacitor.
You need wires to transfer energy from the source to the capacitor and if you also add an LED to the wires it makes no much difference other than reduce the speed at with the energy is transferred to the capacitor.
Is sort of like you want to believe that electrical energy can pass through in order to deny that energy storage exist.
There was a good reason I replaced the energy source with a charged capacitor as I was thinking one way or the other you can understand that capacitor is an energy storage device.
Electrical power is product of electrical current and electrical potential. I think is clear that you can not have electric current flow through air at least not with 20V and 1m of distance so there can not be any power.
You only have an electric field inside the capacitor once you have an imbalance of charge and not the other way around.
All this are well known facts but thanks in part to Derek facts are no longer relevant if you can make absurd claims like "energy doesn't flow in wires" and back them with no evidence. Show a waveform that you do not understand and say here is the proof.
His channel will more appropriately be named mendacium even if he's intent is veritas.
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Show a waveform that you do not understand and say here is the proof.
Walking away from the bench with the middle capacitor of three in series, charged with stored energy is all the proof you should need.
If energy was only able to flow in wires, and capacitors only stored energy and prevented it from passing through, then that would be impossible.
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Walking away from the bench with the middle capacitor of three in series, charged with stored energy is all the proof you should need.
If energy was only able to flow in wires, and capacitors only stored energy and prevented it from passing through, then that would be impossible.
Proof for what ?
I explained that electrons from the plate of once capacitor will be pushed in the other capacitor plate through wire when you have multiple capacitors in series.
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I explained that electrons from the plate of once capacitor will be pushed in the other capacitor plate
Across 1m of air? What is it that the plus side can cause the negative side to rise to the same voltage and put out some current?
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Electrical power is product of electrical current and electrical potential. I think is clear that you can not have electric current flow through air at least not with 20V and 1m of distance so there can not be any power.
What happens when the electric field isn't conservative? Does current always involve electrons?
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I explained that electrons from the plate of once capacitor will be pushed in the other capacitor plate
Across 1m of air? What is it that the plus side can cause the negative side to rise to the same voltage and put out some current?
No electrons and thus no energy will travel that 1m gap.
When you have two capacitors in series connected to a supply each capacitor will have a plate connected to supply one to the positive and the other to the negative terminal of the supply.
Those plates will act as if it was a single capacitor then the other two plates that are connected with a wire between them will always have the same number of electrons when discharged and when charged is just that when charged electrons from one plate travels through the wire to the other plate as capacitors are charged.
If you have a lamp there in series there will just be extra energy wasted as capacitors charge since is like wire has higher resistance to electron flow.
Seen from the outside as a black box with just two wires the two capacitors in series will look like a single capacitor with lower capacity.
There is no difference between one capacitor or two or 3 capacitors in series.
From outside you will see that the amount of energy going in to black box will be equal with the amount going out plus the energy lost as heat on the DC ESR and the DC ESR includes the wire connecting the two capacitors in series inside the black box. If you add a lamp in that black box that will just further contribute to increase in the DC ESR and you will not be able to know form the outside that there is a lamp in the box or if there are two capacitors in series or just one.
You will not be able to know that from outside of the black box because no matter what measurements you make there will be no difference from a single capacitor or multiple in series and or lamps or resistors added to that in series.
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What happens when the electric field isn't conservative? Does current always involve electrons?
In an isolated system the field will be conservative. Electric current will always require some sort of charged particle either electrons or ions. In a wire the electrons will be the ones that move. In a battery the ions will move from one plate to another.
So to transport electrical energy from one plate to another you need electron flow usually in a wire but with very high voltages the electrons can also travel through air but not the case here with 20V and 1m of air.
So electric and magnetic field are conservative in Derek's experiment.
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No electrons and thus no energy will travel that 1m gap.
When you have two capacitors in series connected to a supply each capacitor will
Hey, please try not to introduce dead cats. Just stick with the ultra-simple circuit we were discussing: PSU->cap->res->return.
So, if no energy is passed from one side to the other, what is the resistor burning?
Hang on... O. M. G.... Maybe you are converting to aetherwind?
Joking aside, explain how the resistor can show a voltage across it and sink some current. Where is that coming from?
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""Electric current will always require some sort of charged particle either electrons or ions""
Yes but the charged particles need not "transfer" to transmit energy.
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Hey, please try not to introduce dead cats. Just stick with the ultra-simple circuit we were discussing: PSU->cap->res->return.
So, if no energy is passed from one side to the other, what is the resistor burning?
Hang on... O. M. G.... Maybe you are converting to aetherwind?
Joking aside, explain how the resistor can show a voltage across it and sink some current. Where is that coming from?
No energy passes from one side of the capacitor to the other. Energy is being stored in the capacitor.
Current will from from the source in to capacitor through wires and obviously through the resistor that is also a wire.
Current will decrease as the capacitor is charged until no current flows as capacitor is full.
Say capacitor when discharged start with 1000 free electrons on one plate and 1000 electrons on the other plate.
When battery is connected to capacitor if you connect just the positive nothing will happen same if you connect just the negative.
When battery is connected to capacitor on both sides electrons will flow from battery in to the wire and when that happen there will be an electron wave that will push out electrons on the other side of the wire in to the capacitor plate while at the same time electrons from the other plate will get out of the capacitor plate in to the wire and that will push electrons in to battery.
At the end when capacitor potential equal battery potential current flow stops.
At this point you may have 1100 electrons on one plate and 900 electrons on the opposite plate.
The 100 extra electrons are due to battery pushing those through the wire and the deficit of electrons on the other plate when trough wire in to battery.
No electrons have traveled through the capacitor dielectric.
Now the capacitor contains energy that can be used to do work. So you will say energy flowed in the capacitor where it is stored not through capacitor.
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""Electric current will always require some sort of charged particle either electrons or ions""
Yes but the charged particles need not "transfer" to transmit energy.
What do you mean by "transfer"
Charged particles as electrons repel each other thus if you have a region with higher electron density it will want to migrate to a region with lower density.
Since there is not enough energy for them to travel through air even for a few mm distance between switch contacts the contacts need to be close enough so that 20V in this case has enough energy to push an electron from the negatively charged region in to the region with deficit of electrons.
So what you likely call "transfer" has to do with repelling forces that negative charged particles feel relative to one another. That is not high enough at 20V to have electrons fly from the wire trough 1m of air to the other wire and for that you will need way higher voltage about 3400000V
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No energy passes from one side of the capacitor to the other. Energy is being stored in the capacitor.
Let's stop bringing up the energy stored in the capacitor. Everyone knows energy is stored in capacitors. Everyone agrees energy is stored in capacitors. Nobody has ever denied that energy is stored in capacitors. Stop acting like this a big mystery.
If we look at an energy balance on the system, we have:
1. The energy leaving the source (battery) is ES
2. The energy stored in the capacitor is EC
3. The energy consumed (dissipated) by the load resistor/LED is EL
Since you tell us that energy is always conserved, we know that ES = EC + EL
Clearly some energy from ES ended up as EL, which means some energy was transferred from the battery to the load, and some energy was stored in the capacitor.
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Lets take the issue to the instantaneous case.
So the usual battery , switch, capacitor, resistor all in series.
Lets say batter V = 12V
Capacitor is 1000 uf.
Resistor is 10 ohms.
Initially capacitor is discharged.
At the instant the switch is closed we have.
I = 120 ma
Resistor is dissipating power of 12*12/ 10 = 14.4 watts.
capacitor energy = ce = 0.5C * Vc^2
diff that by time
Dce/Dt = Vc
Vc is zero at the beginning. Capacitor is storing No energy.
In this instantaneous point would you say that the 14.4 watts the resistor is dissipating is traveling "through" the capacitor??
If not how is it getting there?
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Lets take the issue to the instantaneous case.
So the usual battery , switch, capacitor, resistor all in series.
Lets say batter V = 12V
Capacitor is 1000 uf.
Resistor is 10 ohms.
Initially capacitor is discharged.
At the instant the switch is closed we have.
I = 120 ma
Resistor is dissipating power of 12*12/ 10 = 14.4 watts.
capacitor energy = ce = 0.5C * Vc^2
diff that by time
Dce/Dt = Vc
Vc is zero at the beginning. Capacitor is storing No energy.
In this instantaneous point would you say that the 14.4 watts the resistor is dissipating is traveling "through" the capacitor??
If not how is it getting there?
As soon as an electron passes through the resistor an electron will also end up in the capacitor so voltage across the capacitor is no longer zero (it will be super close to zero but not zero).
There are capacitors everywhere not just that huge by comparison 1000uF capacitor. The wires and even resistor will for a capacitor with the return wire.
So if 1 million electrons (absolute random number but can be calculated exactly) end up stored in the capacitor then that number has passed through wires and the resistor witch is also a wire.
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Let's stop bringing up the energy stored in the capacitor. Everyone knows energy is stored in capacitors. Everyone agrees energy is stored in capacitors. Nobody has ever denied that energy is stored in capacitors. Stop acting like this a big mystery.
If we look at an energy balance on the system, we have:
1. The energy leaving the source (battery) is ES
2. The energy stored in the capacitor is EC
3. The energy consumed (dissipated) by the load resistor/LED is EL
Since you tell us that energy is always conserved, we know that ES = EC + EL
Clearly some energy from ES ended up as EL, which means some energy was transferred from the battery to the load, and some energy was stored in the capacitor.
You claim that everyone understand that energy is stored in the capacitor but if that was to be true there will have been no discussion about energy passing through capacitor. Including you based on this post you just made.
And is not me that is saying energy is conserved that is a fact that has never been disproven.
Yes energy was transferred only through wires and some of that energy ended up as lost energy in the wires and load resistor that is also a wire just with higher resistance.
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As soon as an electron passes through the resistor an electron will also end up in the capacitor
What causes that (and all the other) electrons to pass through the resistor just when voltage is applied to the capacitor and not at other times?
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As soon as an electron passes through the resistor an electron will also end up in the capacitor
What causes that (and all the other) electrons to pass through the resistor just when voltage is applied to the capacitor and not at other times?
Like I mentioned many times before. Electrons are charged particles with negative charge so they repel each other.
You can say they are under pressure if you compare this with gas molecules.
Say a discharged capacitor that is basically an energy storage device has 500 free electrons on both plates that means 0V so discharged capacitor and then say capacitor is charged so you have 750 electrons on one side and 250 on the other side. What will happen if capacitor terminal are shorted ?
The wire shorting also has free electrons so the electrons on the plate with 750 electrons when in contact with the wire (closed loop) will want to go to the lowest energy state so electrons will be pushed out of that plate in to the wire and the other end of the wire will release an electron in the capacitor plate that has a deficit thus 749 and 251 will be the total after the first electron did the work.
In the end plates will get back to 500 and 500 electrons so no more stored energy and all that stored energy will end up as heat in the wire (wire temperature will increase with the exact amount that energy stored contained).
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OK, forget it. It's a lost cause and I am out.
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Energy is conserved. At the instant the switch is closed the battery is providing 14.4 watts of power.
That 14.4 watts need to be accounted for everywhere. Assuming zero resistance wires. So for the capacitor we have Joules / second = watts and same for the resistor.
The capacitor is gaining zero energy at the instant the switch is closed.
The capacitor IS passing energy through it. Suppose the capacitor was infinite in capacitance. No energy would ever be stored in it but load current could flow into one plate an out of the other plate with no transfer across the dielectric.
Maybe we are saying the same thing.
What is happening is that the for the plates of the capacitor to "charge" relative to each other current must flow into and out of the plates but NOT across the dielectric barrier.
And instantaneously the energy balance at the start is that no energy is being stored in the capacitor. And somehow energy shows up at the resistor.
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OK, you're on.
Here is the waveform and the circuit that created it. You may have seen it before.
Now, the green line is the voltage applied to the circuit. Clearly, it is the cause of red line, which is the voltage across the resistor. (Or, alternatively, the voltage over the resistor - shown by the red line - is caused by the voltage applied to the circuit.)
There's the proof. Now, perhaps you can say just how that happens? How does the one cause the other?
You forgot to draw the source that is also connected to GND. Maybe that is what confuses you.
electrons from the source will flow in to resistor and in to the capacitor plate while on the other side that other capacitor plate will send the exact same number of electrons from that plate to the source.
Resistor is nothing else than a wire helping to complete the circuit. You need a closed loop to charge that capacitor and maybe because that is not seen in your schematic you are confused about how the energy is stored in to capacitor.
You also understand that red waveform also represents the current through the resistor and as you can see it will fast trend to zero meaning once the capacitor plate has enough electrons on one side with similar deficit on the other the capacitor is fully charged and at same voltage as the source thus no more energy transfer from the source to capacitor.
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Energy is conserved. At the instant the switch is closed the battery is providing 14.4 watts of power.
That 14.4 watts need to be accounted for everywhere. Assuming zero resistance wires. So for the capacitor we have Joules / second = watts and same for the resistor.
The capacitor is gaining zero energy at the instant the switch is closed.
The capacitor IS passing energy through it. Suppose the capacitor was infinite in capacitance. No energy would ever be stored in it but load current could flow into one plate an out of the other plate with no transfer across the dielectric.
Maybe we are saying the same thing.
What is happening is that the for the plates of the capacitor to "charge" relative to each other current must flow into and out of the plates but NOT across the dielectric barrier.
And instantaneously the energy balance at the start is that no energy is being stored in the capacitor. And somehow energy shows up at the resistor.
Exact moment that switch is closed the electron from the tip of the switch will jump to the other side on the wire so there will be no power through the resistor and no energy in the capacitor.
You will also need to consider the energy stored in wire inductance as wire has some inductance that opposes electron flow converting that in to magnetic field around the conductor.
At about the speed of light there will be an electron passing through resistor and arriving in the capacitor at witch point you can no longer say there is no energy delivered to capacitor.
Yes the initial part is super inefficient as say capacitor is at 0.001V and battery 12V that means 11.999V drops across the resistor and that means terrible charging efficiency with a small fraction of a percent efficiency and it will take the capacitor to get to 6V for the efficiency to get to 50% still horrible but not as crazy.
Even if you do not have the capacitor there just the resistor with just one terminal connect to battery the other terminal just not connected to anything and then on the other side of the battery a switch and a piece of wire again that wire not connected to anything due to the capacitance between the resistor and that not connected wire after the switch you will have some capacitance maybe a few pF but still enough so that when you close the switch energy will flow through the resistor for a very short period of time.
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OK, forget it. It's a lost cause and I am out.
It is a lost cause for you as you are the one that has no understanding of the subject.
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OK, forget it. It's a lost cause and I am out.
It is a lost cause for you as you are the one that has no understanding of the subject.
As a rough rule, would you say that capacitors allow voltage & current transients to pass through, and but do not pass steady voltages and currents? (deciding what is a transient and what is a steady voltage can be computed based on the Rs Cs & Ls in the circuit)
So at DC they have infinite resistance, and with signals of high enough frequency they offer very, very little resistance?
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The elementary features I was taught in the first week of electronics 101 were that you can't change the voltage across a capacitor instantaneously, and you can't change the current through an inductor instantaneously.
Analysis of most simple AC and pulse circuits can start from this, and then introduce the finite resistances and time constants.
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OK, forget it. It's a lost cause and I am out.
It is a lost cause for you as you are the one that has no understanding of the subject.
Yes, my lost cause is getting you to a) understand the question and b) answer that without going off on a tangent to confuse things.
Tell you what, just say which of these facts you think are not actually facts:
1. The supply provides energy to the circuit.
2. The resistor consumes energy in the circuit.
Now, everything and anything between those to is completely irrelevant. It doesn't matter if there are electrons, electons, pixies, BFO Tesla(tm) batteries, whatever. None of that matter a jot to those two facts.
3. There's a 1m air gap in that black box between the supply and resistor.
OK? Which of 1-3 do you think is not a fact?
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As a rough rule, would you say that capacitors allow voltage & current transients to pass through, and but do not pass steady voltages and currents? (deciding what is a transient and what is a steady voltage can be computed based on the Rs Cs & Ls in the circuit)
So at DC they have infinite resistance, and with signals of high enough frequency they offer very, very little resistance?
No it will be wrong to say that electrical current passes through a capacitor even as transient. What you think of as transient current passing through is actually current going in not passing trough.
With say AC you have current going in as capacitor is charged and going out as capacitor is discharged.
At no point there is any current going trough the capacitor thus no energy goes through capacitor.
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The elementary features I was taught in the first week of electronics 101 were that you can't change the voltage across a capacitor instantaneously, and you can't change the current through an inductor instantaneously.
Analysis of most simple AC and pulse circuits can start from this, and then introduce the finite resistances and time constants.
Yes that will be basic introduction to have some roule to memorize. But the reason voltage cannot change instantaneously across a capacitor is because energy will be going in the capacitor and stored.
Same thing with inductor where energy is also stored.
All energy that went in except for the amount that was lost in the process as heat will come out of this energy storage devices.
So there is no electrical enenergy that pases trough a capacitor.
Electrical energy is the integral of electrical power over time. To have power you need to have a current different from zero and you can not have current through a dielectric like plastic or air.
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Yes, my lost cause is getting you to a) understand the question and b) answer that without going off on a tangent to confuse things.
Tell you what, just say which of these facts you think are not actually facts:
1. The supply provides energy to the circuit.
2. The resistor consumes energy in the circuit.
Now, everything and anything between those to is completely irrelevant. It doesn't matter if there are electrons, electons, pixies, BFO Tesla(tm) batteries, whatever. None of that matter a jot to those two facts.
3. There's a 1m air gap in that black box between the supply and resistor.
OK? Which of 1-3 do you think is not a fact?
What you call me going on a tangent is essential in understanding what happens. Energy going trough a capacitor will involve electrons traveling through an electrical insulator like air and that is not the case.
1. Yes.
2. Yes.
3. OK
This part is wrong:
"Now, everything and anything between those to is completely irrelevant. It doesn't matter if there are electrons, electons, pixies, BFO Tesla(tm) batteries, whatever. None of that matter a jot to those two facts.
"
It does matter if there are electrons traveling through that 1m air gap as if they where then you and Derek will be right in saying that energy travels outside the wire but for that you need as I mentioned almost three and a half million volt so that air becomes a conductor.
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This part is wrong:
"Now, everything and anything between those to is completely irrelevant. It doesn't matter if there are electrons, electons, pixies, BFO Tesla(tm) batteries, whatever. None of that matter a jot to those two facts.
"
It does matter if there are electrons traveling through that 1m air gap as if they where then you and Derek will be right in saying that energy travels outside the wire but for that you need as I mentioned almost three and a half million volt so that air becomes a conductor.
No, it doesn't matter. We are not interested in the how at this point, only whether it does or not. If there are electrons magically jumping the gap that's fine. It's also fine if they don't. All we need to agree on is that energy is provided on one side, consumed on the other, and there is significant clear air between.
Are you happy agreed with that?
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The elementary features I was taught in the first week of electronics 101 were that you can't change the voltage across a capacitor instantaneously, and you can't change the current through an inductor instantaneously.
Analysis of most simple AC and pulse circuits can start from this, and then introduce the finite resistances and time constants.
Yes that will be basic introduction to have some roule to memorize. But the reason voltage cannot change instantaneously across a capacitor is because energy will be going in the capacitor and stored.
Same thing with inductor where energy is also stored.
All energy that went in except for the amount that was lost in the process as heat will come out of this energy storage devices.
So there is no electrical enenergy that pases trough a capacitor.
Electrical energy is the integral of electrical power over time. To have power you need to have a current different from zero and you can not have current through a dielectric like plastic or air.
Actually, the point behind the circuit rule that voltage across a capacitor cannot change instantaneously is that it would require infinite current. Similarly, infinite voltage for the inductor.
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No, it doesn't matter. We are not interested in the how at this point, only whether it does or not. If there are electrons magically jumping the gap that's fine. It's also fine if they don't. All we need to agree on is that energy is provided on one side, consumed on the other, and there is significant clear air between.
Are you happy agreed with that?
It does matter if you want to claim that energy flows outside the wires.
This is the main point of the entire discussion.
Electrons flow in to wire but do not jump that gap and since electron flow is electrical current and electrical current multiplied with voltage is power and power integrated over time is energy it means energy flows only in wire.
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No, it doesn't matter. We are not interested in the how at this point, only whether it does or not. If there are electrons magically jumping the gap that's fine. It's also fine if they don't. All we need to agree on is that energy is provided on one side, consumed on the other, and there is significant clear air between.
Are you happy agreed with that?
It does matter if you want to claim that energy flows outside the wires.
This is the main point of the entire discussion.
Electrons flow in to wire but do not jump that gap and since electron flow is electrical current and electrical current multiplied with voltage is power and power integrated over time is energy it means energy flows only in wire.
See, this is the problem. You just won't agree with a simple thing without going off on one about something else. Not even a simple fact that is indisputable - you just can't cope with saying, "Yes, OK, that is how it is".
Which of those facts are you disputing? The energy source, the energy sink, or the clear air? Presumably it's the air, but that's actually how it is! Are you going to say the video was photoshopped or something, and that your capacitor calculator (which you chose) is lying or something else?
So, is there or is there not a 1m air gap between the energy source and the energy consumer? Just a straight yes or no, please.
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I've been reading this discussion and I'm intrigued that pages and pages have gone by and the two words - "displacement current" haven't been mentioned (if they have I missed it!).
If one thinks currents only exist where charge flows, then one would be taking issue with Maxwells Eqs:
https://en.wikipedia.org/wiki/Displacement_current
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See, this is the problem. You just won't agree with a simple thing without going off on one about something else. Not even a simple fact that is indisputable - you just can't cope with saying, "Yes, OK, that is how it is".
Which of those facts are you disputing? The energy source, the energy sink, or the clear air? Presumably it's the air, but that's actually how it is! Are you going to say the video was photoshopped or something, and that your capacitor calculator (which you chose) is lying or something else?
So, is there or is there not a 1m air gap between the energy source and the energy consumer? Just a straight yes or no, please.
The problem is yours as you fail to acknowledge the real question.
Does the energy travels outside the wires ? so trough that 1m air gap.
If you answer with yes then you need to show how electrons travel from one wire to the other trough that 1m gap.
If no electrons travel through that 1m air gap then no energy travels through that gap.
So it is not me that needs to prove anything it is you that need to prove that electrons travel outside the wire in Derek's experiment.
The experiment was properly done and the results are not disputed here. What it is disputed is the wrong conclusion Derek got from those results.
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I've been reading this discussion and I'm intrigued that pages and pages have gone by and the two words - "displacement current" haven't been mentioned (if they have I missed it!).
If one thinks currents only exist where charge flows, then one would be taking issue with Maxwells Eqs:
https://en.wikipedia.org/wiki/Displacement_current
You will need to understand Maxwell's equations to understand what they say. They do not say energy flows outside the wires.
Electrons accumulate in to the wires (capacitor plates) and energy is stored that way. The higher the capacitance the more electrons you can push on to plates (conductors/wires).
If you disconnect the battery the potential difference will stay there as stored energy and if you short the two plates electrons will flow through wires from one plate to another discharging the capacitor while energy will end up as heat in the wire as that is the space energy traveled through the wire and not the air space between the plates.
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Electrons flow in to wire but do not jump that gap and since electron flow is electrical current and electrical current multiplied with voltage is power and power integrated over time is energy it means energy flows only in wire.
By your own logic, if a wire has 1A flowing it, but almost zero voltage drop across the wire, then integrating the power over time for that wire proves that there is minimal energy in that wire.
If you do the same calculation for the resistor you get reliable results, so why not for a wire? What is so special about it?
And where along the transition between "resistor" and "wire" does this specialness happen? at what resistance or current does your own math become invalid?
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By your own logic, if a wire has 1A flowing it, but almost zero voltage drop across the wire, then integrating the power over time for that wire proves that there is minimal energy in that wire.
If you do the same calculation for the resistor you get reliable results, so why not for a wire? What is so special about it?
And where along the transition between "resistor" and "wire" does this specialness happen? at what resistance or current does your math become invalid?
A wire is a resistor with lower resistance so there is no difference.
Not quite sure you know what you are asking. Can you be more exact maybe give a proper example with values.
Yes 1A drop on a 0.1Ohm resistor will result in a 0.1V drop across the resistor/wire and thus 0.1W of power lost on the wire as heat.
Same 1A on a 1kOhm wire/resistor will result in 1000V drop thus 1000W of power loss on that wire/resistor.
Where do you see any difference or problem between the two examples ?
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By your own logic, if a wire has 1A flowing it, but almost zero voltage drop across the wire, then integrating the power over time for that wire proves that there is minimal energy in that wire.
If you do the same calculation for the resistor you get reliable results, so why not for a wire? What is so special about it?
And where along the transition between "resistor" and "wire" does this specialness happen? at what resistance or current does your math become invalid?
A wire is a resistor with lower resistance so there is no difference.
Not quite sure you know what you are asking. Can you be more exact maybe give a proper example with values.
Yes 1A drop on a 0.1Ohm resistor will result in a 0.1V drop across the resistor/wire and thus 0.1W of power lost on the wire as heat.
Same 1A on a 1kOhm wire/resistor will result in 1000V drop thus 1000W of power loss on that wire/resistor.
Where do you see any difference or problem between the two examples ?
None, as you say
electrical current multiplied with voltage is power
Sure, a good wire has 1A flowing in it, and 0V measured across either end, so has 0W
and power integrated over time is energy
Sure, let's integrate 0W for as long as we want.... it's 0 J.
it means energy flows only in wire.
This doesn't follow - the result of your calculation is 0 J.
Well, following your own calculations, there was 0J in that wire.
If it was a 1 ohm resistor with 1A
"electrical current multiplied with voltage is power" - 1A at 1V is 1 W
" and power integrated over time is energy" - let's integrate 1W for a second.... it's 1 J.
Yep, 1J as expected. That checks out.
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Sure, a good wire has 1A flowing in it, and 0V measured across either end, so has 0W
Sure, let's integrate 0W for as long as we want.... it's 0 J.
:) good wire ? You mean superconductor.
Yes you can have 1A induced in a superconductor ring and have that 1A flowing forever in there with no energy loss.
A good wire that is not a superconductor will have a resistance low enough that loss is not to high at 1A so that wire will not melt and it is acceptable in your application.
So a resistor is a wire and vice versa thus absolutely no difference between the two.
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See, this is the problem. You just won't agree with a simple thing without going off on one about something else. Not even a simple fact that is indisputable - you just can't cope with saying, "Yes, OK, that is how it is".
Which of those facts are you disputing? The energy source, the energy sink, or the clear air? Presumably it's the air, but that's actually how it is! Are you going to say the video was photoshopped or something, and that your capacitor calculator (which you chose) is lying or something else?
So, is there or is there not a 1m air gap between the energy source and the energy consumer? Just a straight yes or no, please.
The problem is yours as you fail to acknowledge the real question.
Does the energy travels outside the wires ? so trough that 1m air gap.
If you answer with yes then you need to show how electrons travel from one wire to the other trough that 1m gap.
If no electrons travel through that 1m air gap then no energy travels through that gap.
So it is not me that needs to prove anything it is you that need to prove that electrons travel outside the wire in Derek's experiment.
The experiment was properly done and the results are not disputed here. What it is disputed is the wrong conclusion Derek got from those results.
Just one more time: is there or is there not an air gap between the energy source and the energy consumer?
That is the question I am asking. I am not asking the question you would prefer to think I should ask, I am asking this specific and very simple one. Are you afraid to answer it or what? Well, no matter, because this is why we don't get anywhere, and it is why I am done with you.
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Just one more time: is there or is there not an air gap between the energy source and the energy consumer?
That is the question I am asking. I am not asking the question you would prefer to think I should ask, I am asking this specific and very simple one. Are you afraid to answer it or what? Well, no matter, because this is why we don't get anywhere, and it is why I am done with you.
Yes there is a an air gap.
Now you answer this simple question. Is there any electric current passing through this gap ?
If you say no then you also say no to there is no energy delivered through that gap and all energy travels in wire.
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Sure, a good wire has 1A flowing in it, and 0V measured across either end, so has 0W
Sure, let's integrate 0W for as long as we want.... it's 0 J.
:) good wire ? You mean superconductor.
Yes you can have 1A induced in a superconductor ring and have that 1A flowing forever in there with no energy loss.
A good wire that is not a superconductor will have a resistance low enough that loss is not to high at 1A so that wire will not melt and it is acceptable in your application.
So a resistor is a wire and vice versa thus absolutely no difference between the two.
And yet when you say a wire has 0V measured between its ends you insist that it has energy flowing inside it. You wouldn't say that of a resistor!
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And yet when you say a wire has 0V measured between its ends you insist that it has energy flowing inside it. You wouldn't say that of a resistor!
Yes energy can travel through a superconductor and there will be no losses. It is a resistor with zero resistance to current flow.
While superconductors seems to be a theoretical ideal case they exist.
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And yet when you say a wire has 0V measured between its ends you insist that it has energy flowing inside it. You wouldn't say that of a resistor!
Yes energy can travel through a superconductor and there will be no losses. It is a resistor with zero resistance to current flow.
While superconductors seems to be a theoretical ideal case they exist.
Why do you keep bringing up superconductors? :-//
What's wrong with copper?
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Why do you keep bringing up superconductors? :-//
What's wrong with copper?
??? You did that when you mentioned a wire with zero resistance.
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Why do you keep bringing up superconductors? :-//
What's wrong with copper?
??? You did that when you mentioned a wire with zero resistance.
Well let's pick 1m of 2mm copper wire. That's about about 0.005 ohms.
By your math...
...and since electron flow is electrical current and electrical current multiplied with voltage is power and power integrated over time is energy it means energy flows only in wire.
"electrical current multiplied with voltage is power" - 1A at 0.005V is 0.005 W
" and power integrated over time is energy" - let's integrate 1W for a second.... it's 0.005 J.
So for a 1A in that 1m of 2mm copper wire (a voltage drop of 0.005V), your math calculates that only 0.005 J of energy flowed in that wire during that second.
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Well let's pick 1m of 2mm copper wire. That's about about 0.005 ohms.
By your math...
"electrical current multiplied with voltage is power" - 1A at 0.005V is 0.005 W
" and power integrated over time is energy" - let's integrate 1W for a second.... it's 0.005 J.
So for a 1A in that 1m of 2mm copper wire (a voltage drop of 0.005V), your math calculates that only 0.005 J of energy flowed in that wire during that second.
So ? Do you see any problems with that ?
Yes one meter of wire with just 5mOhm resistance will have a voltage drop of just 5mV at 1A.
That 5mJ is the energy that got lost as heat in that wire.
Assuming you had 5V supply the entire circuit should have had 5Ohm for 1A to flow through it so that 1m of wire is just a part of the total circuit.
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Just one more time: is there or is there not an air gap between the energy source and the energy consumer?
That is the question I am asking. I am not asking the question you would prefer to think I should ask, I am asking this specific and very simple one.[
Now you answer this simple question. Is there any electric current passing through this gap ?
If you say no then you also say no to there is no energy delivered through that gap and all energy travels in wire.
So what is you answer? Yes or No?
Are you sure you're an Engineer and not a Politician?
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What happens when the electric field isn't conservative? Does current always involve electrons?
In an isolated system the field will be conservative. Electric current will always require some sort of charged particle either electrons or ions. In a wire the electrons will be the ones that move. In a battery the ions will move from one plate to another.
So to transport electrical energy from one plate to another you need electron flow usually in a wire but with very high voltages the electrons can also travel through air but not the case here with 20V and 1m of air.
So electric and magnetic field are conservative in Derek's experiment.
Wrong and wrong. Do you even know what a conservative field is? Have you heard of displacement current?
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Just one more time: is there or is there not an air gap between the energy source and the energy consumer?
That is the question I am asking. I am not asking the question you would prefer to think I should ask, I am asking this specific and very simple one. Are you afraid to answer it or what? Well, no matter, because this is why we don't get anywhere, and it is why I am done with you.
Yes there is a an air gap.
Now you answer this simple question. Is there any electric current passing through this gap ?
If you say no then you also say no to there is no energy delivered through that gap and all energy travels in wire.
Thank you!
Your question: I don't know. That's what we are ultimately trying to determine, but at this point all we know is that there is an energy transfer. What form that energy takes is something to be figured out.
Hopefully you now agree there is energy transfer across that gap. If not, how can one side insert some and the other side use some? We are still not talking about how it transfers, only that there is a transfer.
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Assuming you had 5V supply the entire circuit should have had 5Ohm for 1A to flow through it so that 1m of wire is just a part of the total circuit.
So you will agree for the attached schematic, that the instant that SW1 is closed, the initial voltage over R1 will be 5V and 1A will be flowing through it? (and of course it will fall away as the capacitors charge)
And by integrating the V*A over time in the resistor we can find out how much energy has been transferred from the battery to the resistor, because of the connecting wires, switch and capacitors?
How does that energy get there, as you say energy only flows in wires and not through a capacitor?
...and since electron flow is electrical current and electrical current multiplied with voltage is power and power integrated over time is energy it means energy flows only in wire.
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You will need to understand Maxwell's equations to understand what they say. They do not say energy flows outside the wires.
They absolutely say that. That's what led Maxwell to unify the electric and magnetic fields into electromagnetism... and it's what led Hertz to try transmitting an electrical signal from one radiator to another without any wires whatsoever.
This statement of yours earlier in the thread,
So to transport electrical energy from one plate to another you need electron flow
is absolutely wrong. You seem to believe the only kind of current that exists is conduction current and that only this current transmits energy and the only way for energy to get from one plate to another is if an electron, somehow, crosses the gap. No. No. NO!
Do you know the difference between conduction current and displacement current?
You can start here:
https://youtu.be/SS4tcajTsW8?t=1157
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Now you answer this simple question. Is there any electric current passing through this gap ?
If you say no then you also say no to there is no energy delivered through that gap and all energy travels in wire.
Don't anyone fall for this question. It's a trick question. He wants you to believe that if conduction current isn't flowing through the gap then no energy is flowing through the gap.
This is against Ampere's Law with Maxwell's addition of Displacement Current. This statement would get an F in any basic physics course.
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Sorry be posting 3 comments in a row but enough nonsense has been perpetrated here about conduction current being the only way that energy can move from A to B in space. Here is a very simple experiment measuring displacement current as equal to the conduction current but there are NO electrons moving across any gaps whatsoever:
https://www.youtube.com/watch?v=EqufEpWaKXw (https://www.youtube.com/watch?v=EqufEpWaKXw)
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Sorry be posting 3 comments in a row but enough nonsense has been perpetrated here about conduction current being the only way that energy can move from A to B in space. Here is a very simple experiment measuring displacement current as equal to the conduction current but there are NO electrons moving across any gaps whatsoever:
https://www.youtube.com/watch?v=EqufEpWaKXw (https://www.youtube.com/watch?v=EqufEpWaKXw)
Interesting.
1. I see that the induced wave on the left of the capacitor was stronger than on the right of the capacitor.
2. And i suppose that if he removed the right half of the capacitor including its wire connection then he would have gotten an even stronger wave on the left part of the circuit (stronger than with the right side in place).
3. Here (1) & (2) support my new (elekton) elekticity.
4. The electricity on the rhs of the capacitor is an (induced) electron electricity.
5. On the lhs its an elekton elekticity.
6. Tween the plates there is no electricity, no current, there is a radio signal, ie em radiation, which induces an electron electric current on the rhs.
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And i suppose that if he removed the right half of the capacitor including its wire connection then he would have gotten an even stronger wave on the left part of the circuit (stronger than with the right side in place).
Easy to suppose, but unless you can show that then "Here (1) & (2) support my new (electon) electricity" is just building on quicksand.
Hell, we can all do that. I suppose that if he removed the right side the left side would have doubled. No, tripled! Now just let me make up a theory to account for that...
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And i suppose that if he removed the right half of the capacitor including its wire connection then he would have gotten an even stronger wave on the left part of the circuit (stronger than with the right side in place).
Easy to suppose, but unless you can show that then "Here (1) & (2) support my new (electon) electricity" is just building on quicksand.
Hell, we can all do that. I suppose that if he removed the right side the left side would have doubled. No, tripled! Now just let me make up a theory to account for that...
Not so fast. We already have some data. The lhs was stronger than the rhs.
How does old electron electricity explain that?
How was there more change (in current) on the lhs?
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Yes, I saw that. What I'm disputing is that you can extrapolate and say you would see anything with the right side missing. Maybe you would, but that wasn't shown and we can't tell from the experiment.
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About the capacitor. Here is what happens when the switch is closed.
Charge accumulates on one plate of the capacitor. The electric field created by this charge pushes some charge OUT of the other plate of the capacitor and through the load.
Transferring energy TO the resistor.
Who can dispute this?
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Thank you!
Your question: I don't know. That's what we are ultimately trying to determine, but at this point all we know is that there is an energy transfer. What form that energy takes is something to be figured out.
Hopefully you now agree there is energy transfer across that gap. If not, how can one side insert some and the other side use some? We are still not talking about how it transfers, only that there is a transfer.
You are trying to determine that as for me is very clear. The energy is not transferred across the gap it is transferred in to capacitor so charge particles accumulate on one wire / plate and equal amount of charges leave the other wire / plate.
Hope you agree that no electrons jump that gap. Since current is defined as flow of electric charged particles and since there are no electrons or ions traveling through that gap there is no energy transfer through that gap.
In order to charge a capacitor you will need to move charge particles (electrons) to and from plates (you move electrons to one plate and simultaneously remove electrons from the other). The electrons you remove from that other plate are there when capacitor is charged and do not come from the other side.
This was the reason I chose to use the two parallel capacitors example as there you have nothing else just capacitors one it is charged and the other is not so you have two gaps in the loop and no electrons travel across any of the two gaps yet there is an electric current in the wire so electrons move from the plate of one capacitor to the plate of the other capacitor through the wire when switch is closed.
On the charged capacitor there is excess of electrons on one plate and deficit of electrons on the other plate.
When you connect the other identical but discharged capacitor in parallel electrons will flow through the wire from one capacitor to the other.
So from the plate with excess electrons on the charged capacitor the electrons will travel to the plate with neutral charge on the discharged capacitor and at the same time electrons from the neutral charge on the discharged capacitors will move to the plate with deficit of electrons on the charged capacitor.
So if you were not to look at what happens to electrons in the capacitor plates you will think that those electrons must have jumped the gap but that is not the case.
In a circuit with resistance energy will be lost when part of the charge is moved from one capacitor to another.
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How does that energy get there, as you say energy only flows in wires and not through a capacitor?
See the replay #680 that I just made. No energy flows through capacitor. All electrons that move through the resistor where already on that side of the gap so electrons that were already on the capacitor plate connected to one side of the resistor will move through the resistor to the other capacitor plate also connected with wires to resistor.
You will end up with a plate that has a deficit of electrons and one that has an excess of electrons.
If you now open the switch and remove the battery that charge imbalance will remain as it is stored energy in the capacitors and so if you connect a wire instead of the battery and close the switch current will flow in the opposite direction powered by energy that you stored in the capacitors and so the electrons will not travel from the plate with excess electrons to the one with deficit until both plates have equal amounts of electrons so they are neutral and capacitors are discharged.
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Thank you!
Your question: I don't know. That's what we are ultimately trying to determine, but at this point all we know is that there is an energy transfer. What form that energy takes is something to be figured out.
Hopefully you now agree there is energy transfer across that gap. If not, how can one side insert some and the other side use some? We are still not talking about how it transfers, only that there is a transfer.
You are trying to determine that as for me is very clear. The energy is not transferred across the gap it is transferred in to capacitor so charge particles accumulate on one wire / plate and equal amount of charges leave the other wire / plate.
You're jumping ahead. Regardless of the how, energy is either transferred across or it is not. Didn't we agree that there is insertion of energy on one side and extraction of same on the other? So therefore energy MUST have been transferred. The only thing in doubt is the exact mechanism.
Are you now disputing this already agreed fact (that there is a transfer of energy from PSU to resistor)?
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Sorry be posting 3 comments in a row but enough nonsense has been perpetrated here about conduction current being the only way that energy can move from A to B in space. Here is a very simple experiment measuring displacement current as equal to the conduction current but there are NO electrons moving across any gaps whatsoever:
Sorry but that is an absolute crap video and the guy that made the video same as you has no idea what displacement current actually means.
He can not measure a current through the gap with that coil because as he correctly mentioned in the video there are no electrons jumping the gap and the only reason he sees something is because coil is huge and the gap is super small so there is current flowing in the wires and plates very close to that hughe coil.
There is no electrical energy traveling through that gap.
You can not have exceptions to the rule so no electron flow through the gap no electrical current and thus no energy.
That energy transferred to the plates stay there as stored energy that is why after the capacitor is charged there is no longer a current flow.
Current will flow into the plates and remain there. If you move the plates closer then some extra current will flow from the battery to capacitor as the capacity has now increased and if you move the capacitor plates further apart current from capacitor will flow back in to battery as capacity decreased and so there is excess charge.
By moving capacitor plates back and forth electrons will move from battery to capacitor and vice versa and since wires have some resistance energy will be wasted every time capacitor plates are moved until you can discharge the entire battery.
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1. I see that the induced wave on the left of the capacitor was stronger than on the right of the capacitor.
2. And i suppose that if he removed the right half of the capacitor including its wire connection then he would have gotten an even stronger wave on the left part of the circuit (stronger than with the right side in place).
3. Here (1) & (2) support my new (electon) electricity.
4. The electricity on the rhs of the capacitor is an (induced) electron electricity.
5. On the lhs its an electon electricity.
6. Tween the plates there is no electricity, no current, there is a radio signal, ie em radiaton, which induces an electron electric current on the rhs.
That is a crap experiment and explanation.
There is inductive coupling between his coil and capacitor circuit as the coil is connected to common GND through the oscilloscope that is why he will see a stronger signal when he gets closer to the positive wire than the negative one.
There is nothing worth seeing in that video.
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You're jumping ahead. Regardless of the how, energy is either transferred across or it is not. Didn't we agree that there is insertion of energy on one side and extraction of same on the other? So therefore energy MUST have been transferred. The only thing in doubt is the exact mechanism.
Are you now disputing this already agreed fact (that there is a transfer of energy from PSU to resistor)?
Energy is stored not transferred. Can you just not imagine energy being stored ?
Even after you disconnect the battery there will still be energy stored in there and if you close the circuit by connecting a wire in the place of the battery current from capacitor will flow now in the opposite direction and all that stored energy will be doing work while capacitor is discharged.
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Since current is defined as flow of electric charged particles and since there are no electrons or ions traveling through that gap there is no energy transfer through that gap.
If you say that there is no energy transfer across an air gap, then you must say that radio does not work, that wi-fi does not work, that cellphones do not work, that wireless charging does not work, that television does not work, in fact the whole modern world does not work. How can this be?
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Energy is stored not transferred. Can you just not imagine energy being stored ?
Some energy was observed to go from the source to the load, so some energy was transferred by definition. Some energy was transferred, some energy was stored.
The transferred energy cannot be the same as the stored energy, or there would be something for nothing. In fact, you would have invented a free energy device. Do you wish to claim free energy?
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If you say that there is no energy transfer across an air gap, then you must say that radio does not work, that wi-fi does not work, that cellphones do not work, that wireless charging does not work, that television does not work, in fact the whole modern world does not work. How can this be?
All those work just fine but not the way you think.
The transmit and receive antenna are each the plate of a large gap capacitor and the circuit is closed through the ground (earth) as the common conductor.
You are charging and discharging that capacitor thus you see a current on both sides as loop is closed through earth.
You can have one transmit antenna and multiple receive antenna is like having a capacitor with one plate on one side and multiple plates on the other side.
If you made communication using lasers then yes energy will be transferred from transmitter to receiver as you are sending photons.
Also with lasers you do not need the return like ground as you get the photons.
There are no particles exchanged between the two plates thus no energy passes through that gap.
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Some energy was observed to go from the source to the load, so some energy was transferred by definition. Some energy was transferred, some energy was stored.
The transferred energy cannot be the same as the stored energy, or there would be something for nothing. In fact, you would have invented a free energy device. Do you wish to claim free energy?
The load in this case is just a conductor. If your main goal was to charge the capacitor then you will call this energy as lost.
There is absolutely no energy that is not accounted for. You seems to be the ones that try to ignore stored energy.
When you charge a capacitor from a source the energy provided by the source will be equal with stored energy and energy loss on the wire.
When you discharge that stored energy you get the same loss again with current flow in the opposite direction.
You will have a problem adding some extra energy that crosses the gap as there is nothing more than stored and lost on the wire coming from the source.
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There are no particles exchanged between the two plates thus no energy passes through that gap.
If no energy passes through that gap then radio does not work, wi-fi does not work, cellphones do not work, wireless charging does not work, television does not work. All of those require energy to cross the gap.
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The load in this case is just a conductor. If your main goal was to charge the capacitor then you will call this energy as lost.
There is absolutely no energy that is not accounted for. You seems to be the ones that try to ignore stored energy.
When you charge a capacitor from a source the energy provided by the source will be equal with stored energy and energy loss on the wire.
When you discharge that stored energy you get the same loss again with current flow in the opposite direction.
You will have a problem adding some extra energy that crosses the gap as there is nothing more than stored and lost on the wire coming from the source.
All of these comments apply to charge, not energy.
If you talk about electrical charge, then everything is fine.
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There are no particles exchanged between the two plates thus no energy passes through that gap.
If no energy passes through that gap then radio does not work, wi-fi does not work, cellphones do not work, wireless charging does not work, television does not work. All of those require energy to cross the gap.
I already answered this question a few minutes ago. They work you do not understand how.
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All of these comments apply to charge, not energy.
If you talk about electrical charge, then everything is fine.
You don't quite understand what energy is and in this particular case electrical energy.
a) Is electrical energy the integral over time of electrical power ?
b) Is electrical power the product of electrical potential and electrical current?
c) Is electrical current defined as a flow of electrical charged particles? In this particular cases we are discussing flow of electrons.
Let me know if you answer as No on any of the 3 questions above.
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What about series compensation of power lines? Isn't that energy being transferred through capacitors?
How do you start a single-phase induction motor without a capacitor? Another example of energy being transferred through a capacitor.
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a) Is electrical energy the integral over time of electrical power ?
b) Is electrical power the product of electrical potential and electrical current?
c) Is electrical current defined as a flow of electrical charged particles? In this particular cases we are discussing flow of electrons.
Let me know if you answer as No on any of the 3 questions above.
a) Not always
b) Not always
c) Not always
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a) Not always
b) Not always
c) Not always
??? OK give an example. I need to admit I did not expected this answer. Waiting for your clarification.
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You're jumping ahead. Regardless of the how, energy is either transferred across or it is not. Didn't we agree that there is insertion of energy on one side and extraction of same on the other? So therefore energy MUST have been transferred. The only thing in doubt is the exact mechanism.
Are you now disputing this already agreed fact (that there is a transfer of energy from PSU to resistor)?
Energy is stored not transferred. Can you just not imagine energy being stored ?
Even after you disconnect the battery there will still be energy stored in there and if you close the circuit by connecting a wire in the place of the battery current from capacitor will flow now in the opposite direction and all that stored energy will be doing work while capacitor is discharged.
I don't dispute that a capacitor can store energy. As can a battery, and many other things. But that isn't relevant here - what is relevant is that there is a input on one side and an output on the other. The bits inbetween are a black box, and all we know is that there is no continuous transfer, but there is some.
So, are you disputing that the energy input on one side causes the transfer that allows consumption on the other side?
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I don't dispute that a capacitor can store energy. As can a battery, and many other things. But that isn't relevant here - what is relevant is that there is a input on one side and an output on the other. The bits inbetween are a black box, and all we know is that there is no continuous transfer, but there is some.
So, are you disputing that the energy input on one side causes the transfer that allows consumption on the other side?
You need to have a circuit so obviously that as many electrons will exit on one side of the battery as they enter the other side else there is just no energy transfer.
You can not just connect one terminal of the battery say the negative to a circuit and expect electrons to flow. You need a complete loop for that.
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I don't dispute that a capacitor can store energy. As can a battery, and many other things. But that isn't relevant here - what is relevant is that there is a input on one side and an output on the other. The bits inbetween are a black box, and all we know is that there is no continuous transfer, but there is some.
So, are you disputing that the energy input on one side causes the transfer that allows consumption on the other side?
You need to have a circuit so obviously that as many electrons will exit on one side of the battery as they enter the other side else there is just no energy transfer.
You can not just connect one terminal of the battery say the negative to a circuit and expect electrons to flow. You need a complete loop for that.
So you are saying that what we see in that circuit is false? That the input energy from the PSU does not cause the resistor to consume energy?
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"The transmit and receive antenna are each the plate of a large gap capacitor and the circuit is closed through the ground (earth) as the common conductor."
absolutely incorrect. You know we receive radio waves from space right ?
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So you are saying that what we see in that circuit is false? That the input energy from the PSU does not cause the resistor to consume energy?
What I'm saying is that you can not connect just one end of a resistor to one terminal of the battery and expect a current except for the one needed to charge the newly created capacitor that is made up by the other battery terminal and the resistor with air being the dielectric.
It is about how current flows in to a capacitor when being charged and not through a capacitor.
Say you have an ideal voltage source 12V so there is no internal series resistance considered for this ideal voltage source.
Then say you have two wires each 0.5Ohms connected to this 12V supply and then you short the ends of those wires you will get 12V/ (0.5+0.5) =12A of current. This 12V * 12A will result in 144W energy wasted as heat so each second 144Ws (144J same thing) or 40mWh again same thing.
So 144Ws delivered by the source and 144Ws end as heat nothing stored.
Then say you connect a 5Ohm resistor and energy will go through wires and through resistor toal 6Ohms so 2A * 12V = 24W in one second it will be 24Ws delivered by the source and 24Ws all end as heat dissipated by wires and by resistor as energy travels through them.
Now say you leave the wires and instead of a resistor you connect a capacitor say it is 1000uF = 1mF and say internal DC ESR is basically zero to keep things simple.
a) How much energy will power supply deliver in one second after the discharged capacitor was connected ?
b) How much energy will end up as heat dissipated by wires?
c) How much energy is stored in the capacitor after this second?
d) How much energy is delivered by the source in the next second? so after first second ended and another second from that point.
Answer correctly to all this questions and you should understand what happens. Answer wrong and I may be able to help you understand what part you got wrong.
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absolutely incorrect. You know we receive radio waves from space right ?
I know but do you know how that works ?
If you think you do give an answer to the above problem the one in post #701
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I know but do you know how that works ?
Yes, Maxwell's equations predict that a time varying field produces a propagating wave. Just like shaking your hand on the surface of a pool creates a propagating wave on the surface of the water.
So wave a magnet about and you have in fact sent a tiny propagating wave out to the universe.
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if current flows into one terminal of a capacitor and out the other terminal I think most of us here call that "through" the capacitor.
Are you saying that this situation is not correctly described by using the word "through" to describe the currents ?
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if current flows into one terminal of a capacitor and out the other terminal I think most of us here call that "through" the capacitor.
Are you saying that this situation is not correctly described by using the word "through" to describe the currents ?
You are calling it through but that is incorrectly. Saying electrical current flows through dielectric is incorrect and you know that.
Try and solve the problem in post #701 and see if going "through" is helpful.
Electric current flows through a resistor but it will not flow through a capacitor but in and out as it is stored there not doing any work.
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Then say you connect a 5Ohm resistor and energy will go through wires and through resistor toal 6Ohms so 2A * 12V = 24W in one second it will be 24Ws delivered by the source and 24Ws all end as heat dissipated by wires and by resistor as energy travels through them.
Now say you leave the wires and instead of a resistor you connect a capacitor say it is 1000uF = 1mF and say internal DC ESR is basically zero to keep things simple.
No, this is the wrong question. Keep the resistor there and insert the capacitor as well, so you have both the resistor and the capacitor in series in the circuit. Now answer your questions below:
a) How much energy will power supply deliver in one second after the discharged capacitor was connected ?
b) How much energy will end up as heat dissipated by wires and resistor?
c) How much energy is stored in the capacitor after this second?
d) How much energy is delivered by the source in the next second? so after first second ended and another second from that point.
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a) How much energy will power supply deliver in one second after the discharged capacitor was connected ?
b) How much energy will end up as heat dissipated by wires?
c) How much energy is stored in the capacitor after this second?
d) How much energy is delivered by the source in the next second? so after first second ended and another second from that point.
The circuit you describe is basically 12V zero impedance source, 1 ohm resistor ( both 0.5 ohm wires ) , and 1000 u capacitor all in series.
Vc = 12 * ( 1 - exp(-R*C* t))
Vc = 12 * (1 - exp(- 1 * 1000e-6* t)
plot of the energy stored in the capacitor vs time
[attachimg=1]
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a) How much energy will power supply deliver in one second after the discharged capacitor was connected ?
b) How much energy will end up as heat dissipated by wires and resistor?
c) How much energy is stored in the capacitor after this second?
d) How much energy is delivered by the source in the next second? so after first second ended and another second from that point.
No, this is the wrong question. Keep the resistor there and insert the capacitor as well, so you have both the resistor and the capacitor in series in the circuit. Now answer your questions below:
Wires already have a resistance 1Ohm in total so it will not make any difference if you keep the 5Ohm resistor in series. If you want keep the resistor in series and solve the problem.
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a) How much energy will power supply deliver in one second after the discharged capacitor was connected ?
b) How much energy will end up as heat dissipated by wires?
c) How much energy is stored in the capacitor after this second?
d) How much energy is delivered by the source in the next second? so after first second ended and another second from that point.
The circuit you describe is basically 12V zero impedance source, 1 ohm resistor ( both 0.5 ohm wires ) , and 1000 u capacitor all in series.
Vc = 12 * ( 1 - exp(-R*C* t))
Vc = 12 * (1 - exp(- 1 * 1000e-6* t)
plot of the energy stored in the capacitor vs time
(Attachment Link)
The graph looks wrong (x axis) just answer the questions. Those are us so 5000us not 5000ms
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Yes the xaxis should be us
here are the other questions.
[attachimg=1]
[attachimg=2]
These are the complete answers to your questions.
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Yes the xaxis should be us
here are the other questions.
These are the complete answers to your questions.
I want you to provide the numbers not the graphs only. After you provide the numbers you can tell me exactly how much energy passes through capacitor.
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""You are calling it through but that is incorrectly. Saying electrical current flows through dielectric is incorrect and you know that.""
This is the crux of the discussion. Just simply an obscure definition of "through"
Electrodous... What word would you use to describe the power delivered to the other side of the capacitor. You do agree that power is found on the other side of the capacitor ?
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""You are calling it through but that is incorrectly. Saying electrical current flows through dielectric is incorrect and you know that.""
This is the crux of the discussion. Just simply an obscure definition of "through"
Electrodous... What word would you use to describe the power delivered to the other side of the capacitor. You do agree that power is found on the other side of the capacitor ?
There is no energy delivered to the other side. Just answer the questions in that problem and you will see that is the case.
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So you are saying that what we see in that circuit is false? That the input energy from the PSU does not cause the resistor to consume energy?
What I'm saying is that you can not connect just one end of a resistor to one terminal of the battery and expect a current except for the one needed to charge the newly created capacitor that is made up by the other battery terminal and the resistor with air being the dielectric.
It is about how current flows in to a capacitor when being charged and not through a capacitor.
Say you have an ideal voltage source 12V so there is no internal series resistance considered for this ideal voltage source.
Then say you have two wires each 0.5Ohms connected to this 12V supply and then you short the ends of those wires you will get 12V/ (0.5+0.5) =12A of current. This 12V * 12A will result in 144W energy wasted as heat so each second 144Ws (144J same thing) or 40mWh again same thing.
So 144Ws delivered by the source and 144Ws end as heat nothing stored.
Then say you connect a 5Ohm resistor and energy will go through wires and through resistor toal 6Ohms so 2A * 12V = 24W in one second it will be 24Ws delivered by the source and 24Ws all end as heat dissipated by wires and by resistor as energy travels through them.
Now say you leave the wires and instead of a resistor you connect a capacitor say it is 1000uF = 1mF and say internal DC ESR is basically zero to keep things simple.
a) How much energy will power supply deliver in one second after the discharged capacitor was connected ?
b) How much energy will end up as heat dissipated by wires?
c) How much energy is stored in the capacitor after this second?
d) How much energy is delivered by the source in the next second? so after first second ended and another second from that point.
Answer correctly to all this questions and you should understand what happens. Answer wrong and I may be able to help you understand what part you got wrong.
Jesus F Christ!
A simple yes or no question results in waffling that isn't relevant yet again! You're incapable of following a logical process and always jump to the conclusion before working back, then just fill up reams of inappropriate gumpf to support that foregone concluision instead of just answering the damn question as stated.
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Jesus F Christ!
A simple yes or no question results in waffling that isn't relevant yet again! You're incapable of following a logical process and always jump to the conclusion before working back, then just fill up reams of inappropriate gumpf to support that foregone concluision instead of just answering the damn question as stated.
Not sure what that fictional character has to do with this problem. Other that he is as fictional as you energy going through a capacitor.
Answer the question and you will see the answer for yourself.
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""You are calling it through but that is incorrectly. Saying electrical current flows through dielectric is incorrect and you know that.""
This is the crux of the discussion. Just simply an obscure definition of "through"
Electrodous... What word would you use to describe the power delivered to the other side of the capacitor. You do agree that power is found on the other side of the capacitor ?
It's worse than that. electrodacus doesn't understand displacement current.
He keeps saying "electrical current" hoping this smokescreen will distract from his ignorance of there being TWO terms for current in Ampere's Law - the conduction current AND the displacement current. In the region between two capacitors, there is no conduction current, but there IS displacement current.
Asserting that displacement current propagates no energy is anti-Maxwell.
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In profane usage, the middle initial is "H".
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It's worse than that. electrodacus doesn't understand displacement current.
He keeps saying "electrical current" hoping this smokescreen will distract from his ignorance of there being TWO terms for current in Ampere's Law - the conduction current AND the displacement current. In the region between two capacitors, there is no conduction current, but there IS displacement current.
Asserting that displacement current propagates no energy is anti-Maxwell.
Provide an answer to the problem then let me know the amount of energy that passed "through" capacitor.
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You will need to understand Maxwell's equations to understand what they say. They do not say energy flows outside the wires.
They absolutely say that. That's what led Maxwell to unify the electric and magnetic fields into electromagnetism... and it's what led Hertz to try transmitting an electrical signal from one radiator to another without any wires whatsoever.
No they don't. They are equations giving the curl/divergence of two functions, and none of the two are energy.
Also, "displacement current" is a concept made up by Maxwell, grouping various effects into one.
Much like you can use "magnetic monopoles current/density" and it can simplify computations, it does NOT mean magnetic monopoles exist.
Asserting that displacement current propagates no energy is anti-Maxwell.
1) How is it a problem.
2) Asserting it does is anti-Poynting.
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It's worse than that. electrodacus doesn't understand displacement current.
He keeps saying "electrical current" hoping this smokescreen will distract from his ignorance of there being TWO terms for current in Ampere's Law - the conduction current AND the displacement current. In the region between two capacitors, there is no conduction current, but there IS displacement current.
Asserting that displacement current propagates no energy is anti-Maxwell.
Provide an answer to the problem then let me know the amount of energy that passed "through" capacitor.
It's the same energy you believe only exists when conduction current flows - this is what Ampere's Law is trying to tell you.
@Naej
No they don't. They are equations giving the curl/divergence of two functions, and none of the two are energy.
https://web.mit.edu/6.013_book/www/chapter11/11.2.html
The electric and magnetic fields are confined to the free space regions. Thus, power flow and energy storage pictured in terms of these variables occur entirely in the free space regions.
The Poynting Vector is derived from Maxwell's Eqs. If Maxwell's Eqs are true - then so is Poynting.
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In profane usage, the middle initial is "H".
Ah, but this was crude profanity >:D
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Ahite. I got your numbers. And discovered how rusty i am at this stuff. skool was a long time ago
a) How much energy will power supply deliver in one second after the discharged capacitor was connected ? -> 0.1439 Joules
b) How much energy will end up as heat dissipated by wires? -> 0.0719 joules
c) How much energy is stored in the capacitor after this second? -> 0.0720 Joules
d) How much energy is delivered by the source in the next second? so after first second ended and another second from that point. -> 0 ( time constant is 1 ms )
What does this mean?
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I tell ya what I think it means.
The resistance of the wires got to dissipate as much energy as is stored in the capacitor.
And the current that made this possible went into one terminal of the capacitor and out the other terminal. I call this "through" what do you call it ?
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No they don't. They are equations giving the curl/divergence of two functions, and none of the two are energy.
https://web.mit.edu/6.013_book/www/chapter11/11.2.html
The electric and magnetic fields are confined to the free space regions. Thus, power flow and energy storage pictured in terms of these variables occur entirely in the free space regions.
The Poynting Vector is derived from Maxwell's Eqs. If Maxwell's Eqs are true - then so is Poynting.
These are 2 completely different things: Maxwell's equations describe an approximation of physics (a very good one when the number of photons is large, and their energy small compared to electrons); Poynting's theorem is well, a theorem (not an approximation), which is true if you assume Maxwell's equations are correct. Maxwell had nothing do to with ExH, and there's no dD/dt in ExH.
The important part of the quote is what I put in bold, it's a picture, and you might as well take S=JV.
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I tell ya what I think it means.
The resistance of the wires got to dissipate as much energy as is stored in the capacitor.
And the current that made this possible went into one terminal of the capacitor and out the other terminal. I call this "through" what do you call it ?
Thanks for doing the calculation and providing the numbers.
What it means is that energy from the source was split in two with one part ending as heat in wire/resistor and the other as stored energy in capacitor thus no energy passed through capacitor.
You can prove that that amount of energy 72mJ is stored in capacitor either by calculation or by testing.
If any energy will have been going "through" capacitor it will have done some work so end up as heat instead it was all stored so going in to capacitor.
An example where energy was going through something will be if you replaced the capacitor with a fuse that say melted with 72mJ but then all 144mJ provided by the source will have ended as heat nothing stored with 72mJ passing through wires and another 72mJ passing trough fuse before it melted all ending up as heat so 144mJ of heat.
If energy was to pass through capacitor then you could not store anything. It will mean your capacitor is defective and plates are shorted.
But if plates where shorted you will have dissipated way more energy as heat in one second on the wires 144J instead of 2000x less at 72mJ
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Sorry be posting 3 comments in a row but enough nonsense has been perpetrated here about conduction current being the only way that energy can move from A to B in space. Here is a very simple experiment measuring displacement current as equal to the conduction current but there are NO electrons moving across any gaps whatsoever:
https://www.youtube.com/watch?v=EqufEpWaKXw (https://www.youtube.com/watch?v=EqufEpWaKXw)
Interesting.
1. I see that the induced wave on the left of the capacitor was stronger than on the right of the capacitor.
2. And i suppose that if he removed the right half of the capacitor including its wire connection then he would have gotten an even stronger wave on the left part of the circuit (stronger than with the right side in place).
3. Here (1) & (2) support my new (elekton) elekticity.
4. The electricity on the rhs of the capacitor is an (induced) electron electricity.
5. On the lhs its an elekton elekticity.
6. Tween the plates there is no electricity, no current, there is a radio signal, ie em radiation, which induces an electron electric current on the rhs. [/quote]
And i suppose that if he removed the right half of the capacitor including its wire connection then he would have gotten an even stronger wave on the left part of the circuit (stronger than with the right side in place).
Easy to suppose, but unless you can show that then "Here (1) & (2) support my new (elekton) elekticity" is just building on quicksand.
Hell, we can all do that. I suppose that if he removed the right side the left side would have doubled. No, tripled! Now just let me make up a theory to account for that...
Not so fast. We already have some data. The lhs was stronger than the rhs.
How does old electron electricity explain that?
How was there more change (in current) on the lhs?
Yes, I saw that. What I'm disputing is that you can extrapolate and say you would see anything with the right side missing. Maybe you would, but that wasn't shown and we can't tell from the experiment.
Yes. And i said that my new (elekton) elekticity could explain. But that old (electron) electricity couldn’t explain. But praps old (electron) electricity can explain. Lemmeseenow.
Old (electron) electricity would have to say that there is electric energy lost in the air gap, whilst the electric energy was going from left to right.
Which raises a question – is the AC energy going from left to right? I reckon that it is.
If the AC energy is alternating & coming from the left & then from the right etc then i can't say that energy is lost in the air gap – koz if alternating then the signal would have the same strength on lhs & rhs.
[I doubt that electrodacus is correct that the energy loss across the air gap is due to a problem with scope grounding.]
Anyhow, old (electron) electricity might be correct if it says that energy is flowing from left to right & if it says that much energy is lost in the air gap (or including the plates too).
The em radiation around a wire has intrinsic energy (which can be called Poynting Field energy if u like) in the non-transient electricity case, & the em radiation (Poynting Field if u like) transmits energy from the wire in the transient case.
All of this energy has come from or is coming from the source (eg battery). Via wires (via elektons & electrons).
Old (electron) electricity (i reckon) demands an energy loss in the air gap.
My new (elekton) elekticity duznt need an energy loss, it just needs an inefficient energy transmission across the air gap. This inefficient process involves elektons on one plate (lhs) inducing an electron voltage on the other plate (rhs).
In old (electron) electricity the process would involve electrons (mainly surface electrons i reckon)(not so much internal electrons, which old (electron) electricity reckons) on both plates.
A large energy loss in the air gap (including the plates) would demand a rise in temperature, & or some kind of new radiation.
Not forgetting that there is a small energy loss in every inch of every wire all the time anyhow, including gaps.
What everyone around here (cept me) probly duznt know, is that the em radiation around a wire duznt accord with conservation of energy. The field is due to the radiation from elektons & electrons. The radiation from elektons & electrons is perpetual. Standard science has no explanation for this. It duznt have an explanation koz it duznt need an explanation. It duznt need an explanation koz it duznt recognise the question. Aether theory has the answer.
The above duznt apply to transient electricity. Transient electricity involves the em radiation, but it also involves conservation of energy.
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I'm afraid to use a light switch now. Will the light turn on? Will the universe split in two?
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If any energy will have been going "through" capacitor it will have done some work so end up as heat instead it was all stored so going in to capacitor.
You have a problem with your terminology.
In thermodynamics, the First Law can be roughly paraphrased to say: Energy = Heat + Work
So the energy dissipated in the resistor was Heat, and the energy stored in the capacitor was Work. It takes work to store energy, because that's where the stored energy comes from, it is the accumulation of work done.
If the energy stored in the capacitor were not Work, then it would be Heat, and the capacitor would get hot. The capacitor does not get hot, so the electricity is not generating Heat, it is doing Work.
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You have a problem with your terminology.
In thermodynamics, the First Law can be roughly paraphrased to say: Energy = Heat + Work
So the energy dissipated in the resistor was Heat, and the energy stored in the capacitor was Work. It takes work to store energy, because that's where the stored energy comes from, it is the accumulation of work done.
If the energy stored in the capacitor were not Work, then it would be Heat, and the capacitor would get hot. The capacitor does not get hot, so the electricity is not generating Heat, it is doing Work.
Energy stored in capacitor is not work.
It can do work as any other stored energy if you want to use it but it is not work.
So in my example capacitor stored 72mJ so half of what the source provided the other half was converted in to heat.
You can then remove the source and just add a wire where the source was since that way energy stored in capacitor can be converted to heat in that resistor so another 72mJ and now all the energy that source provided was used.
And the word trough means something completely different than in / out.
If I say water went in to bucket or through the bucket is not the same thing.
Where through the bucket will mean bucket is damaged has a hole so water goes through instead of in.
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And the word trough means something completely different than in / out.
Suppose you have a pipe with a diaphragm separating the left from right sides, both filled with water. A piston on the left side is pushed in, and as a consequence a piston on the right is pushed out.
Water is pushed in, but no water transfers from left water to right water. However there is energy transfer through from one piston to the other.
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And the word trough means something completely different than in / out.
Suppose you have a pipe with a diaphragm separating the left from right sides, both filled with water. A piston on the left side is pushed in, and as a consequence a piston on the right is pushed out.
Water is pushed in, but no water transfers from left water to right water. However there is energy transfer through from one piston to the other.
That is a bad example.
I already provided a better example but as any analogy will still have limitations.
A cylinder separated with a diaphragm but filled with air or any other gas.
If pressure is equal on both side it will be equivalent to a discharged capacitor so there is no stored energy.
You can install a pump that takes air molecules from one side and moves them in to the other side.
Say you start with 1000 molecules on each side so 2000 molecules in total. Then you use the pump so you need to put in some mechanical energy to push air from one side in to the other side charging this energy storage device with energy.
Say you move 500 air molecules from one side to the other so now you have 500 on one side and 1500 on the other.
Now say you want to charge an identical discharged cylinder from this charged one.
You will connect two hoses (hoses also have air in them just at normal pressure) so when you connect to the discharged cylinder
pressure will want to equalize.
You start with an excess of 500 air molecules one one side and a deficit of 500 molecules on the other side so 250 molecules will be transferred for the pressure to equalize.
You end up with 1250 / 750 in both cylinders. So was there any work done other than the heat loss due to friction of moving the molecules from one cylinder to another ?
But as mentioned you can not push any analogy to far.
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Energy stored in capacitor is not work.
Work has to be done to store energy in a capacitor. So the energy stored in a capacitor results from the electricity doing work.
There is either heat or work. If the electricity is not generating heat then it is doing work. If it is not doing work then it is generating heat.
So in my example capacitor stored 72mJ so half of what the source provided the other half was converted in to heat.
Half of what the source provided was used to do work on the capacitor, the other half was converted to heat.
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Work has to be done to store energy in a capacitor. So the energy stored in a capacitor results from the electricity doing work.
There is either heat or work. If the electricity is not generating heat then it is doing work. If it is not doing work then it is generating heat.
Half of what the source provided was used to do work on the capacitor, the other half was converted to heat.
You are confusing work with stored energy.
In fact you completely ignore energy storage as if it was not still there.
Out of the 144mJ of energy from the supply half was converted to heat the other half was stored in the same form that it was delivered so electrical energy.
That other half is stored in the capacitor and you can do whatever you want with it as it was not been used to do any work.
Not sure why is so hard to get that energy is in the same form electrical energy and it is available it was not used.
The capacitor is now a source containing 72mJ.
I think you accused me of violating the energy conservation but it looks like that is you.
I say 144mJ delivered by source = 72mJ as heat in the wires + 72mJ stored as electrical energy.
You say 144mJ = 72mJ as heat in the wires + 72mJ work + 72mJ stored ???
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I think you accused me of violating the energy conservation but it looks like that is you.
I say 144mJ delivered by source = 72mJ as heat in the wires + 72mJ stored as electrical energy.
You say 144mJ = 72mJ as heat in the wires + 72mJ work + 72mJ stored ???
You didn't read what I said. You must learn to read.
Energy = Heat + Work
144 mJ (Energy) = 72 mJ (Heat) + 72 mJ (Work)
It's not complicated, is it?
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I think you accused me of violating the energy conservation but it looks like that is you.
I say 144mJ delivered by source = 72mJ as heat in the wires + 72mJ stored as electrical energy.
You say 144mJ = 72mJ as heat in the wires + 72mJ work + 72mJ stored ???
You didn't read what I said. You must learn to read.
Energy = Heat + Work
144 mJ (Energy) = 72 mJ (Heat) + 72 mJ (Work)
It's not complicated, is it?
You're incredibly patient. I don't quite know what dacus's problem is, but he's long past the point of entertaining for me.
It's like telling him that it takes time to charge a capacitor and he replies that time is not energy. :-// |O :-DD
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You didn't read what I said. You must learn to read.
Energy = Heat + Work
144 mJ (Energy) = 72 mJ (Heat) + 72 mJ (Work)
It's not complicated, is it?
So there is no stored energy ? A 1000uF capacitor with 12V across contains no stored energy ?
What did that work you are talking about did ?
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You didn't read what I said. You must learn to read.
Energy = Heat + Work
144 mJ (Energy) = 72 mJ (Heat) + 72 mJ (Work)
It's not complicated, is it?
So there is no stored energy ? A 1000uF capacitor with 12V across contains no stored energy ?
What did that work you are talking about did ?
The battery. Those electrons didn't just get into the capacitors by themselves now did they?
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The battery. Those electrons didn't just get into the capacitors by themselves now did they?
Losing half the energy as heat is not enough ? You can say they went downhill.
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You're incredibly patient. I don't quite know what dacus's problem is, but he's long past the point of entertaining for me.
It's like telling him that it takes time to charge a capacitor and he replies that time is not energy. :-// |O :-DD
It's fascinating seeing how many different ways he can come up with to argue that black is white, but yes, it does get boring after a while.
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The battery. Those electrons didn't just get into the capacitors by themselves now did they?
Losing half the energy as heat is not enough ? You can say they went downhill.
Oh, on second read your question it makes no sense.... I missed the extra "did" on the end.
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Oh, on second read your question it makes no sense.... I missed the extra "did" on the end.
What question are you referring to?
Storing 72mJ in the capacitor resulted in 72mJ lost as heat so you can say charge efficiency is 50%
This charge efficiency can be improved to at least 90% if if you add as a minimum an inductor and diode as demonstrated in the past.
Also that 50% loss can be converted to something other than heat like maybe visible light by adding a lamp in series (just one example of many posible).
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It's fascinating seeing how many different ways he can come up with to argue that black is white, but yes, it does get boring after a while.
You are making the black white.
Energy stored is just that and not work.
Claiming a charged capacitor has no stored energy is just absurd.
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Claiming a charged capacitor has no stored energy is just absurd.
Of course. That is why nobody has ever claimed that.
Energy stored is just that and not work.
On the contrary, work is energy stored, heat is energy lost.
Here's a reference:
https://openstax.org/books/university-physics-volume-2/pages/8-3-energy-stored-in-a-capacitor
To move an infinitesimal charge dq from the negative plate to the positive plate [of a capacitor] (from a lower to a higher potential), the amount of work dW that must be done on dq is dW=Vdq=(q/C)dq.
This work becomes the energy stored in the electrical field of the capacitor. In order to charge the capacitor to a charge Q, the total work required is...
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It's fascinating seeing how many different ways he can come up with to argue that black is white, but yes, it does get boring after a while.
You are making the black white.
Energy stored is just that and not work.
Claiming a charged capacitor has no stored energy is just absurd.
I don't think anyone claimed that.
Does a capacitor store time?
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Of course. That is why nobody has ever claimed that.
On the contrary, work is energy stored, heat is energy lost.
You are very confused.
Doing work means using energy not storing energy.
Once you have done the work that energy to do the work is used so no longer available.
As an example a vehicle using 1kWh to travel 5km is work done.
That 1kWh is no longer available it was used to cover friction thus lost as heat.
You either admit energy is stored in the capacitor or you can claim that some work was done. You can not claim both are true at the same time as that is not how this universe works.
Getting those 72mJ in the capacitor was already very inefficient at just 50% with another 72mJ lost as heat.
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Of course. That is why nobody has ever claimed that.
On the contrary, work is energy stored, heat is energy lost.
You are very confused.
Doing work means using energy not storing energy.
Once you have done the work that energy to do the work is used so no longer available.
As an example a vehicle using 1kWh to travel 5km is work done.
That 1kWh is no longer available it was used to cover friction thus lost as heat.
You either admit energy is stored in the capacitor or you can claim that some work was done. You can not claim both are true at the same time as that is not how this universe works.
Getting those 72mJ in the capacitor was already very inefficient at just 50% with another 72mJ lost as heat.
Wikipedia disagrees... see "Work (physics)"
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Of course. That is why nobody has ever claimed that.
On the contrary, work is energy stored, heat is energy lost.
You are very confused.
You appear to be surrounded by confused people. What's the common factor? It's you!
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Doing work means using energy not storing energy.
That is correct. For example, moving charge from a battery (energy storage, yes?) to a capacitor. Now the energy is in the capacitor, that took work.
Sort of like slogging through your obtuse ramblings.
Does a capacitor store time?
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That is correct. For example, moving charge from a battery (energy storage, yes?) to a capacitor. Now the energy is in the capacitor, that took work.
Sort of like slogging through your obtuse ramblings.
Does a capacitor store time?
The work that was done to move that charge from the voltage source to capacitor resulted in 72mJ of heat on the wires.
So efficiency of moving electrical energy from the ideal voltage source to capacitor was 50%
Thus you end up with 72mJ of heat due to work you performed to move the charge and 72mJ of electrical energy stored in to capacitor.
The 72mJ stored in the capacitor did not do any work but they are there to do whatever work you want.
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You appear to be surrounded by confused people. What's the common factor? It's you!
Where is the confusion for you?
Do you understand the difference between energy that was stored and energy that did work ?
The process of storing energy in capacitor was in this case 50% efficient meaning that the 50% that was lost as heat on the wire is gone and the 50% stored in the capacitor is just that stored energy it did not do any work but it can be used to do work.
If instead of capacitor you will have had an electric motor that use 72mJ to move a weight from point A to point B then that will have been energy that has done work and it will have been gone no more 72mJ for you to use as you want.
If someone can not understand the difference between stored energy and energy that did work then I can not call him or her anything other than confused.
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If someone can not understand the difference between stored energy and energy that did work then I can not call him or her anything other than confused.
You seem to lack basic high-school knowledge of physics and electricity. That's why you don't understand what these people are talking about. Then you project your confusion on them, saying they are wrong.
Maybe it's time to revisit the basic concepts and try to understand what they're saying.
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You seem to lack basic high-school knowledge of physics and electricity. That's why you don't understand what these people are talking about. Then you project your confusion on them, saying they are wrong.
Maybe it's time to revisit the basic concepts and try to understand what they're saying.
Or maybe high-school physics and even university level is badly done.
School in general seems to be tailored to those that can memorize a lot of facts with no effort put into understanding them.
Here is an example to illustrate the difference between energy used to do work and stored energy.
So say you are the source of energy and you are pushing an electric vehicle on a flat road.
Vehicle is in neutral but there is of course friction loss so you put in 200Wh worth of energy and push the vehicle for 1km.
Now this 200Wh of energy did work.
Then at the same time I also push an identical vehicle and also expend 200Wh on the same flat road but I moved the vehicle just 100m = 0.1km because the regenerative brakes on the vehicle where engaged and 180Wh ended as energy stored in the battery.
When you look at this two systems the input energy was exactly the same 200Wh but in first case all energy input in the system did work while in the second case only 10% was used to do work the other 90% was stored as electricity and that stored energy can be used to do work but at this point in time it is stored energy and anything can be done with that not necessarily move the vehicle the other 900m but maybe just use the energy to listen to music on the car radio.
In this example there was an energy conversion from mechanical to electrical before storing unlike the capacitor problem where no conversion was done.
The 144Ws of energy delivered by the source was used to charge the capacitor with an fairly bad efficiency of just 50% so you end up with 72Ws stored in the capacitor but those 72Ws in the capacitor are still there in the original form of electrical energy ready to be used to do work.
Capacitor is an energy storage device that will not let current pass through thus no energy is passing through (the displacement current is just a mathematical concept not a physical current).
Something like displacement current or displacement energy is a fictional math concept for calculation purposes as maybe energy storage was harder to explain.
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You seem to lack basic high-school knowledge of physics and electricity. That's why you don't understand what these people are talking about. Then you project your confusion on them, saying they are wrong.
Maybe it's time to revisit the basic concepts and try to understand what they're saying.
Or maybe high-school physics and even university level is badly done.
School in general seems to be tailored to those that can memorize a lot of facts with no effort put into understanding them.
Here is an example to illustrate the difference between energy used to do work and stored energy.
So say you are the source of energy and you are pushing an electric vehicle on a flat road.
Vehicle is in neutral but there is of course friction loss so you put in 200Wh worth of energy and push the vehicle for 1km.
Now this 200Wh of energy did work.
Then at the same time I also push an identical vehicle and also expend 200Wh on the same flat road but I moved the vehicle just 100m = 0.1km because the regenerative brakes on the vehicle where engaged and 180Wh ended as energy stored in the battery.
When you look at this two systems the input energy was exactly the same 200Wh but in first case all energy input in the system did work while in the second case only 10% was used to do work the other 90% was stored as electricity and that stored energy can be used to do work but at this point in time it is stored energy and anything can be done with that not necessarily move the vehicle the other 900m but maybe just use the energy to listen to music on the car radio.
In this example there was an energy conversion from mechanical to electrical before storing unlike the capacitor problem where no conversion was done.
The 144Ws of energy delivered by the source was used to charge the capacitor with an fairly bad efficiency of just 50% so you end up with 72Ws stored in the capacitor but those 72Ws in the capacitor are still there in the original form of electrical energy ready to be used to do work.
Capacitor is an energy storage device that will not let current pass through thus no energy is passing through (the displacement current is just a mathematical concept not a physical current).
Something like displacement current or displacement energy is a fictional math concept for calculation purposes as maybe energy storage was harder to explain.
You do realize that you will be pushing 10x harder in the second case? 0.1 times the distance means 10x the force...
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You do realize that you will be pushing 10x harder in the second case? 0.1 times the distance means 10x the force...
I already mentioned that same amount of energy will be used 200Wh so the fact that you need to push harder is obvious.
In both cases the same amount of energy was used just that in first case all that energy was used to do work while in second case small amount was used to do work the rest was stored.
The stored energy can not be called work.
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The person pushing the car has done the same amount of work. Just in the first case more got 'lost' to the environment.
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The person pushing the car has done the same amount of work. Just in the first case more got 'lost' to the environment.
We are not discussing the source.
Yes 144mWs where provided by the source in capacitor charging example that is the equivalent to the person providing 200Wh in both cases.
The result was 72mWs of heat loss in the wires transporting the energy to capacitor and 72mWs of stored energy in the capacitor.
The result for vehicle examples was in first case 200Wh of work for moving the vehicle 1km
For second example there was 20Wh of work to move the vehicle 100m and 180Wh of energy stored in the EV battery.
Those 72mWs stored in capacitor and those 180Ws stored in battery are are not work. The energy stored there can be used to do work at any time but at that moment it is just stored electrical energy which is not work.
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Are you really arguing that work is not required to store energy in capacitor (or in this case a cars battery)?
Are we in Humpty Dumpty land again, where words only mean what you say the do?
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Are you really arguing that work is not required to store energy in capacitor (or in this case a cars battery)?
Are we in Humpty Dumpty land again, where words only mean what you say the do?
If transfer could be done at 100% efficiency then yes no work will be done to transfer energy from source to capacitor.
In the particular example efficiency was just 50% so half of the energy ended as heat during transportation.
You can use a DC-DC converter and transport the energy with about 90% efficiency so that only 10% is lost for moving the charge from a source to a capacitor (energy storage device).
What part exactly is not clear.
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What part exactly is not clear
How the resistor manages to consume energy without there being any transfer from one side to the other.
This diversion to how capacitors charge and at what cost, etc, is simply your smokescreen that you hope will obscure your inability to deal with the real issue.
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Are you really arguing that work is not required to store energy in capacitor (or in this case a cars battery)?
Are we in Humpty Dumpty land again, where words only mean what you say the do?
If transfer could be done at 100% efficiency then yes no work will be done to transfer energy from source to capacitor.
In the particular example efficiency was just 50% so half of the energy ended as heat during transportation.
You can use a DC-DC converter and transport the energy with about 90% efficiency so that only 10% is lost for moving the charge from a source to a capacitor (energy storage device).
What part exactly is not clear.
I'm not quite clear how you reconcile that with the actual formal definition of work for an electric field:
https://en.wikipedia.org/wiki/Work_(electric_field)
But then again, since you reject the existence of electric fields, that would hardly be surprising.
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What part exactly is not clear
How the resistor manages to consume energy without there being any transfer from one side to the other.
This diversion to how capacitors charge and at what cost, etc, is simply your smokescreen that you hope will obscure your inability to deal with the real issue.
You are charging an energy storage device.
There are electrons flowing through wires that have resistance in to the capacitor plates. This electrons remain on the plate do not jump the dielectric and this electrons can flow back delivering energy when you want to use that stored energy.
Will thinking at a rechargeable battery be easier ?
When you charge your phone battery say a LiCoO2 type that is charged from 3V to 4.2V trough a linear regulator can even be a resistor form a 5V supply.
Will you not have a part energy dissipated as heat on the wire resistance and series linear regulator and one part stored energy in the battery.
The energy that went in to battery did no work it is stored energy and can be used latter to supply your phone.
Capacitor same as the rechargeable battery are energy storage devices so energy will not flow through them but in them.
When Maxwell was alive he had no idea such things as electrons exist. He was also a mathematician not a physicist. He made an important contribution at that time showing electricity and magnetism are related.
Is the word store not different to you from the word work?
If you stored some quantity of energy then it is stored it did not do any work but it is ready to do work when you want to use it.
In the process of moving that energy from a source to the storage device some energy will be lost and in this case just happened to be 50% lost as heat but it could have been 20% lost as heat and 30% used to move an object so do work and then the other 50% stored.
It could also have been 10% lost as heat and 90% stored if you will have used a DC-DC constant current regulator.
So stored just means moving energy available in one place to another place and that amount of energy that was stored has done no work and in this case energy is in the exact same form as the original not converted.
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I'm not quite clear how you reconcile that with the actual formal definition of work for an electric field:
https://en.wikipedia.org/wiki/Work_(electric_field)
But then again, since you reject the existence of electric fields, that would hardly be surprising.
That definition you linked to has little to do with what we are discussing.
In this case a particle moves from source to capacitor (energy storage device) through a wire that has a resistance.
When I say the electron moves from the source to capacitor that will be incorrect as it is not the same electron that exits the source with the particle that enters the capacitor plate.
It is like a domino effect that happens at the speed of light but in a wire that has resistance this creates loss of energy so the electron that exits the source has more energy than the electron entering the capacitor with the difference resulting in heat.
So from source 144mWs worth of energy will leave with half that ending as heat due to wire resistance and the other half ends stored in the capacitor.
I think the main problem is to understand what stored means and it is not the same thing with work (quite the opposite). Stored energy has the ability to do work but it has not done so as long as it remains stored.
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What part exactly is not clear
How the resistor manages to consume energy without there being any transfer from one side to the other.
This diversion to how capacitors charge and at what cost, etc, is simply your smokescreen that you hope will obscure your inability to deal with the real issue.
You are charging an energy storage device.
Actually, I am heating up a resistor. Where did that energy come from, and how did it get there? The only energy source available is the power supply upwind of the capacitor. So, from what I can gather from your 'explanation' it goes something like this:
PSU: "Hey Cap+, gonna shove you some lovely joules."
Cap+: "Eh? Oh, hang on a mo..."
Cap+: Oy, Cap-, you wanna dump your old electrons and get some new ones?"
Cap-: "Sounds fab to me."
Cap+: "OK, let's have them then"
PSU: "Here you go. Enjoy!"
Cap-: "Hey, resitor, 50gazillion elect(r)ons coming your way!"
Resistor: "WTF? Now?!?! Shit, have to burn them up."
I don't think energy storage has meaning in this context. The cap is series with the resister and you talk as if it's in parallel and providing the juice. It isn't - the PSU is. If anything, the cap is acting as a blockage.
Bringing in a battery complicates things, just as inserting a diode would. Let's keep it simple with the minimal parts necessary to show the problem.
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Actually, I am heating up a resistor. Where did that energy come from, and how did it get there? The only energy source available is the power supply upwind of the capacitor. So, from what I can gather from your 'explanation' it goes something like this:
PSU: "Hey Cap+, gonna shove you some lovely joules."
Cap+: "Eh? Oh, hang on a mo..."
Cap+: Oy, Cap-, you wanna dump your old electrons and get some new ones?"
Cap-: "Sounds fab to me."
Cap+: "OK, let's have them then"
PSU: "Here you go. Enjoy!"
Cap-: "Hey, resitor, 50gazillion elect(r)ons coming your way!"
Resistor: "WTF? Now?!?! Shit, have to burn them up."
I don't think energy storage has meaning in this context. The cap is series with the resister and you talk as if it's in parallel and providing the juice. It isn't - the PSU is. If anything, the cap is acting as a blockage.
Bringing in a battery complicates things, just as inserting a diode would. Let's keep it simple with the minimal parts necessary to show the problem.
What do you think the major difference is between a capacitor and a rechargeable battery ?
Thy are both energy storage devices.
Yes 144mJ where provided by the source out of this 50% ended up stored in capacitor and the other 50% ended up heating the wires.
How come you did not used Res+ and Res- just Cap+ and Cap- ?
To have a current flow you need a close circuit. Connecting just the positive wire of a supply to a resistor or a capacitor will have no effect.
Maybe one of the people that read this posts can do a better job than me explaining the difference between energy that did work and stored energy.
Ideal case with no resistance anywhere in the circuit 72mJ will flow from the ideal source in to capacitor and system form an energy balance point of view has no charge.
Financial equivalent will me moving $72 from one bank account (yours) to another of your back accounts with no transaction fee. In the end you still have $72 is just in another bank account but other than that there is no difference.
But if transferring $72 costs you another $72 in fees then you pay $144 from your first bank account and you end up with $72 in another bank account thus you lost $72 just to move $72 from one account to another.
You still have $72 but you used to have $144. You are acting like all $144 were lost.
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"half that ending as heat due to wire resistance"
It took power to heat the wire. That power came from the flow of electrons into and out of the capacitor wires.
We consider the heat in the wires to be useful work.
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Are we still really arguing that power can be transferred through a capacitor???
If anybody says NO then would they please explain how this circuit gets power out the right side when connected to other power source through only a capacitor.
https://en.wikipedia.org/wiki/Capacitive_power_supply
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"half that ending as heat due to wire resistance"
It took power to heat the wire. That power came from the flow of electrons into and out of the capacitor wires.
We consider the heat in the wires to be useful work.
Yes electrons flowed in one side of the capacitor and other unrelated electrons flowed out of the other side but this means energy was stored in the capacitor.
When you will discharge the capacitor say by shorting the terminals of that capacitor electrons that flowed in will flow out and in to the side where electrons flowed out resulting in 72mJ of heat (the amount that was stored in the capacitor).
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"Yes electrons flowed in one side of the capacitor and other unrelated electrons flowed out of the other side but this means energy was stored in the capacitor."
And that energy that heated the wires flowed because of charge moving into and out of the capacitor.
Do you agree that a capacitor can not only store energy but can also pass through energy?
If not how does the capacitor power supply above work?
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And that energy that heated the wires flowed because of charge moving into and out of the capacitor.
Do you agree that a capacitor can not only store energy but can also pass through energy?
If not how does the capacitor power supply above work?
Yes wires had resistance to current flow so part of total energy 144mJ ended as heat. 72mJ to be exact ended as heat and the other 72mJ ended as stored energy in the capacitor.
No energy passed through capacitor as there is nothing left 144mJ = 72mJ lost as heat + 72mJ stored in the capacitor.
A capacitive dropper works because capacitor is always charged and discharged.
For example let say you want to power an incandescent lamp in the example I provided before.
You just add a lamp in series with the ideal DC voltage supply and the capacitor and say the lamp resistance is 9 Ohms.
Then you will have 144mJ from the DC supply form that 72mJ will end up stored in the capacitor
The other 72mJ will be lost on the wire's and lamp total 10Ohms = (9Ohm + 0.5Ohm + 0.5Ohm)
So 90% will be delivered to the incandescent lamp 64.8mJ (mostly infrared photons and some visible light photons) and the other 10% will end as heat loss on the wires 7.2mJ
But that is all done in a few ms then you can wait for an hour and there will be no more energy flowing just this 144mJ from which half was stored and half was used.
But if you reverse the polarity of the power supply or you even just remove the supply and close the circuit the capacitor will deliver the stored 72mJ to the lamp and wires then you can repeat this.
So with DC you have just one pulse as the capacitor chargers and with AC you constantly charge and discharge the capacitor.
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A capacitive dropper works because capacitor is always charged and discharged.
You are confusing charge with energy.
No energy passed through capacitor
On the contrary, you mean no charge passed through the capacitor. But power went around the the whole circuit for the duration that the capacitor charged.
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""So with DC you have just one pulse as the capacitor chargers and with AC you constantly charge and discharge the capacitor..""
Yes and in that one charging pulse there was a finite amount of energy available to drive a load on the other side of the capacitor.
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A capacitive dropper works because capacitor is always charged and discharged.
You are confusing charge with energy.
No energy passed through capacitor
On the contrary, you mean no charge passed through the capacitor. But energy went around the the whole circuit for the duration that the capacitor charged.
Charging a capacitor means storing energy in to it.
No energy went around the capacitor but in to capacitor.
So when you are charging your phone battery is the energy going around you battery or in to your phone battery ?
The concept of energy storage seems to be completely missing from your understanding.
You have the concept that you can transfer electrical energy with just one wire so no closed loop.
You are probably thinking at a compressed air tank and how you take air from environment and push it in storing energy.
A more accurate way to think will be to have the tank split in two isolated chambers and you are moving air from one half of the cylinder to the other half to charge it and then to discharged it so use the energy you do the reverse let the air from the pressurized half go back in to the half with lower pressure.
This last example is much more accurately describing a rechargeable battery or capacitor.
For that air pump to work you need to connect both halves of the cylinder so you need to have a closed loop.
Now say you have a similar compressed air tank with two chambers that is identical but discharged so both sides contain air but at same pressure.
If you now connect this charged cylinder to the discharged one with two hoses the pressure delta will be lower and equal in both cylinders and now you have lost half of the energy as you only have a quarter of initial energy in each tank because this sort of transfer was 50% efficient.
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""So with DC you have just one pulse as the capacitor chargers and with AC you constantly charge and discharge the capacitor..""
Yes and in that one charging pulse there was a finite amount of energy available to drive a load on the other side of the capacitor.
I'm trying very hard to understand what sort of thing you imagine happens.
Why was the amount of energy available finite ?
Why do you say the other side of the circuit ? It is irrelevant on what side of the circuit you load is.
You can connect the lamp on any side of the capacitor it will work exactly the same.
With that example that I provided you will have 9.5Ohms on the side you connect the lamp 9Ohms lamp + 0.5Ohms wire and on the other side you only have the 0.5Ohms wire.
On one side you have 72mJ * 0.95 = 68.4mJ and on the other side you have 3.6mJ so loss on wires as heat is now just 7.2mJ in total the rest is useful work assuming you wanted an incandescent lamp (you can replace the lamp with an electric motor or any other load you like).
There is a big difference between the 64.8mJ that ended on the lamp and the 72mJ that ended in capacitor.
The 64.8mJ already converted that energy in something that you wanted (light) while the 72mJ in the capacitor did nothing it is still available for you in original form as stored electrical energy.
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Charging a capacitor means storing energy in to it.
No energy went around the capacitor but in to capacitor.
So what caused the resistor to heat up? If it had been an LED it would emit light. If it were a motor it would turn. You cannot charge that capacitor without those things happening and yet, according to you, there was no energy available to them. They would be perpetual motion machines running on nothing!
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Charging a capacitor means storing energy in to it.
No energy went around the capacitor but in to capacitor.
So what caused the resistor to heat up? If it had been an LED it would emit light. If it were a motor it would turn. You cannot charge that capacitor without those things happening and yet, according to you, there was no energy available to them. They would be perpetual motion machines running on nothing!
I do not get where did I say anything that looked like a perpetual motion machine ? It is a supper inefficient circuit only 50% efficient.
Supply needs to deliver 144mJ just to charge the capacitor with 72mJ so the other 50% 72mJ are lost on the wires or if you add a lamp or motor they do some work.
Just wires and capacitor 144mJ = 72mJ + 72mJ
Adding a 9Ohm incandescent lamp 144mJ = 64.8mJ + 7.2mJ + 72mJ
Where do you see a perpetual motion machine ? 144mJ are provided and some of that is lost as heat the rest is doing some work and half of it in this case remains stored in the capacitor.
I need to know where do you see the perpetuum mobile ?
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Lets do a simple question.
the circuit here
https://en.wikipedia.org/wiki/Capacitive_power_supply
If a DC voltage ( 20 V) is switched into X1 and X1 and left connected indefinitely will there be any current through R4 and R5 temporarily?
Simple yes /no question,
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Lets do a simple question.
the circuit here
https://en.wikipedia.org/wiki/Capacitive_power_supply
If a DC voltage ( 20 V) is switched into X1 and X1 and left connected indefinitely will there be any current through R4 and R5 temporarily?
Simple yes /no question,
Yes there will be a short current spike trough R4 and R5 as the C1 is charged but current will be very small maybe almost impossible to measure because there is a much larger capacitor C2 that will charge while being discharged across R3,R4 and R5
And there is that R2 that will continue to let some small current pass even after this initial spike.
If you remove R2 then it will be just that initial spike.
Why use that much more complex example when you have the super simple capacitor + wires + voltage source that we are discussing ?
If energy was to go through an ideal capacitor instead of in to it when charging then voltage across the capacitor will remain zero. We know voltage across capacitor increases while it is being charged as all energy goes into the capacitor and remains there as stored energy.
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Well I used it cause the simple example seems to be mixed up with a lot of phoohey.
So we know ... and you agreed .. that some energy made it through the circuit to R4 and R5 with the application of a dc source to the circuit input.
Do you dispute this ?
This energy seems to have "gotten by" the capacitor. It is a series element in the circuit.
So what do you say about this energy. We know that if an AC source is connected the circuit functions as a power supply allowing energy to be extracted from X3 and X4. A suddenly applied DC signal has some AC in it.
If you still say that NO energy is transferred by the capacitor where did the energy for R4 and R5 come from?
Where does it come from if AC is applied to X3 and X4.
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Well I used it cause the simple example seems to be mixed up with a lot of phoohey.
So we know ... and you agreed .. that some energy made it through the circuit to R4 and R5 with the application of a dc source to the circuit input.
Do you dispute this ?
This energy seems to have "gotten by" the capacitor. It is a series element in the circuit.
So what do you say about this energy. We know that if an AC source is connected the circuit functions as a power supply allowing energy to be extracted from X3 and X4. A suddenly applied DC signal has some AC in it.
If you still say that NO energy is transferred by the capacitor where did the energy for R4 and R5 come from?
Where does it come from if AC is applied to X3 and X4.
Those resistors are part of the return for the capacitor that is being charged.
You can not charge that capacitor by just connecting one of the terminals from a battery or ideal voltage supply.
You can not have just electrons leaving the battery without same number returning and for that to happen you need a circuit.
How come that after capacitor is charged no more current is flowing ? Why no energy can go "through" the capacitor after the first few ms ?
The answer should be obvious and that is energy went in to capacitor not trough.
All evidence points to this correct conclusion but for some reason you chose to ignore all evidence.
Once you have 72mJ stored in the 1000uF capacitor there is no more current flow and if you take the capacitor out of the circuit you can still measure 12V across and can calculate that 72mJ are contained as stored energy and you can also discharge that energy and measure that is all there all 72mJ.
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How come that after capacitor is charged no more current is flowing ?
The cap is fully charged so cannot be charged more.
The answer should be obvious and that is energy went in to capacitor not trough.
The CHARGE went into the capacitor. The energy went around the circuit to allow charging. Once the capacitor is charged there can be no more circulating power, so no more energy. But if there is charge going INTO the capacitor then there is power going AROUND the circuit. If that wasn't so you wouldn't NEED a circuit!
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What difference does it make if the resistors are part of the return circuit of the capacitor. Nobody would argue that.
What we know is that being in that return circuit for the "Capacitor" they get to absorb some of the charging energy.
This charging energy is a result of electrons going into one terminal of the capacitor and out the other.
So we have electrons entering and leaving the terminals of the capacitor that go on to impart some energy in other things in the "return path"
And if we keep switching the polarity of the input we can keep getting power out in the return path.
This power will not flow without a route into and out of the capacitor terminals. the capacitor in integral to the transfer of power to the load. All this is very simple.
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The cap is fully charged so cannot be charged more.
The CHARGE went into the capacitor. The energy went around the circuit to allow charging. Once the capacitor is charged there can be no more circulating power, so no more energy. But if there is charge going INTO the capacitor then there is power going AROUND the circuit. If that wasn't so you wouldn't NEED a circuit!
Yes capacitor is charged and can not charge more.
Energy and more specifically electrical energy went into capacitor not around it. If it will have been going around it then less energy will have been needed from the supply.
144mJ worth of energy left the supply.
72mJ where dissipated as heat in the wires
72mJ where stored in the capacitor.
For a capacitor:
Charge is C * V
Energy is C * V * V * 0.5
Can you understand the relation and also the difference ?
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This charging energy is a result of electrons going into one terminal of the capacitor and out the other.
So we have electrons entering and leaving the terminals of the capacitor that go on to impart some energy in other things in the "return path"
And if we keep switching the polarity of the input we can keep getting power out in the return path.
This power will not flow without a route into and out of the capacitor terminals. the capacitor in integral to the transfer of power to the load. All this is very simple.
The electrons going in to one plate will never get on the other side. Those electrons that exit the other side where already there in that plate.
This excess of electrons on one plate and deficit on the other plate is the stored energy.
After 144mJ left the source and you get to steady state. You remove the terminal from the supply and connect them with reverse polarity.
What will happen then ? How much energy will be dissipated on the 1Ohm wires ?
Initially it was 72mJ but how much will it be when you change the polarity ?
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Can you understand the relation and also the difference ?
Can you not understand that to charge the capacitor you need a completed circuit for SOMETHING to go around. If you leave off the return path then nothing will happen. Therefore something must be going down that return path.
Are you saying that this is wrong? That you don't need a completed circuit for the capacitor to charge?
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Can you understand the relation and also the difference ?
Can you not understand that to charge the capacitor you need a completed circuit for SOMETHING to go around. If you leave off the return path then nothing will happen. Therefore something must be going down that return path.
Are you saying that this is wrong? That you don't need a completed circuit for the capacitor to charge?
Yes that is right. There is no complete circuit that is why energy stops flowing once the capacitor is fully charged.
The circuit is one source supplying a energy storage with limited capacity so current will flow from the source through the wires in the energy storage device (capacitor). No energy flows through the capacitor at any point in time.
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so current will flow from the source through the wires in the energy storage device (capacitor)
And the same current will flow back to the source from the other side on your energy storage device. If not, explain WTF is going through that resistor, and why you need a return path.
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so current will flow from the source through the wires in the energy storage device (capacitor)
And the same current will flow back to the source from the other side on your energy storage device. If not, explain WTF is going through that resistor, and why you need a return path.
See the problem I asked snarkysparky to solve in post 784. Result form that will blow your minds.
I asked to reverse the polarity of the voltage source after the capacitor has charged so circuit is in steady state.
I can provide the answer but likely you won't believe if you do not do that yourself.
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so current will flow from the source through the wires in the energy storage device (capacitor)
And the same current will flow back to the source from the other side on your energy storage device. If not, explain WTF is going through that resistor, and why you need a return path.
See the problem I asked snarkysparky to solve in post 784. Result form that will blow your minds.
I asked to reverse the polarity of the voltage source after the capacitor has charged so circuit is in steady state.
I can provide the answer but likely you won't believe if you do not do that yourself.
I am wrong there - got carried away. But there is A current going through that resistor isn't there? Go on, you have to admit to that. Let me remind you with the simulation:
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I am wrong there - got carried away. But there is A current going through that resistor isn't there? Go on, you have to admit to that. Let me remind you with the simulation:
Of course there will be a current through the resistor as you are charging the capacitor. Energy needs to travel through wire and the resistor is a wire.
But the new question is what happens if after you charged the capacitor so no current but capacitor contains those 72mJ and you reverse the polarity on the source.
How much energy will be delivered by the source and how much of that will end up in the resistor ?
I'm fairly certain you will be surprised by the result as it will be very different from initial result because the capacitor is already charged.
The problem data is a 12V ideal voltage source 0.5Ohm resistance for each wire so 1Ohm in total and 1000uF capacitor.
Initial connection when capacitor was discharged is 144mJ supplied by the ideal voltage source = 72mJ dissipated on wires as heat and 72mJ stored in the 1000uF capacitor.
Then after this is done just remove the supply and connect it again with reverse polarity.
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But the new question is what happens if after you charged the capacitor so no current but capacitor contains those 72mJ and you reverse the polarity on the source.
No, you're jumping ahead and going off on a diversion again. We have yet to determine exactly what is going through that resistor since you insist it can't be energy.
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No, you're jumping ahead and going off on a diversion again. We have yet to determine exactly what is going through that resistor since you insist it can't be energy.
Of course energy goes through resistor. When have I ever mentioned that is not ?
It is not going through capacitor but in to capacitor.
The entire point I want to make is that energy travels through wires and wires and resistors are the same thing.
If you reverse the polarity after capacitor was charged you will see that all energy now delivered by the supply ends up as heat on the resistor. All of it not just half as at initial connection when capacitor was discharged.
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Are you really arguing that work is not required to store energy in capacitor (or in this case a cars battery)?
Are we in Humpty Dumpty land again, where words only mean what you say the do?
If transfer could be done at 100% efficiency then yes no work will be done to transfer energy from source to capacitor.
In the particular example efficiency was just 50% so half of the energy ended as heat during transportation.
You can use a DC-DC converter and transport the energy with about 90% efficiency so that only 10% is lost for moving the charge from a source to a capacitor (energy storage device).
What part exactly is not clear.
It is clear that you are completely wrong.
When there is 0V across a 1F capacitor it takes a very tiny bit of work to remove a charge from the - plate, and add on to the + one. It doesn't matter if it is the same electron or not - it's a little bit of work to pull the electron out of one plate, leaving it slightly positively charged, and a little bit of work to push it into the other plate, which was quite happily neutral and now won't be.
But for the next time there is 1/(6.24 x 10^18) of a V between the plates, so that requires more work. If you do the math it will be 4x that of the first electron.
But for the next time there is 2/(6.24 x 10^18) of a V between the plates, so that requires even more work than the previous one. Once again, if you do the math it will be 9x that of the first electron.
For the (6.24 x 10^18)th electron there is now (6.24 x 10^18)/(6.24 x 10^18) = 1V between the plates. It is now taking a whopping (6.24 x 10^18)^2 times the energy it took to move the first electron.
In fact if you sum up all the work done on each charge it will equal 0.5 * C * Vfinal^2.
And all work being performed by your source capacitor.
The inductor in the DC-DC configuration when the switch is closed it is taking lower current at the higher voltage to make establish a magnetic field, and then when the switch opens it the collapsing magnetic field moves higher current at a lower voltage, at just slightly higher than the voltage of the uncharged capacitor.
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It is clear that you are completely wrong.
When there is 0V across a 1F capacitor it takes a very tiny bit of work to remove a charge from the - plate, and add on to the + one. It doesn't matter if it is the same electron or not - it's a little bit of work to pull the electron out of one plate, leaving it slightly positively charged, and a little bit of work to push it into the other plate, which was quite happily neutral and now won't be.
But for the next time there is 1/(6.24 x 10^18) of a V between the plates, so that requires more work. If you do the math it will be 4x that of the first electron.
But for the next time there is 2/(6.24 x 10^18) of a V between the plates, so that requires even more work than the previous one. Once again, if you do the math it will be 9x that of the first electron.
For the (6.24 x 10^18)th electron there is now (6.24 x 10^18)/(6.24 x 10^18) = 1V between the plates. It is now taking a whopping (6.24 x 10^18)^2 times the energy it took to move the first electron.
In fact if you sum up all the work done on each charge it will equal 0.5 * C * Vfinal^2.
And all work being performed by your source capacitor.
The inductor in the DC-DC configuration when the switch is closed it is taking lower current at the higher voltage to make establish a magnetic field, and then when the switch opens it the collapsing magnetic field moves higher current at a lower voltage, at just slightly higher than the voltage of the uncharged capacitor.
You forget one super important thing.
That is reversible so energy storage and not work.
The work will be done when you discharge that capacitor.
So all that energy you put in the capacitor 0.5 * C * V2 is still there in the capacitor ready to do work.
In my most recent example with 12V ideal voltage source 1Ohm total circuit resistance 0.5Ohm for each wire and the 1000uF capacitor.
Energy stored in capacitor was 0.5 * 0.001F * 12V2 = 72mWs
But the ideal voltage source needed to supply 2x as much 144mWs in order to cover the losses in transportation of that energy that happens to also be 72mWs drop on that 1Ohm resistance as energy traveled through wires to get from voltage source to capacitor.
Now try to calculate what happens if you after the capacitor is charged you disconnect the voltage source and connect it back but with reverse polarity.
What you will notice is that now all the energy delivered by the voltage source ends up as heat in the wire (all of it not just half like when capacitor was discharged).
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It is clear that you are completely wrong.
When there is 0V across a 1F capacitor it takes a very tiny bit of work to remove a charge from the - plate, and add on to the + one. It doesn't matter if it is the same electron or not - it's a little bit of work to pull the electron out of one plate, leaving it slightly positively charged, and a little bit of work to push it into the other plate, which was quite happily neutral and now won't be.
But for the next time there is 1/(6.24 x 10^18) of a V between the plates, so that requires more work. If you do the math it will be 4x that of the first electron.
But for the next time there is 2/(6.24 x 10^18) of a V between the plates, so that requires even more work than the previous one. Once again, if you do the math it will be 9x that of the first electron.
For the (6.24 x 10^18)th electron there is now (6.24 x 10^18)/(6.24 x 10^18) = 1V between the plates. It is now taking a whopping (6.24 x 10^18)^2 times the energy it took to move the first electron.
In fact if you sum up all the work done on each charge it will equal 0.5 * C * Vfinal^2.
And all work being performed by your source capacitor.
The inductor in the DC-DC configuration when the switch is closed it is taking lower current at the higher voltage to make establish a magnetic field, and then when the switch opens it the collapsing magnetic field moves higher current at a lower voltage, at just slightly higher than the voltage of the uncharged capacitor.
You forget one super important thing.
That is reversible so energy storage and not work.
The work will be done when you discharge that capacitor.
:-// Oh come on, you know the formula: work = force x distance,
If you or I push down on a lever, lifting a weight, it is still work, even though the weight on the far end can move the lever back to the original position, undoing our efforts.
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:-// Oh come on, you know the formula: work = force x distance,
If you or I push down on a lever, lifting a weight, it is still work, even though the weight on the far end can move the lever back to the original position, undoing our efforts.
You will be storing potential energy.
So say you put in 144mWs by dropping a weight from some height and it is connected through a pulley to another weight that is lifted (this will be the input).
Say lifting that other weight gained you 72mWs in potential gravitational energy and 72mWs were wasted as heat due to lifting mechanism friction.
Now all you have lost is the 72mWs due to friction the other 72mWs are still available in the same form as the original so you can say you moved potential gravitational energy from one weight to another not very efficiency just 50% efficiency.
I guess the above is a decently accurate analogy to what happens with the capacitor circuit.
Best case ideal case when you have no friction you are able to lift the same weight as the one providing the input to the exact same height.
So will you say that you did some work or did you just transferred the potential kinetic energy from one weight to another ?
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Here is the mechanical analogy to the two capacitor problem that actually fits.
Springs can store energy, Stored energy is proportional to the 'stretch' squared - just like a how capacitors store energy. In this analogy, the spring is our capacitor.
Flywheels can store kinetic energy. They also have very good bearings that cause minimal energy loss. This is our deliberately introduced inductance.
Ropes can transfer the forces (admittedly only when it tension, but we will make sure they stay in tension). Here I am making no statement if these are the wires, or the electric field. These are just a mechanical way to transfer the forces. When the rope is moving it is assumed to have minimal momentum. Consider the momentum of the rope as self-inductance - if the flywheel is very light enough (or things are arranged so no flywheel is needed), this might becomes important in in how the system behaves. But with any sensible flywheel it becomes unimportant.
The flywheel has a clamp on it. When the clamp is on the rope can't move, so no force can be transferred through a clamped flywheel. This is our switch.
The setup
See the lefthand side of the attached image.
We have two springs bolted to the ground, and a very heavy flywheel attached to the roof. We tie a rope to the lefthand spring, wrap it around the flywheel a couple of times so it can't slip, and tie it to the righthand spring, so it is nice and snug, but with minimal stretch on either spring. We check that the springs are both at equal height, and then clamp the flywheel in place.
This is our "zero energy in either capacitor" state.
Charging
With the flywheel clamped, we shorten the righthand rope till we get one unit of stretch in it. We tie the rope off, and trim it off nice and neat.
All of the energy is now in the righthand spring and the apparatus is armed and the experiment can proceed.
The experiment.
We release the clamp. The extra tension of the right hand side causes the flywheel to spin up in the clockwise direction, and the left-hand spring start to stretch.
What I expect to see, in an 'idea' world
When the springs get to equal length, in a perfect world half the energy in the flywheel, so it keeps the rope moving, slowly spinning down, finally coming to a stop when the LH spring now has ALL the energy in it. The cycle reverses, and continues on forever - i.e. the flywheel start turning counterclockwise, and ALL the energy is transferred back to the RH spring.
What I expect to see, in the real world
Energy gets lost due to friction and other effects (like rope slippage and stretch). Regardless of the weight and quality of the flywheel, springs and rope the system will settle down with only half the initial energy stored, shared equally between both springs. The only thing that is different is how long it takes to settle down.
That is it. there is no more to the "Two capacitor problem" than that.
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Electrodacus has agreed that there is power available to do work in the return circuit. And that this current flowed in and out of the capacitor terminals. As to whether this power went "through" the capacitor i guess we will never agree.
Am I correct ?
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That appears to be the gist of it.
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Here is the mechanical analogy to the two capacitor problem that actually fits.
Springs can store energy, Stored energy is proportional to the 'stretch' squared - just like a how capacitors store energy. In this analogy, the spring is our capacitor.
Flywheels can store kinetic energy. They also have very good bearings that cause minimal energy loss. This is our deliberately introduced inductance.
Ropes can transfer the forces (admittedly only when it tension, but we will make sure they stay in tension). Here I am making no statement if these are the wires, or the electric field. These are just a mechanical way to transfer the forces. When the rope is moving it is assumed to have minimal momentum. Consider the momentum of the rope as self-inductance - if the flywheel is very light enough (or things are arranged so no flywheel is needed), this might becomes important in in how the system behaves. But with any sensible flywheel it becomes unimportant.
The flywheel has a clamp on it. When the clamp is on the rope can't move, so no force can be transferred through a clamped flywheel. This is our switch.
You got to complicated and the analogy is wrong.
I guess you wanted to discuss the two parallel capacitor problem but more recently we are discussing a ideal voltage source and a 1000uF capacitor charged by that.
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Electrodacus has agreed that there is power available to do work in the return circuit. And that this current flowed in and out of the capacitor terminals. As to whether this power went "through" the capacitor i guess we will never agree.
Am I correct ?
The energy will go in a capacitor when it is charged and out of the capacitor when it is discharged. There will never be a case where energy goes through capacitor unless the capacitor is defective in the sense that plates are shorted.
Energy flows through a resistor / wire but flows in or out of a capacitor.
As I see that problem was not solved I will provide the correct results.
12V ideal voltage power supply no internal resistance
2x 0.5Ohm wires connecting a 1000uF capacitor also with no significant DC ESR compared to wire resistance so we ignore for simplicity.
Initial connection we get:
The supply providing 144mWs in order to charge the capacitor with 72mWs worth of energy 0.5 * 0.001F * 12V2
The other 72mWs end as heat on the wires that have 1Ohm of resistance.
If now after capacitor is charged we reverse the polarity of the power supply:
The supply will provide 288mWs and all of that will end up as heat on the wires so 288mWs of heat on wires while capacitor will still have the 72mWs stored from the initial connection.
So since capacitor was already fully charged even if it will be discharged and charged with reverse polarity all energy provided by the source will end up as heat on the wires.
If you disagree with this results we can discuss in more details what happens.
This is for the questions about AC supplies that a few of you asked. So this will be the first cycle.
After this first cycle if power supply keeps switching polarity (AC) then all energy will end up as heat/doing work as capacitor remains with same charged energy as it is discharged and then charged with reverse polarity for each half cycle.
So no energy passes through the capacitor. The current in the wires is due to capacitor being discharged (energy out) and charged (energy in) on each half cycle in case of AC.
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What happens to the energy when you charge a capacitor, then move the plates apart?
What about when charging an inductor?
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For me the question is entirely about DC and energy inside vs outside the wire. Nothing to do with switches, transmission lines, capacitors, inductors, transformer theory, antenna theory etc etc.
Which begs the question, is really such a thing as DC? For practical purposes we treat DC as a current that has existed since before the beginning of the universe and will last forever, but in reality whatever the current, it started at some definite point in time in the past and will stop in a foreseeable future. So, what we call DC is in fact just a very wide rectangular pulse. It is essentially transient as anything else, because impermanence seems to be the norm in nature.
If that is so, Heaviside must have thought that, well, if whatever rectangular pulse sent through a transmission line seems to show that energy is conveyed through the space between the wires, why not DC?
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Which begs the question, is really such a thing as DC? For practical purposes we treat DC as a current that has existed since before the beginning of the universe and will last forever, but in reality whatever the current, it started at some definite point in time in the past and will stop in a foreseeable future. So, what we call DC is in fact just a very wide rectangular pulse. It is essentially transient as anything else, because impermanence seems to be the norm in nature.
If that is so, Heaviside must have thought that, well, if whatever rectangular pulse sent through a transmission line seems to show that energy is conveyed through the space between the wires, why not DC?
Energy will not travel through the space between wires no matter if you have DC, or AC.
It is very easy to see that with DC as the energy storage is not involved.
At power up the energy storage is charged and at power down the energy storage is discharged. At all times electrical energy travels inside the wires.
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Can energy (or power) at an appropriate frequency propagate down a waveguide, inside the metal walls?
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What happens to the energy when you charge a capacitor, then move the plates apart?
What about when charging an inductor?
This is not that relevant to current discussion as distance between wires in Derek experiment is not changing.
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If you have a two-plate vacuum-dielectric capacitor, charge it up to a finite voltage, disconnect it, and leave it floating electrically, the charge remains constant when the plates move.
If, mechanically through insulating rods, you pull the plates apart, the capacitance decreases, and the stored energy E = Q2/(2C) increases. Therefore, you need to do work by pulling on the rods.
If, the "brake" on the rods releases, the plates will move towards each other and the rods can do work on an external mechanical system.
This is consistent with the well known effect that unlike charges (on the two plates) attract each other: Left to their own devices, the two plates would attract and crash into each other, and discharge themselves.
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Can energy (or power) at an appropriate frequency propagate down a waveguide, inside the metal walls?
We are getting way out of subject.
Nothing moves in Derke's experiment.
We have a long transmission line that is charged during transient thus the energy seen through lamp that is in series with the charging capacitors.
Energy flows in capacitors and not through capacitors. Energy flows through wires and resistors (that are also wires).
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We are getting way out of subject.
Nothing moves in Derke's experiment.
We have a long transmission line that is charged during transient thus the energy seen through lamp that is in series with the charging capacitors.
Energy flows in capacitors and not through capacitors. Energy flows through wires and resistors (that are also wires).
In Derek's experiment, energy or power reached the load before any electrical signals had time to propagate down the wires. Therefore, by experiment, it is proven that energy reached the load by some other path than through the wires.
Whether you like that result or not is irrelevant. The experiment stands. You may wish to come up with your private explanation for that result, and the rest of us may wish to adopt a different explanation. That is OK, but do not expect to persuade anyone to accept your view unless it conforms with the regularly understood laws of physics.
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In Derek's experiment, energy or power reached the load before any electrical signals had time to propagate down the wires. Therefore, by experiment, it is proven that energy reached the load by some other path than through the wires.
Whether you like that result or not is irrelevant. The experiment stands. You may wish to come up with your private explanation for that result, and the rest of us may wish to adopt a different explanation. That is OK, but do not expect to persuade anyone to accept your view unless it conforms with the regularly understood laws of physics.
That is the exact claim Derek made but it is wrong.
He completely ignores the transmission line energy storage properties.
I never claimed the result is incorrect and I showed the same result based on simulation.
The claim I made is that the conclusion he got that energy travels outside the wire is completely wrong.
What he is basically saying is that electrical energy travels through the 1m air gap and that is not true.
Electrical energy is electrical power integrated over time.
Electrical power is the product of electrical potential (voltage) and electrical current.
Electrical current is defined as flow of electrical charged particles (electrons in this case).
All of you seem to agree there are no electrons flying through that 1m gap from one wire to another.
So all of you agree that there is no electrical current through that 1m air gap.
Many of you mentioned the displacement current but that is a mathematical concept so the equations work not a real current.
At the time Maxwell (mathematician not physicist) has come up with the concept (fictional) displacement current nobody had any clue including Maxwell about the existence of electrons.
All energy in Derek's experiment traveled through wires and that should not be of any surprise to anyone that understands what a transmission line is.
Energy is not flowing through a capacitor but in or out of capacitor (in when charging and out when discharging).
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If you write words that agree with the textbooks, then you are wasting your time because we can all read the textbooks.
If you write words that disagree with the textbooks, then you are wasting your time because you won't convince anyone else and apparently you don't wish to learn.
Either way, you are wasting your time posting.
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If you write words that agree with the textbooks, then you are wasting your time because we can all read the textbooks.
If you write words that disagree with the textbooks, then you are wasting your time because you won't convince anyone else and apparently you don't wish to learn.
Either way, you are wasting your time posting.
Show me the textbooks that say "energy doesn't travel inside the wire" as that is exact quote Derek made.
Why anyone will use wires to transfer energy if they are not needed ?
Also a switch is a capacitor before is being closed so why will you need to close a switch if energy can be transferred through a capacitor.
I do see I'm wasting my time but the alternative is to let humanity fall even more in to idiocracy.
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""Many of you mentioned the displacement current but that is a mathematical concept so the equations work not a real current.""
Yes it is a real current.
In a parallel plate capacitor when charge accumulates on one plate the rising electric field ( result of the charge accumulation ) causes charge to leave the other plate (if there is a return path)
This is exactly the mechanism of displacement current. It requires a continuous charge buildup between the plates and this is current entering / leaving the plates.
http://teacher.pas.rochester.edu/phy122/lecture_notes/Chapter35/chapter35.html (http://teacher.pas.rochester.edu/phy122/lecture_notes/Chapter35/chapter35.html)
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""Many of you mentioned the displacement current but that is a mathematical concept so the equations work not a real current.""
Yes it is a real current.
In a parallel plate capacitor when charge accumulates on one plate the rising electric field ( result of the charge accumulation ) causes charge to leave the other plate (if there is a return path)
This is exactly the mechanism of displacement current. It requires a continuous charge buildup between the plates and this is current entering / leaving the plates.
http://teacher.pas.rochester.edu/phy122/lecture_notes/Chapter35/chapter35.html (http://teacher.pas.rochester.edu/phy122/lecture_notes/Chapter35/chapter35.html)
Maxwell had no idea what an electron is ?
The displacement current is fictional (to make observed results fit the math).
If energy was to pass through a capacitor as it does through a wire (with or without resistance) then no energy could ever be stored in a capacitor.
Those 72mJ did not pass through the 1000uF capacitor but in to it and will remain there as stored energy.
That is why when you reverse the voltage supply wires you no longer get anything as it is already charged and all energy delivered by the supply 288mJ will end up dissipated as heat in the wires.
So before capacitor is charged
144mJ = 72mJ on wires as heat + 72mJ in the capacitor as stored energy.
After capacitor is charged there is no energy transfer and if you reverse the supply polarity you have
288mJ = 288mJ on wires as heat and that is all then no more energy flow until you reverse the wires again and then it will again be 288mJ = 288mJ as heat.
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Those 72mJ did not pass through the 1000uF capacitor but in to it and will remain there as stored energy.
Where in the capacitor is the energy stored? What is it's physical location?
288mJ will end up dissipated as heat in the wires.
I believe you have said that this heat energy is radiated away in the form of infrared radiation. What is the difference between infrared radiation and radio waves?
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Where in the capacitor is the energy stored? What is it's physical location?
Will that even be relevant or is the important part that energy is stored ?
The electron imbalance between the two plates is the stored energy. You start with equal amount of electrons on both plates and that is a discharged capacitor and in a charged capacitor you will have excess of electrons "parked" in one plate and deficit of electrons in the other plate.
Even when you remove the capacitor from the circuit this imbalance remains and that is how energy is stored.
To discharge you can just short the two plates and electrons form the plate with excess will travel through the wire to the plate with deficit until number of electrons is again equal on both plates.
I believe you have said that this heat energy is radiated away in the form of infrared radiation. What is the difference between infrared radiation and radio waves?
That heat is radiated by a resistive device like the wires in this example and a pure resistor is not storing electrical energy.
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Can energy (or power) at an appropriate frequency propagate down a waveguide, inside the metal walls?
Yes.
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Where in the capacitor is the energy stored? What is it's physical location?
I believe you have said that this heat energy is radiated away in the form of infrared radiation. What is the difference between infrared radiation and radio waves?
You didn't answer either question. Are you saying there is no physical location?
You didn't answer the second question at all either.
Do you know what a discussion is?
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Can energy (or power) at an appropriate frequency propagate down a waveguide, inside the metal walls?
Yes.
Is it a trick question? Inside .. ???
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You didn't answer either question. Are you saying there is no physical location?
You didn't answer the second question at all either.
Do you know what a discussion is?
??? Have I not mentioned electrons that are a physical particle and they have a charge.
It is almost like you are denying that capacitors are an energy storage device.
We are discussing electrical energy not thermal energy.
The infrared photons traveling from one wire to another (or one capacitor plate to another) have nothing to do with electric current.
Yes a few infrared photons from one wire will get to the other wire and slightly heat the other wire but is completely irrelevant for this discussion.
Even if the wire is so hot that it glows in visible light it will not transfer electrical energy to the other wire. It will need to be UV or above in order to displace electrons.
Radio waves are on the other side of the infrared closer to DC.
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??? Have I not mentioned electrons that are a physical particle and they have a charge.
It is almost like you are denying that capacitors are an energy storage device.
So charge equates to energy?
We are discussing electrical energy not thermal energy.
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Radio waves are on the other side of the infrared closer to DC.
I sense a contradiction. Sounds like infrared and radio waves are both in the realm of electrical energy.
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??? Have I not mentioned electrons that are a physical particle and they have a charge.
It is almost like you are denying that capacitors are an energy storage device.
So charge equates to energy?
We are discussing electrical energy not thermal energy.
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Radio waves are on the other side of the infrared closer to DC.
I sense a contradiction. Sounds like infrared and radio waves are both in the realm of electrical energy.
Charge Q=C*V and energy is 0.5 * Q * V
If you have an ideal circuit meaning no resistance then a 12V ideal voltage source charging an 1000uF ideal capacitor will result in 72mJ being transferred from source to capacitor with no loss so there is no infrared. Total electrical energy in a closed system before charging the capacitor and after is the same and no work was done.
With resistance half of the energy so another 72mJ will be lost as heat so 50% efficiency.
With a more efficient DC-DC circuit to transfer energy you can get to say 90% efficiency so 72mJ in capacitor plus 8mJ wasted as heat.
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Charge held by a capacitor is potential energy.
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Can energy (or power) at an appropriate frequency propagate down a waveguide, inside the metal walls?
Yes.
Is it a trick question? Inside .. ???
I think it was, but a correct answer is that current/energy is propagating inside the metal walls. Now the skin depth is typically very small, but it's still inside.
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Can energy (or power) at an appropriate frequency propagate down a waveguide, inside the metal walls?
Yes.
Is it a trick question? Inside .. ???
I think it was, but a correct answer is that current/energy is propagating inside the metal walls. Now the skin depth is typically very small, but it's still inside.
My statement was ambiguous: I suggested that power propagates down the length of a waveguide, in the air/vacuum that is included in the guide's cross-section, inside the metal walls, in the form of traveling fields.
At those frequencies, the skin depth is very thin, and current does flow within that depth in the walls to establish the boundary conditions.
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When you believe that light transferring energy and momentum is the biggest lie told by mainstream science, you'll obviously need a wire to transfer your energy. And when the wire ends, how does the energy go from one wire to the next? Magic, pure magic!
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When you believe that light transferring energy and momentum is the biggest lie told by mainstream science, you'll obviously need a wire to transfer your energy. And when the wire ends, how does the energy go from one wire to the next? Magic, pure magic!
Electrical energy will only pass through a gap in circuit if there is enough potential energy (high enough voltage) so that electrons can jump that gap.
At 1m air gap you will need almost 3 and a half million volt.
You confuse energy storage in a capacitor with energy going through a capacitor.
When you connect two wires to the battery so not close circuit just two parallel wires unconnected to echoder each wire on one terminal of the battery the wires will experience an charge imbalance with one wire getting extra electrons from battery and the other donating the same amount on the other terminal.
If you disconnect this wires from battery they will keep their charge imbalance and the battery will now have less stored energy than it had before.
Keeping the wires after the charge was transferred (a few nanoseconds) will not make any change.
Connecting the wires again will also not make any change unless you short the wires for a moment so that excess electrons from one wire are donated to the wire that has the deficit basically discharging the stored energy.
Is there any part from the above that you think is untrue ?
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If you write words that agree with the textbooks, then you are wasting your time because we can all read the textbooks.
If you write words that disagree with the textbooks, then you are wasting your time because you won't convince anyone else and apparently you don't wish to learn.
Either way, you are wasting your time posting.
Show me the textbooks that say "energy doesn't travel inside the wire" as that is exact quote Derek made.
Here ya go.
It is evident that the power flow is through the empty space surrounding the circuit, the conductors of the circuit acting as guiding elements. From the circuit point of view we usually think of the power as flowing through the wires but this is an oversimplification and does not represent the actual situation.
J.D. Kraus, Electromagnetics Chapter 10, P.418
http://xn--webducation-dbb.com/wp-content/uploads/2019/09/McGraw-Hill-electrical-and-electronic-engineering-series-John-D.-Kraus-Keith-R.-Carver-Electromagnetics-McGraw-Hill-1981.pdf (http://xn--webducation-dbb.com/wp-content/uploads/2019/09/McGraw-Hill-electrical-and-electronic-engineering-series-John-D.-Kraus-Keith-R.-Carver-Electromagnetics-McGraw-Hill-1981.pdf)
Let me guess? You're gonna start railing about how Kraus, THE Kraus, 'doesn't understand capacitors and energy storage'? :-DD :-DD :-DD
You really ought to familiarize yourself with the textbooks Derek is using as his source material. Kraus' diagram of the Poynting energy flow on P.417 is very standard in advanced electromagnetics.
https://en.wikipedia.org/wiki/John_D._Kraus (https://en.wikipedia.org/wiki/John_D._Kraus)
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Electrical energy will only pass through a gap in circuit if there is enough potential energy (high enough voltage) so that electrons can jump that gap.
Is there any part from the above that you think is untrue ?
Do you honestly think that a microwave oven cooks your food by making electrons jump the gap from the magnetron to a potato? How does the electrical energy get there across the air gap?
http://ffden-2.phys.uaf.edu/104_spring2004.web.dir/arts_mcnulty/howmicrowaveovenswork.htm (http://ffden-2.phys.uaf.edu/104_spring2004.web.dir/arts_mcnulty/howmicrowaveovenswork.htm)
There is plenty of electrical energy crossing gaps but not an electron in sight - except inside the potato. There is a thing called... di... duh...diissssplacement currrrreeennntttt. Electromagnetic radiation? Photons? Ever heard of them? >:D
Now, of course you might try to say that microwave radiation doesn't belong in circuit theory... but why not? Maxwell's Equations are perfect descriptors of both phenomena. The consequence is freedom from the confines of wires. It's okay though - Tesla was going wireless before the world had wires. :P
https://www.youtube.com/watch?v=FWCN_uI5ygY (https://www.youtube.com/watch?v=FWCN_uI5ygY)
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When you believe that light transferring energy and momentum is the biggest lie told by mainstream science, you'll obviously need a wire to transfer your energy. And when the wire ends, how does the energy go from one wire to the next? Magic, pure magic!
When you believe that wires transferring energy and momentum is the biggest lie told by mainstream science, you'll obviously need vacuum to transfer your energy. And when the vacuum ends, how does the energy go from one vacuum to the next? Magic, pure magic!
Yes science looks like magic to people without knowledge. It's ok. You can learn if you want to, but you have to stop mocking it first.
It is evident that the power flow is through the empty space surrounding the circuit, the conductors of the circuit acting as guiding elements. From the circuit point of view we usually think of the power as flowing through the wires but this is an oversimplification and does not represent the actual situation.
J.D. Kraus, Electromagnetics Chapter 10, P.418
a) He didn't justify, in any way, how exactly it is an oversimplification. Sure, it's simpler.
b) He didn't justify, in any way, how it does not represent the actual situation.
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Yes science looks like magic to people without knowledge. It's ok. You can learn if you want to, but you have to stop mocking it first.
You are the antimatter version of the Aetherist. He thinks that without an ether locality is violated. You, on the other hand, think that space is empty and needs some kind of material support for locality to exist.
If you some day meet him, you'll annihilate each other and release at least two very energetic and momentous photons.
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J.D. Kraus, Electromagnetics Chapter 10, P.418
http://xn--webducation-dbb.com/wp-content/uploads/2019/09/McGraw-Hill-electrical-and-electronic-engineering-series-John-D.-Kraus-Keith-R.-Carver-Electromagnetics-McGraw-Hill-1981.pdf (http://xn--webducation-dbb.com/wp-content/uploads/2019/09/McGraw-Hill-electrical-and-electronic-engineering-series-John-D.-Kraus-Keith-R.-Carver-Electromagnetics-McGraw-Hill-1981.pdf)
Let me guess? You're gonna start railing about how Kraus, THE Kraus, 'doesn't understand capacitors and energy storage'? :-DD :-DD :-DD
You really ought to familiarize yourself with the textbooks Derek is using as his source material. Kraus' diagram of the Poynting energy flow on P.417 is very standard in advanced electromagnetics.
https://en.wikipedia.org/wiki/John_D._Kraus (https://en.wikipedia.org/wiki/John_D._Kraus)
The guys that lived way before electron was discovered (John Henry Poynting 1884) could only guess what happens and it guessed wrong.
Derk has significant lack in understanding energy and energy storage as it clearly showed in the video about faster than wind direct downwind vehicle and now here with the transmission line.
Derek only so called "proof" was to show that some current flows through the load before it had time to travel around the wire. But that is no proof if you understand that two parallel wires have capacitance and understand that capacitor is a energy storage device.
Electrical current in modern times (after the electron was discovered so after both Poynting and Maxwell where alive) is the flow of a charged particle (electron in this particular case but can also be ions).
As there is no electron flow outside the wires in this example at 20V and 1m of air between wires all energy both at constant DC but also during transient is transported through wires only.
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For all those that think that energy is traveling outside wires I have a question.
Can you shield against electric and magnetic fields ?
If yes then can I shield the battery or the load so that energy can no longer get to load ?
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It's been two threads and things are still running in circles.
While I do not agree with everything electrodacus has written, I understand his question here: I don't think I've really seen a proper answer (except possibly in the early stages of the first thread), neither on here or in Veritasium's videos, about the role of wires and why we actually needed them (but we do for sure.)
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It's been two threads and things are still running in circles.
While I do not agree with everything electrodacus has written, I understand his question here: I don't think I've really seen a proper answer (except possibly in the early stages of the first thread), neither on here or in Veritasium's videos, about the role of wires and why we actually needed them (but we do for sure.)
Is all this about why we need wires to transmit energy? I haven't followed along enough to know what the hell is going on, so I probably should keep my mouth shut. :-DD
But no, we don't need wires to get energy from point A to point B. You can do it wirelessly. Say you have an electric lawn mower: you need electric energy to power it. You can get it from a battery on-board (short wires). Or you can use long wires and use an extension cord back to the mains outlet.
Or ...
You (in theory) could use a high powered rf beam of a certain frequency to get the energy to the mower. You'd either have to blanket the yard with rf, (very inefficient) or have a tracking system so both antennas have the maximum gain aligned all the time you are mowing. Then an on-board power supply system to turn the rf into DC. We're forgetting about what the high power rf beam will do to your body at the moment. ???
Yea it's possible but go ahead and try to build the wireless lawn mower power system and see how long it is before you go back to good old wires ...
:palm:
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One reason I mentioned waveguides as a way of transferring power:
Another group at my university needed to transfer power to the top of a 1 MV accelerator, to operate the cathode and auxiliary electronics referenced to the cathode.
They used a dielectric waveguide: like a metal one, but with a high dielectric constant ceramic as the envelope. It was long enough to hold off the 1 MV DC potential difference between the terminal and ground.
A CW magnetron down at ground potential launched power up the guide to a receiving antenna, where a simple (high speed) rectifier provided DC power for the terminal.
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It's been two threads and things are still running in circles.
While I do not agree with everything electrodacus has written, I understand his question here: I don't think I've really seen a proper answer (except possibly in the early stages of the first thread), neither on here or in Veritasium's videos, about the role of wires and why we actually needed them (but we do for sure.)
Wires are wave guides.
Fields emanate from charges - charges live in conductors and the field disturbances carry the energy which influences nearby charges or charges in other conductors. I posted a video a few replies ago that illustrates this. And Kraus in Chapter 10 makes this explicit when he describes the Poynting Vector in circuit theory terms in a chapter about wave guides (including showing what happens when an infinite conducting sheet is placed between battery and load and how this changes the shape of the Poynting vector field).
And Feynman includes a lecture on this:
https://www.feynmanlectures.caltech.edu/II_24.html (https://www.feynmanlectures.caltech.edu/II_24.html)
In the last chapter we studied what happened to the lumped elements of circuits when they were operated at very high frequencies, and we were led to see that a resonant circuit could be replaced by a cavity with the fields resonating inside. Another interesting technical problem is the connection of one object to another, so that electromagnetic energy can be transmitted between them. In low-frequency circuits the connection is made with wires, but this method doesn’t work very well at high frequencies because the circuits would radiate energy into all the space around them, and it is hard to control where the energy will go. The fields spread out around the wires; the currents and voltages are not “guided” very well by the wires. In this chapter we want to look into the ways that objects can be interconnected at high frequencies. At least, that’s one way of presenting our subject.
The 'energy always travels in wires' people cannot hope to design, let alone explain, how a high-frequency waveguide works and how to CONTROL where energy is going.
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It's been two threads and things are still running in circles.
While I do not agree with everything electrodacus has written, I understand his question here: I don't think I've really seen a proper answer (except possibly in the early stages of the first thread), neither on here or in Veritasium's videos, about the role of wires and why we actually needed them (but we do for sure.)
You need the wires to guide the energy from the source to the load.
For the case of the battery powering the light, there is very little potential drop along the good conductor wires, so they set up the potential across the load.
The power dissipated in the load is VI. For DC, based on conservation of charge, the current has to be the same everywhere in the circuit. So the wires are also needed to provide a circuit for current to flow.
This is why you need the wires for DC. It says nothing about where the energy is "flowing". Based on conservation of energy, it flows out of the volume of space surrounding the battery and into the volume of space surrounding the load.
The energy flow could be inside the wires or through the empty space.
For AC it's a slightly different story.
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The 'energy always travels in wires' people cannot hope to design, let alone explain, how a high-frequency waveguide works and how to CONTROL where energy is going.
A lot of people still live in the 1800's when electrons did not exist.
We are discussing a DC transmission line yet Derek prefered to just concentrate at the first 65ns of transient that he has no understanding of.
Even if we ignore (because you do not understand) the energy storage that is responsible for the transient part what happens at steady state after that transient as there is no longer any fluctuating electric or magnetic field they are just constant. Yet lamp/resistor is using energy.
Answer the earlier question about shielding against magnetic and electric fields either the battery or the load so that no energy can be transferred ?
I can simply made a cut / gap in the conductor 1mm will be more than sufficient for 20V and I can stop the energy flow between the battery and lamp.
My 1mm gap example should pe proof enough for any sane person that energy travels through wires but somehow with no proof you like to think there energy magically travels outside the wire.
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A lot of people still live in the 1800's when electrons did not exist.
We are discussing a DC transmission line yet Derek prefered to just concentrate at the first 65ns of transient that he has no understanding of.
You still don't get the point of the experiment? To demonstrate the radiation of energy during a transient getting there faster than it could traveling through the wire alone which means some energy was radiated from the switch to the lamp. Ben Watson's simulation demonstrates this explicitly. I'm pretty sure the people who designed HFSS know about electrons... :-DD :-DD :-DD
Even if we ignore (because you do not understand) the energy storage that is responsible for the transient part what happens at steady state after that transient as there is no longer any fluctuating electric or magnetic field they are just constant. Yet lamp/resistor is using energy.
The Poynting vector is still established. See Kraus Chapter 10.
Answer the earlier question about shielding against magnetic and electric fields either the battery or the load so that no energy can be transferred ?
I can simply made a cut / gap in the conductor 1mm will be more than sufficient for 20V and I can stop the energy flow between the battery and lamp.
My 1mm gap example should pe proof enough for any sane person that energy travels through wires but somehow with no proof you like to think there energy magically travels outside the wire.
And you'd be utterly mystified if the frequency of the voltage source changed in any way and the lamp could still illuminate.
I have another way. Put an infinite conducting sheet between the source and the load. This is actually a far more effective shielding mechanism. E-fields can still cross your gap. E-fields cannot get through the infinite sheet. ExH is zero. No energy flow. Again, see Kraus Chapter 10.
I would never use an air gap if, hypothetically, I wanted to guarantee no energy ever crosses from one circuit to another. How do you think coaxial cables or rectangular waveguides work?!?! :o
The only thing special about DC is that the waveguide has to guide the wave the whole way at steady-state.
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You need the wires to guide the energy from the source to the load.
For the case of the battery powering the light, there is very little potential drop along the good conductor wires, so they set up the potential across the load.
The power dissipated in the load is VI. For DC, based on conservation of charge, the current has to be the same everywhere in the circuit. So the wires are also needed to provide a circuit for current to flow.
This is why you need the wires for DC. It says nothing about where the energy is "flowing". Based on conservation of energy, it flows out of the volume of space surrounding the battery and into the volume of space surrounding the load.
The energy flow could be inside the wires or through the empty space.
For AC it's a slightly different story.
There is no difference between AC and DC in the sense that energy travels through wires. Else why will you invest in massive transmission lines for AC.
In a battery ions are involved so is maybe best to use a charged capacitor as the energy source to better understand what happens.
The capacitor plates are basically wires and you have excess of electrons in one plate and deficit of electrons on the other plate.
If you connect a wire between the two capacitor plates you allow the electrons to travel from the plate with excess of electrons to the plate with deficit so that they both become neutral as stored energy is discharged.
That flow of electrons is electrical current and the density of excess electrons is the electrical potential (voltage). The product of this two is electrical power and the integral of electrical power over time is electrical energy.
Since you can not have electron flow outside wires so no electrical current you can not have energy flow outside wires.
Let me know what part of my explanation you think is wrong.
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What can we actually measure?
We can measure fields based on what they do to charges. We can make tiny power sensors with resistors and thermistors for example and measure what happens to them when we position them at a point in space. If they get hot, we can say there is energy at that point in space.
So we can develop a theory based on the concept that there is an electromagnetic energy density at any given point in space, based on how much energy is required to move charges or currents to this point in space from infinity. We can calculate what this energy density would be based on the E and B field. We can measure it with the thermistor power sensor or something similar and convince ourselves that the measurements agree with the theory.
Based on this theory and the definition of electromagnetic energy density, we can measure energy outside of a wire conducting a time varying current. Similarly, we can measure energy in the empty space of a waveguide. We can measure the energy of a radio signal in free space.
We can extend this to a theory of the "flow" of electromagnetic energy and verify it by measurement.
For a wire carrying a DC current it's a slightly different story.
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You still don't get the point of the experiment? To demonstrate the radiation of energy during a transient getting there faster than it could traveling through the wire alone which means some energy was radiated from the switch to the lamp. Ben Watson's simulation demonstrates this explicitly. I'm pretty sure the people who designed HFSS know about electrons... :-DD :-DD :-DD
You made no mention about energy storage ? Do you understand that two parallel conductors are a capacitor? and that a capacitor is an energy storage device?
Have you see the Spice simulation I made for a transmission line where I turn the switch ON for just 30ns then turn it OFF
[attach=1]
The green graph is the power provided by the battery and with magenta power dissipated by the lamp/resistor.
How come total energy provided by battery exactly matches the energy arriving at the lamp (most of it quite some time after the switch was OFF) and the delta in energy is found as heat loss in the wire.
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The 'energy always travels in wires' people cannot hope to design, let alone explain, how a high-frequency waveguide works and how to CONTROL where energy is going.
It's in the conductor, and follows the conductor. And if you put holes, you'll start radiating the energy.
Can the 'energy always travels in vacuum' people explain why you need conductors in waveguides? (in dielectrics, you have polarization current instead of current)
Or, since waveguides are wires, in circuits?
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I think one simple (or simplified) way to look at things is that delivered power is the product of voltage and current. The voltage comes from the E field. A strong E field can be propagated by wires and can give a large power delivery, but it has to travel the length of the wires to arrive. A weaker E field can be propagated across the air gap and can give a much smaller power delivery, but it can arrive faster. This qualitatively aligns with Derek's experiment.
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The 'energy always travels in wires' people cannot hope to design, let alone explain, how a high-frequency waveguide works and how to CONTROL where energy is going.
It's in the conductor, and follows the conductor. And if you put holes, you'll start radiating the energy.
Can the 'energy always travels in vacuum' people explain why you need conductors in waveguides? (in dielectrics, you have polarization current instead of current)
Or, since waveguides are wires, in circuits?
There are dielectric waveguides for microwave frequencies, usually using a high-dielectric-constant ceramic for the walls.
https://farside.ph.utexas.edu/teaching/jk1/Electromagnetism/node118.html
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I think one simple (or simplified) way to look at things is that delivered power is the product of voltage and current. The voltage comes from the E field. A strong E field can be propagated by wires and can give a large power delivery, but it has to travel the length of the wires to arrive. A weaker E field can be propagated across the air gap and can give a much smaller power delivery, but it can arrive faster. This qualitatively aligns with Derek's experiment.
You only have an E field when you have a delta in charge. So you need first the extra electron then you have the E field.
There is a very strong E field between the plates of a charged capacitor so how come that no energy is transferred from one plate to the other basically discharging the capacitor?
What you consider that weeker electric field it is due to delta in charge between the two wires. Those parallel wires form a capacitor that is an energy storage device and energy from battery flows in that capacitor (in and not trough) and since energy travels through wires from battery to that capacitor and the lamp or resistor is also a wire the energy to charge the capacitor will flow through them.
If energy will not have traveled through wire then wire thickness will not have matter. You could transfer a lot of power from battery to lamp/load with a very thin copper wire that just shows the power the direction it needs to travel :)
It is like saying that energy from a compressed air tank is delivered to a compressed air tool outside the hose.
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Those parallel wires form a capacitor that is an energy storage device and energy from battery flows in that capacitor (in and not trough) and since energy travels through wires from battery to that capacitor and the lamp or resistor is also a wire the energy to charge the capacitor will flow through them.
I see. So if you have three capacitors in series, the energy to charge the middle capacitor flows through the outer capacitors. If energy does not flow through the outer capacitors the middle capacitor will not be charged. If energy does flow through the outer capacitors the middle capacitor will be charged.
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I see. So if you have three capacitors in series, the energy to charge the middle capacitor flows through the outer capacitors. If energy does not flow through the outer capacitors the middle capacitor will not be charged. If energy does flow through the outer capacitors the middle capacitor will be charged.
At no point in time any energy flows through the capacitor.
3 or more capacitors in series are no different from a single capacitor.
If all 3 capacitors are 1000uF when connected in series will look like only a single 333uF capacitor.
Imagine a parallel plate capacitor with 5mm between plates and say it has 1uF capacitance.
Then insert a very thin 0.001mm plate between the two plates with same area. This will be the equivalent of two capacitors in series so from outside the capacitor will still look like a single 1uF capacitor (just slightly more due to that 0.001mm reduction in distance between plates).
If that plate is instead 2.5mm thick then from outside it will look like a 2uF capacitor but it is actually two 4uF capacitors in series.
Adding multiple plates will not change the fact that from outside any number of series connected capacitors will appear as a single capacitor and it can be charged.
There will be a current flow inside the middle plate as electrons from one side of the plate travel to the other side (it is a short distance even for the 2.5mm thick plate) but there is still a current flow as electrons that where already there travel from one side of the plate to the other while capacitor is being charged.
This charge separation remains there after you remove the battery so it is stored energy not used energy. Now this capacitor or series capacitors can be used to do any sort of work you want with that stored energy.
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If energy will not have traveled through wire then wire thickness will not have matter. You could transfer a lot of power from battery to lamp/load with a very thin copper wire that just shows the power the direction it needs to travel :)
It is like saying that energy from a compressed air tank is delivered to a compressed air tool outside the hose.
Run your "compressed air" analogy, with a hard vacuum in the pipe instead, supplied from a very large cylinder, with a very good vacuum pump.
The pipes are literally delivering nothing, so energy can be extracted from the air around the pipes.
If your pipe is too tiny, you won't get enough 'nothing' to run your tools efficiently, but it won't matter if the pipe is too large.
Pipes are essential to the system working, even though they carry no usable energy.
The energy is outside of the pipes, but the pipes dictate where you can extract the energy - no pipe, you can't do work.
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Run your "compressed air" analogy, with a hard vacuum in the pipe instead, supplied from a very large cylinder, with a very good vacuum pump.
The pipes are literally delivering nothing, so energy can be extracted from the air around the pipes.
If your pipe is too tiny, you won't get enough 'nothing' to run your tools efficiently, but it won't matter if the pipe is too large.
Pipes are essential to the system working, even though they carry no usable energy.
The energy is outside of the pipes, but the pipes dictate where you can extract the energy - no pipe, you can't do work.
I guess I need to be more explicit as you did not understood any of my compressed air analogies.
Air molecules are the electrons and compressed air cylinder with with two chambers represents the capacitor.
A discharged capacitor is like an air cylinder that has the same pressure in both chambers say atmospheric pressure but it can be any value as long as the same number of air particles are found in both chambers and inside the hose's.
Now you can charge the capacitor using say mechanical energy to pump air from one chamber to the other one. You can use the same pump in reverse to do work with the stored energy.
Since now there are more air molecules (electrons) in one chamber (capacitor plate) than the other one you have stored energy.
If you connect the two chambers through a pipe air molecules (electrons) will flow (electric current) from the higher pressure chamber to the other one until pressure (voltage) is equalized.
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I see. So if you have three capacitors in series, the energy to charge the middle capacitor flows through the outer capacitors. If energy does not flow through the outer capacitors the middle capacitor will not be charged. If energy does flow through the outer capacitors the middle capacitor will be charged.
At no point in time any energy flows through the capacitor.
What I wrote was not a question. It follows directly from what you wrote. If you contradict it, you contradict yourself.
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Now, I am well aware that after the Faraday vs. Kirchhoff threads I am the last person who should point this out but...
This short video summarize perfectly this thread:
https://www.youtube.com/watch?v=LnLDMqPBeKQ (https://www.youtube.com/watch?v=LnLDMqPBeKQ)
As for the ongoing discussion about the role of the wires, I would like to point out the role of surface charges.
To simplify: we need surface charge of different polarities at the resistor ends. The surface charge distribution will cause conduction electrons inside the resistor to be accelerated and then collide with the ion lattice, releasing heat (classical ED). Now, why don't the electrons that arrive on the other side of the resistor cancel the positive charge on the surface on that side? Because we have the wire to scoop them away (and to replenish on the other side before they enter).
The role of the wires is that to provide the boundary conditions that will keep the surface charge distribution at the resistor. And what makes the charge distribute on the surface of the conductors and resistor? The fields all around them. The electric field inside is the result of the surface charge distribution.
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Run your "compressed air" analogy, with a hard vacuum in the pipe instead, supplied from a very large cylinder, with a very good vacuum pump.
The pipes are literally delivering nothing, so energy can be extracted from the air around the pipes.
If your pipe is too tiny, you won't get enough 'nothing' to run your tools efficiently, but it won't matter if the pipe is too large.
Pipes are essential to the system working, even though they carry no usable energy.
The energy is outside of the pipes, but the pipes dictate where you can extract the energy - no pipe, you can't do work.
I guess I need to be more explicit as you did not understood any of my compressed air analogies.
Air molecules are the electrons and compressed air cylinder with with two chambers represents the capacitor.
A discharged capacitor is like an air cylinder that has the same pressure in both chambers say atmospheric pressure but it can be any value as long as the same number of air particles are found in both chambers and inside the hose's.
Now you can charge the capacitor using say mechanical energy to pump air from one chamber to the other one. You can use the same pump in reverse to do work with the stored energy.
Since now there are more air molecules (electrons) in one chamber (capacitor plate) than the other one you have stored energy.
If you connect the two chambers through a pipe air molecules (electrons) will flow (electric current) from the higher pressure chamber to the other one until pressure (voltage) is equalized.
What? This was in reply to your post:
It is like saying that energy from a compressed air tank is delivered to a compressed air tool outside the hose.
I was pointing out a system where a vacuum tank and pipes can deliver energy to an air-powered tool, using the air outside of the hose, by literally delivering nothing in the pipes.
You just don't like it because the open air molecules have no 'special' added energy, and they also have no 'special' added energy when in the vacuum pipes on the way back to the tank & pump, pipes are required, those pipes are empty, and yet somehow still deliver energy to the tool.
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What I wrote was not a question. It follows directly from what you wrote. If you contradict it, you contradict yourself.
It can not follow from what I wrote as I never wrote that energy flows through a capacitor but energy flows in or out of a capacitor.
Is like you are unable to see a certain color (that color will be called energy storage).
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I was pointing out a system where a vacuum tank and pipes can deliver energy to an air-powered tool, using the air outside of the hose, by literally delivering nothing in the pipes.
You just don't like it because the open air molecules have no 'special' added energy, and they also have no 'special' added energy when in the vacuum pipes on the way back to the tank & pump, pipes are required, those pipes are empty, and yet somehow still deliver energy to the tool.
All analogies have limitations and if I used an analogy then it was used to explain just a single aspect and you can not go with the analogy further.
That vacuum pipe will carry air molecules that are taken from ambient so you just reversed where the source of air particle is the polarity.
There is a delta in pressure between the vacuum (absence of air molecules) and ambient. So you vacuum cylinder will be filled with air molecules at same concentration as ambient when all energy is used up.
The vacuum analogy is not great for a capacitor as you can never remove all electrons from a plate not even close.
So your empty pipe is no longer empty when it delivers energy.
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It's been two threads and things are still running in circles.
While I do not agree with everything electrodacus has written, I understand his question here: I don't think I've really seen a proper answer (except possibly in the early stages of the first thread), neither on here or in Veritasium's videos, about the role of wires and why we actually needed them (but we do for sure.)
Wires are wave guides.
Fields emanate from charges - charges live in conductors and the field disturbances carry the energy which influences nearby charges or charges in other conductors. I posted a video a few replies ago that illustrates this. And Kraus in Chapter 10 makes this explicit when he describes the Poynting Vector in circuit theory terms in a chapter about wave guides (including showing what happens when an infinite conducting sheet is placed between battery and load and how this changes the shape of the Poynting vector field).
And Feynman includes a lecture on this:
https://www.feynmanlectures.caltech.edu/II_24.html (https://www.feynmanlectures.caltech.edu/II_24.html)
IIRC, I did post a link to this very lecture quite a while ago.
I suggest anyone not fricking tired of it already to go read the 80 pages of the other thread and then this one just to 1/ learn a few things and 2/ realize how this has been running in circles for ages now. I'm sure it can keep going almost forever though. It's almost as though perpetual motion has finally been resolved.
For a wire carrying a DC current it's a slightly different story.
It is, and as we discussed a long time ago already, the unfortunate part of the whole Veritasium example (not the theory itself) is that the transient phenomenon upon closing the switch and the operation at steady state have been blurred somewhat. I don't think this point has been fully addressed in the second video. But the problem is that having a transient phase is the only straightforward way of making the point. *Insert coin*
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It is, and as we discussed a long time ago already, the unfortunate part of the whole Veritasium example (not the theory itself) is that the transient phenomenon upon closing the switch and the operation at steady state have been blurred somewhat. I don't think this point has been fully addressed in the second video. But the problem is that having a transient phase is the only straightforward way of making the point. *Insert coin*
The claim Derek (Veritasium) made that "energy doesn't travel through wires" is more than ridiculous.
His so called "evidence" was to show that some energy arrives at the lamp before the electron wave has the time to travel the entire wire distance.
For anyone that properly understand what a capacitor is the obvious answer to that small amount energy that arrives earlier is the loss in wires/lamp while charging that capacitance and it is accurately described and calculated by the lumped model.
You can get rid of this capacitance charging through the lamp by shielding just one side of the circuit bypassing the lamp then it can be seen that no energy arrives at the lamp before the electron wave gets there.
This will show that capacitance charging is what produces an electron flow through the lamp and also shows that energy is delivered through wires else no energy could arrive at the lamp as it is shielded from electric field.
I need to apologise for being a bit rude in some of my replayes as most of you rely on you learned from others and unfortunately the way this is teached in schools is not up to current understanding.
I do not think it is OK for Maxwell to still be relevant when at his time the structure of the atom was not discovered. Yes the equations still work but the conclusions that can be drawn from them can be very misleading.
To be fair to schools I do not think any engineering school teaches that energy travels outside the wires but it also not makes that clear enough so that a youtube celebrity can not convince them the opposite is true.
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You can get rid of this capacitance charging through the lamp by shielding just one side of the circuit bypassing the lamp then it can be seen that no energy arrives at the lamp before the electron wave gets there.
You mean "then it should show". Unless you've actually done this then it is supposition, but you continue to pretend things like this are fact. That's one of the things that rubs people up the wrong way.
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""It can not follow from what I wrote as I never wrote that energy flows through a capacitor but energy flows in or out of a capacitor.""
So can energy both flow in and out of a capacitor at the same time ?
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You mean "then it should show". Unless you've actually done this then it is supposition, but you continue to pretend things like this are fact. That's one of the things that rubs people up the wrong way.
I have not done this but I do not need to do something if I understand how it works. You can do accurate predictions about a system if you understand how it works.
You only need to shield one of the two sides of the circuit either left or right so in the Derek example just 10m of shielding from the corner (where the pipe is bent 90 degree) up to that 1.1k Resistor but you connect the shield on the other side of the resistor not on the same side you shielded.
Then there will still be a capacitance between the shield and the lower wire/pipe but the current to charge that capacitance will bypass the resistor as you connect the shield on the other side of the resistor.
So energy will start flowing through resistor but only after about 65ns needed for electron wave to reach the resistor through wire.
Will seeing that the above is true convince you that energy travels through wires ?
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So can energy both flow in and out of a capacitor at the same time ?
No.
Energy flows in capacitor while it is being charged and stays there until you discharge the capacitor then energy flows out.
Even if you have a resistor in parallel with the capacitor energy can only flow in or out of capacitor not both at the same time.
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You can do accurate predictions about a system if you understand how it works.
It's still 'should' and not 'will', particularly when it is a) an off-the-wall suggestion no-one has done before, and b) the whole point of this mega-thread is to determine how exactly what you're going on about works.
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You can do accurate predictions about a system if you understand how it works.
It's still 'should' and not 'will', particularly when it is a) an off-the-wall suggestion no-one has done before, and b) the whole point of this mega-thread is to determine how exactly what you're going on about works.
Not going to claim I'm a grammar expert :) because I'm not. Still I will have used will to show how sure I'm about my prediction.
b) I'm not saying anything controversial unlike Derek. I never heard the claim that "energy doesn't travel through wires" before so he needed extraordinary evidence for such an extraordinary claim. He failed to do that with his only so called evidence the small current flow through load much earlier with not even a mention about line capacitance in first video and some mention in the second but ignored.
Seeing the question snarkysparky just posted I realized that I should have been more clear on what energy in and out means and that it was not in and out as the same time since that will just mean trough.
I meant in when charging and out when discharging and this things never happen at the exact same time.
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How does IanB's middle capacitor get charged if no energy can flow through the outer capacitors.
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How does IanB's middle capacitor get charged if no energy can flow through the outer capacitors.
Two or more capacitors in series are no different from a single capacitor.
You can imagine a two parallel plates capacitor and you insert another plate in between those plates and that plate is not connected to anything and it will not affect the functionality or even capacity of that capacitor if plate is thin enough compared to the gap.
Inserting that unconnected plate will make that system a two capacitor in series setup and adding another plate will be equivalent with three capacitors in series.
a) -| |-
b) -| | |-
c) -| | | |-
a) 100pF single capacitor.
b) two 200pF capacitors in series.
c) three 300pF capacitors in series.
For the unconnected plates in the middle all that will happen is that free electrons already in those plates when capacitor is charged or discharged will move from one side of the plate to the other.
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An interesting variation on case B is if you use double-sided aluminized Mylar for the center plate.
Back in grad school, when some of my buddies were building wire chambers for high-energy physics experiments, we considered using the tooling to build large electrostatic speakers, but the engineer in charge warned us that the double-sided Mylar film used in the chambers would burn out if we tried to use them for the membrane, since charge would have to flow through a very thin film to get from one side to the other as the membrane moved.
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An interesting variation on case B is if you use double-sided aluminized Mylar for the center plate.
Back in grad school, when some of my buddies were building wire chambers for high-energy physics experiments, we considered using the tooling to build large electrostatic speakers, but the engineer in charge warned us that the double-sided Mylar film used in the chambers would burn out if we tried to use them for the membrane, since charge would have to flow through a very thin film to get from one side to the other as the membrane moved.
There will have been no reason for that not to work with low power. The wires mesh plates should have been fairly close to be an effective speaker but he should have allowed you to test this if nothing else will have been damaged other than worst case some inexpensive mylar tape.
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""For the unconnected plates in the middle all that will happen is that free electrons already in those plates when capacitor is charged or discharged will move from one side of the plate to the other.""
Nope. Not the same. The unconnected plates will have no net change in charge levels.
The plates on the middle capacitor will gain charge on one side and lose charge on the other. The proof is that you can remove the middle capacitor after applying voltage to the series set and it will have a charge on it.
How did that energy get there.
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""For the unconnected plates in the middle all that will happen is that free electrons already in those plates when capacitor is charged or discharged will move from one side of the plate to the other.""
Nope. Not the same. The unconnected plates will have no net change in charge levels.
The plates on the middle capacitor will gain charge on one side and lose charge on the other. The proof is that you can remove the middle capacitor after applying voltage to the series set and it will have a charge on it.
How did that energy get there.
Yes the unconnected plates will have no net charge but you can charge the setup c) and after you charged that you can connect a lamp between the two unconnected internal plates and discharge the middle capacitor.
So it is exactly the same thing the difference is that each unconnected plate is in case of capacitors made of two plates connected through a wire so charge moves from one plate to the other trough that wire instead of directly from one side of the plate to the other.
That group of two plates connected by a wire is the same as a single plate and there will be no net charge in either case.
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Not going to claim I'm a grammar expert :) because I'm not. Still I will have used will to show how sure I'm about my prediction.
The problem there is you are using a circular argument: my theory says X will do Y, and X doing Y proves my theory. Which is fine if you show that X does indeed do Y, but otherwise it should do Y. That's still a 100% confidence on your part, but you are accepting reality (that no-one's seen it yet, is it is supposition regardless of how entrenched the idea is).
b) I'm not saying anything controversial unlike Derek. I never heard the claim that "energy doesn't travel through wires" before so he needed extraordinary evidence for such an extraordinary claim.
It's not that controversial - Derek is only putting forward something that has been figured out for a while. But the difference here is that he did the actual experiment and the result was as he predicted. That's a lot better than just saying "it will" and then expecting it to be taken on trust but those that are doubting your theory in the first place.
He failed to do that with his only so called evidence the small current flow through load much earlier with not even a mention about line capacitance in first video and some mention in the second but ignored.
Well, that is what this thread is supposed to figuring out. Simply bashing your opponent with a circular argument before diverting to some irrelevant stuff isn't doing a great job of disproving what he thinks he's shown.
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Just answer this.
In the three capacitor setup where did the energy come from that charged the middle capacitor.
There can be no doubt that disconnecting the circuit and taking the middle capacitor out it will have a charge.
If you remove the two middle plates they will not have a charge. Because they were isolated.
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The problem there is you are using a circular argument: my theory says X will do Y, and X doing Y proves my theory. Which is fine if you show that X does indeed do Y, but otherwise it should do Y. That's still a 100% confidence on your part, but you are accepting reality (that no-one's seen it yet, is it is supposition regardless of how entrenched the idea is).
It is not my theory (I did not come up with it). It is how things work.
Many people (including me) seen this and use this to do shielding from electric fields.
It's not that controversial - Derek is only putting forward something that has been figured out for a while. But the difference here is that he did the actual experiment and the result was as he predicted. That's a lot better than just saying "it will" and then expecting it to be taken on trust but those that are doubting your theory in the first place.
Many people including me could have predicted the result. That result is not representing what he claimed "energy doesn't travel in wires"
Well, that is what this thread is supposed to figuring out. Simply bashing your opponent with a circular argument before diverting to some irrelevant stuff isn't doing a great job of disproving what he thinks he's shown.
So you mean line capacitance is irrelevant ? Or the fact that a capacitor is an energy storage device.
To be clear are you saying that shielding the way I mentioned will not eliminate the energy through lamp/resistor in the first 65ns in Derek's setup ?
If this is demonstrated to you will you then admit that energy doesn't travel outside the wire?
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Just answer this.
In the three capacitor setup where did the energy come from that charged the middle capacitor.
There can be no doubt that disconnecting the circuit and taking the middle capacitor out it will have a charge.
If you remove the two middle plates they will not have a charge. Because they were isolated.
The two or three capacitors in series are just one capacitor.
Yes in case c if you remove the middle plates they will not have a charge but that is not what you are doing when you remove the middle capacitor.
You first isolate the charges by disconnecting the capacitor pins the remove the middle capacitor.
d) -| |_| |_| |-
So d) is the same as c) but you can disconnect one face of the plate from the other by disconnecting the wire connecting the two faces.
Then those isolated middle faces will contain 33.33% of the energy that was put in those 3 capacitors in series.
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To be clear are you saying that shielding the way I mentioned will not eliminate the energy through lamp/resistor in the first 65ns in Derek's setup ?
To be Frank I have no idea since I've given up following your diversions so haven't really looked at it. Also it seems a pretty pointless thing to consider since you're unlikely to replicate Derek's experiment with whatever shielding, so it just comes back to what you say would occur rather than what actually does.
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""The two or three capacitors in series are just one capacitor.""
Factually they are not!
I ask to consider the actual case where there are three distinct capacitors. And the middle one receives energy from somewhere.
If you don't think capacitors can transmit energy you should be able to explain this.
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An interesting variation on case B is if you use double-sided aluminized Mylar for the center plate.
Back in grad school, when some of my buddies were building wire chambers for high-energy physics experiments, we considered using the tooling to build large electrostatic speakers, but the engineer in charge warned us that the double-sided Mylar film used in the chambers would burn out if we tried to use them for the membrane, since charge would have to flow through a very thin film to get from one side to the other as the membrane moved.
There will have been no reason for that not to work with low power. The wires mesh plates should have been fairly close to be an effective speaker but he should have allowed you to test this if nothing else will have been damaged other than worst case some inexpensive mylar tape.
As I remember it, the aluminized Mylar sheets were roughly 1 by 2 m, and the double-sided metallization was extremely thin (< 1 um). Power is a strange concept in electrostatic speakers (capacitive load), but we were planning to drive them with quite high voltages directly from the plates of 6DQ5 tubes. In our initial calculations, we had forgotten that the material available (for free) was double-sided. We were trying to take advantage of the tooling for stringing the wires on a glass-epoxy frame.
In your example case B, when the charge flows from the left to the right face of the central solid conductor, it flows through a huge cross-sectional area (width times height) compared with the Mylar case where it flows through an area that is wide but very thin.
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""The two or three capacitors in series are just one capacitor.""
Factually they are not!
I ask to consider the actual case where there are three distinct capacitors. And the middle one receives energy from somewhere.
If you don't think capacitors can transmit energy you should be able to explain this.
What about two or three capacitors in parallel? They are factually two or three in parallel but is there any difference in functionality compared to a single larger capacity capacitor ?
Same thing applies for two or three in series is just that leakage is a bit of a problem as it will not be equal so they may charge a bit inequal and if used close to the voltage limit you will usually add some parallel resistors if used this way in a circuit.
When connecting identical capacitors in parallel you increase the plate area with distance remaining the same.
When connecting identical capacitors in series you just increase the distance between plates of the equivalent capacitor that is why capacity drops.
Example c) and example d) are identical with the exception that you can disconnect the two middle plates leaving you with 3 capacitors after you do that.
e) -| |_| |_| |-
Say you have e) so 3 capacitors in series but the one in middle has plates closer say 2x the capacity of those at the ends.
When you disconnect those 3 capacitors after you charged them
Do you think:
1) all have the same amount of energy stored equally divided.
2) the one in the middle contains twice as much energy as each of the other ones.
3) the one in the middle contains half of the energy each of the other two contain.
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To be Frank I have no idea since I've given up following your diversions so haven't really looked at it. Also it seems a pretty pointless thing to consider since you're unlikely to replicate Derek's experiment with whatever shielding, so it just comes back to what you say would occur rather than what actually does.
I put in the effort to read what each one of you has to say. If you understand how capacitors and transmission lines work you will not need to even do the experiment to see that what I say holds true.
Likely you never seen any sort of shield that prevents a load getting energy from a source that is connected to load through wires. So you know for sure even without testing that lamp/resistor will receive energy from battery no matter what shield I create around.
The question is only if you think the shield that I proposed or any similar shield can stop energy getting to lamp/resistor in those first 65ns required for the energy to be delivered through the long wires.
Question is if this if proven true to you it is enough or you can still invent something to allow for energy transfer outside wire to still be true?
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As I remember it, the aluminized Mylar sheets were roughly 1 by 2 m, and the double-sided metallization was extremely thin (< 1 um). Power is a strange concept in electrostatic speakers (capacitive load), but we were planning to drive them with quite high voltages directly from the plates of 6DQ5 tubes. In our initial calculations, we had forgotten that the material available (for free) was double-sided. We were trying to take advantage of the tooling for stringing the wires on a glass-epoxy frame.
In your example case B, when the charge flows from the left to the right face of the central solid conductor, it flows through a huge cross-sectional area (width times height) compared with the Mylar case where it flows through an area that is wide but very thin.
That was a large sheet (not something I had in mind) and the double metalization will not have been helpful quite the opposite as mylar as outside layer will have helped reduce the risk of a discharge when sheet got to close to the grid. You also needs some sort of transformer to add audio signal.
Yes in my case b) the heat loss due to conduction is fairly small.
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As I remember it, the aluminized Mylar sheets were roughly 1 by 2 m, and the double-sided metallization was extremely thin (< 1 um). Power is a strange concept in electrostatic speakers (capacitive load), but we were planning to drive them with quite high voltages directly from the plates of 6DQ5 tubes. In our initial calculations, we had forgotten that the material available (for free) was double-sided. We were trying to take advantage of the tooling for stringing the wires on a glass-epoxy frame.
In your example case B, when the charge flows from the left to the right face of the central solid conductor, it flows through a huge cross-sectional area (width times height) compared with the Mylar case where it flows through an area that is wide but very thin.
That was a large sheet (not something I had in mind) and the double metalization will not have been helpful quite the opposite as mylar as outside layer will have helped reduce the risk of a discharge when sheet got to close to the grid. You also needs some sort of transformer to add audio signal.
Yes in my case b) the heat loss due to conduction is fairly small.
Huge sheets: these were large high-energy physics contraptions. Full-size electrostatics (Quad and KLH) were common then, 6DQ5s were still in production, and lots of military-surplus high-voltage transformers were available.
A transformer is not needed if you can couple directly to the plates of the push-pull driver, using the Vbb and another bias voltage.
We were younger then, and full of large ambitions.
I was contrasting this situation to your case B.
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Huge sheets: these were large high-energy physics contraptions. Full-size electrostatics (Quad and KLH) were common then, 6DQ5s were still in production, and lots of military-surplus high-voltage transformers were available.
A transformer is not needed if you can couple directly to the plates of the push-pull driver, using the Vbb and another bias voltage.
We were younger then, and full of large ambitions.
I was contrasting this situation to your case B.
Yes it can work with some high voltage push-pull mosfet or IGBT and optically isolated gate driver but is more complex than just a transformer with center tap.
I did Electrical Engineering so high voltage / high energy stuff and we had laboratories for lightning simulation tho I was more passionate about low voltage automation electronics.
As a hobby I never did projects exceeding 60Vdc open circuit the safe low voltage standard limit.
Case b) and c) where done to show that the internal plates that are fully isolated from supply have no net charge. You need to go to something like d) and then disconnect the two surfaces in order to isolate the sides with different charges.
For example the setup in Derek's test was more like
f) -| |_[resistor]_| |-
So capacitors on each side and resistor/lamp in between (that is the transient part).
So while current flows through resistor it is electrons from one plate moving to the other plate as the capacitors charge from the source.
If source is then removed and you short the capacitors all stored energy will flow back through resistor to have those plates neutral again.
So amount of energy through the resistor is directly proportional with the amount of charge being stored in the capacitors and then when they are discharged the same amount will flow back.
So electrons are the ones that travel through wire and transport the energy.
If we shield one quarter of Derek's transmission line and connect that shied on the opposite side of the resistor this capacitive energy storage will be done around the resistor so no more energy before those 65ns or so but then despite this shielding energy arrives through wires at the resistor/lamp.
And it can be DC or AC it will still get to that lamp resistor through wires despite the shield that removes the capacitive effect of the line.
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I put in the effort to read what each one of you has to say.
That may be, but what you write in response to a post is usually irrelevant. It's just waffle as a diversion. For instance, just above this snarkysparky asked about three capacitors in series and clearly stated that it was the middle one that was interesting. Your response (https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/msg4196629/#msg4196629) is 16 lines rabbiting on about parallel capacitors.
Either you didn't understand their short and to the point post, so just latched onto 'multiple capacitors' and made a guess at what it was about, or you deliberately obfuscated by diverting to a parallel setup (which is completely irrelevant).
Then, just to rub it in you then asked three questions designed to steer the discussion away from the series setup, and you will act miffed when those aren't answered.
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I put in the effort to read what each one of you has to say.
That may be, but what you write in response to a post is usually irrelevant. It's just waffle as a diversion. For instance, just above this snarkysparky asked about three capacitors in series and clearly stated that it was the middle one that was interesting. Your response (https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/msg4196629/#msg4196629) is 16 lines rabbiting on about parallel capacitors.
Either you didn't understand their short and to the point post, so just latched onto 'multiple capacitors' and made a guess at what it was about, or you deliberately obfuscated by diverting to a parallel setup (which is completely irrelevant).
Then, just to rub it in you then asked three questions designed to steer the discussion away from the series setup, and you will act miffed when those aren't answered.
What is the difference between this two ?
1) -| | | |-
2) -| |_| |_| |-
If you can not disconnect those links at 2) it is exactly the same as 1)
At 2) the electrons flow through that wire link to another plate but the net charge is zero as no electrons from battery get there.
As soon as you disconnect the links you have 3 separate and isolated capacitors and if all equal each contains 1/3 of the energy that battery delivered.
But you needed to take the action to cut the connections after the capacitor was charged.
No energy ever flowed from battery to the middle capacitor.
If you get all the middle plates out either on 1) or on 2) (without disconnecting the connections at 2) then the remaining two outer plates will contain the initial amount of energy. We will assume that the dielectric thickness remains about the same so plates have negligible thickness.
And yes I post long replays to explain how things actually work.
You still did not answered the main question for the last few posts. How will energy travel outside wire if you shield the wire from electric fields?
And also avoided to answer if proving that to you will be sufficient evidence or will you find some other invented theory to deny that electrical energy travels through wires.
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And also avoided to answer if proving that to you will be sufficient evidence or will you find some other invented theory to deny that electrical energy travels through wires.
If you can prove it one way or the other I would be happy to accept the definitive explanation.
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[How will energy travel outside wire if you shield the wire from electric fields?
Let's think of the following setup:
1. Place both the battery and bulb in separate sealed boxes made of a perfect conductor.
2. Replace the two cables with coaxial cable made of perfect conductors and a lossless dielectric.
3. Connect the shields of the two coaxial cables to the two sealed boxes. (No pigtails allowed)
4. Connect the inner conductors to the bulb, switch, and battery, like in Vertasium's original video.
How long will it now take for the bulb to turn on?
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[How will energy travel outside wire if you shield the wire from electric fields?
Let's think of the following setup:
1. Place both the battery and bulb in separate sealed boxes made of a perfect conductor.
2. Replace the two cables with coaxial cable made of perfect conductors and a lossless dielectric.
3. Connect the shields of the two coaxial cables to the two sealed boxes. (No pigtails allowed)
4. Connect the inner conductors to the bulb, switch, and battery, like in Vertasium's original video.
How long will it now take for the bulb to turn on?
You did not mentioned connecting the shield to anything.
Just connect the shield to say one side of the lamp.
But you do not even need all that complication to shield everything.
All you need is to shield a quarter of the wire so 10m of shield in Derek's experiment and connect that to the opposite side of the lamp/resistor.
______________________[resistor]_________________________
| ______________________________| |
| |
| |
| |
|___________________/ ____- +____________________________|
Hope you can see the diagram. The shield is only electrically connected to the right side of the resistor/lamp
What will happen is that the shield will now be the second plate of the capacitor formed with the lower wire and current for charging the capacitor will travel through that shield and connected to the opposite side of the resistor.
So there will still be current through the shield on left side and the line on the right side as the line capacitance charge but no more current through the resistor thus no more power and no energy until after the 65ns or so the electron wave needs to get to resistor through wire.
So you need to replace that right side with coaxial cable or just use at least a half pipe (half cylinder or U shaped) to shield the wire form the electric field generated by the lower wire.
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You are answering a different question. The shield is connected to the two boxes. Connect the shield to ground (mother earth) if you want to.
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You are answering a different question. The shield is connected to the two boxes. Connect the shield to ground (mother earth) if you want to.
That will not be called a shield if it is not connected. Earth has nothing to do with this isolated system so connecting the shield to earth will do nothing unless you want to shield against external influence.
The shield I proposed is the simplest that will shield against electric field and so it shows that even with blocking the electric field from getting to Lamp/resistor the energy travels through wires and gets to resistor.
To "shield" against energy traveling through wire all you need to do is cut a gap of say 1mm in the wire :)
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What is your answer to my original question? How long will it take?
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What is your answer to my original question? How long will it take?
If as you say the shield is not connected to any point to the circuit then is not actually a shield so you still get energy through that lamp/resistor well before 65ns.
In simplest form what you say is
-| | |- a capacitor with a plate in the middle so like two capacitors in series. You can not call the middle plate a shield unless is connected to one of the sides.
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What is your answer to my original question? How long will it take?
I wish you luck and great patience :popcorn:
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So you need to replace that right side with coaxial cable or just use at least a half pipe (half cylinder or U shaped) to shield the wire form the electric field generated by the lower wire.
um, what "electric field generated by the lower wire"?
You were arguing that the energy is all in the wires. So why the need to shield from what you consider a non-existent electric field?
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So you need to replace that right side with coaxial cable or just use at least a half pipe (half cylinder or U shaped) to shield the wire form the electric field generated by the lower wire.
um, what "electric field generated by the lower wire"?
You were arguing that the energy is all in the wires. So why the need to shield from what you consider a non-existent electric field?
The lower line and top line form a capacitor that will be charged. There is one capacitor made by the lines on each side.
If you add that shield then the shield will be the other plate of that capacitor.
If you connect that to the other side of the resistor the current to charge the line capacitance will no longer flow through the lamp/resistor.
Since the only so called "proof" Derek had was that energy arriving earlier than 65ns. Adding this shield will get rid of that energy and still energy will get to resistor and travel through wires.
Derek's claim "energy doesn't travel through wires" and I can say just make a cut in the wire and see if any energy gets to lamp/resistor.
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So how does energy transfer from side of a vacuum tube diode to the other?
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So how does energy transfer from side of a vacuum tube diode to the other?
I'm not so old to have ever used a vacuum tube diode but a quick google revealed how that worked.
This is a direct quote that should provide you the answer:
"When the cathode of a vacuum tube is heated, electrons are emitted. If the anode has an electric potential higher than the cathode, the emitted electrons are attracted by it and an electric current start to flow "
(https://www.electronics-notes.com/images/valve-tube-diode-operation-01.svg)
So electrons do travel through vacuum in that particular case.
There are no electrons traveling from one wire to the other 1m apart wire in Derek's circuit thus energy travels through wire.
But the vacuum tube diode is a good example of energy traveling through vacuum so outside of the wire. It just that the vacuum tube diode has nothing to do with Derek's circuit where all energy delivered to lamp travels through wire.
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What is your answer to my original question? How long will it take?
If as you say the shield is not connected to any point to the circuit then is not actually a shield so you still get energy through that lamp/resistor well before 65ns.
In simplest form what you say is
-| | |- a capacitor with a plate in the middle so like two capacitors in series. You can not call the middle plate a shield unless is connected to one of the sides.
A transmission line and a capacitor is not the same thing. Maybe you should think again.
What happens inside a perfectly conducting sealed enclosure stays inside the enclosure, and what happens outside stays outside. Magnetic and electric fields cannot exit or enter the enclosure.
So by shielding the electric and magnetic fields, you are forcing the "electricity" to go all the way round the loop.
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A transmission line and a capacitor is not the same thing. Maybe you should think again.
What happens inside a perfectly conducting sealed enclosure stays inside the enclosure, and what happens outside stays outside. Magnetic and electric fields cannot exit or enter the enclosure.
So by shielding the electric and magnetic fields, you are forcing the "electricity" to go all the way round the loop.
They are basically the same thing. Both the capacitor and the transmission line have resistance, inductance and capacitance just in different proportions.
If the shield as you say is not connected to any part of your circuit it will not do anything for this particular case.
The plate between the two capacitors plates unconnected to anything is useless.
-| | |-
While the simpler smaller shield I proposed will eliminate the energy being delivered to the lamp/resistor in the first 65ns as now the line capacitance is between that shield so current to charge that capacitor will flow through shield and not resistor.
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If the shield as you say is not connected to any part of your circuit it will not do anything for this particular case.
The plate between the two capacitors plates unconnected to anything is useless.
Nonsense!
Do you understand how shielding works? Have you heard of a Faraday cage?
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Nonsense!
Do you understand how shielding works? Have you heard of a Faraday cage?
What you have proposed is a special type of cage, is a rectangular donut :)
But leaving all this aside. Is there a way you can intrerup the energy flow from source to lamp/resistor without making a cut in the wire ?
Because if you can not that is another proof that energy travels through wire.
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So how does energy transfer from side of a vacuum tube diode to the other?
I thought that energy was supposed to flow around the vacuum tube?
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https://www.youtube.com/watch?v=fxCR2bFWHxM (https://www.youtube.com/watch?v=fxCR2bFWHxM)
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Some time back Dave pointed out that for most practical engineers it doesn’t matter. I'd agree with that. But it may or may not matter, it depends on the situation. It matters very much to ic designers and semiconductor physicists.
Its ok to believe in little green men, it doesn’t affect satelite launches. (or does it....)
If the electrons move under influence of the potental difference they would have only Vbat electon volts of energy and get bounced back by the first atoms in the condcutor.
It is the bounce that is observed to travel in a conductor not particles. There is the drift velocity of notional charge carriers- thats really slow. It would take at least a decade or ten before the power got from the power plant to your home using that mechanism. Whats moving in the drift are electronic irregularities in the condcutors crystal lattice. "Things" in the convential sense are not going anywhere.
The thing and the name of a thing are not the same thing.
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Some time back Dave pointed out that for most practical engineers it doesn’t matter. I'd agree with that. But it may or may not matter, it depends on the situation. It matters very much to ic designers and semiconductor physicists.
Its ok to believe in little green men, it doesn’t affect satelite launches. (or does it....)
If the electrons move under influence of the potental difference they would have only Vbat electon volts of energy and get bounced back by the first atoms in the condcutor.
It is the bounce that is observed to travel in a conductor not particles. There is the drift velocity of notional charge carriers- thats really slow. It would take at least a decade or ten before the power got from the power plant to your home using that mechanism. Whats moving in the drift are electronic irregularities in the condcutors crystal lattice. "Things" in the convential sense are not going anywhere.
The thing and the name of a thing are not the same thing.
I think it matters even for the practical engineer. Like for example shielding a circuit from electric fields.
I explained how a simple shield that removes that small current through the lamp/resistor in Dereks experiment.
It is also important to understand that energy flows through a conductor and not outside of it. So in case of a capacitor being charged energy flows into capacitor and not through capacitor.
That old vacuum diode someone used as an example is good for visualization. If the filament on that is not working it acts as a capacitor but having the filament heater it allows electrons to travel through that vacuum from one plate to the other and they (electrons / charged particles) are what transports energy.
There are no electrons traveling through that 1m of air between wires in Derek's experiment meaning no energy travels from source to lamp/resistor outside the wire. All energy both in transient/AC and DC phase travels through wire in Derek's experiment while he claims the opposite with zero proof.
What he claims as proof is that small current in the first 65ns that is due to energy being stored in the transmission line. Adding the electric shield I mentioned will get rid of that and yet after 65ns there will be energy flowing through lamp/resistor and obviously trough wire which is also a resistor.
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Yes, that which we call Ohm's Law is a result of the scattering of electrons by lots of things in a conductor, so that they do not achieve a high velocity from the voltage gradient along the wire.
An example calculation in https://www.macmillanlearning.com/studentresources/college/physics/tiplermodernphysics6e/classial_concept_review/chapter_10_ccr_10_mean_free_path.pdf (https://www.macmillanlearning.com/studentresources/college/physics/tiplermodernphysics6e/classial_concept_review/chapter_10_ccr_10_mean_free_path.pdf) gives a mean free path between scattering events for electrons in copper as 0.39 nm.
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I explained how a simple shield that removes that small current through the lamp/resistor in Dereks experiment.
Here you go again. You make a supposition and then treat it as fact, and then build other stuff on that pseudo-fact. You suggested a simple shield that could/should do what you say. Run the experiment and then you can say it actually does it.
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Here you go again. You make a supposition and then treat it as fact, and then build other stuff on that pseudo-fact. You suggested a simple shield that could/should do what you say. Run the experiment and then you can say it actually does it.
I made multiple correct predictions based on correct understanding of the subject.
If you think that the shielding I mentioned will not work it is your choice to test (run the experiment) and prove me wrong.
Derek made the experiment with all the results being correct but due to him not understanding the subject he came to wrong conclusions.
He noticed some energy flowing through lamp/resistor sooner than 65ns and he wrongly concluded that is because "energy doesn't flow in wire"
He had no idea or ignored the fact that a transmission line has capacitance (energy storage) in first video then when he found out he made a second video just mentioning that but not taking it seriously in consideration.
-Do you agree that making a cut in the wire will result in no energy transfer after the first few ns required to charge the capacitor formed by the parallel wires.
-When you also understand that shielding will get rid of that initial current through lamp/resistor but energy will still start flowing after those first 65ns you will know that energy flows through wires.
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Has "how does noise voltage work then?" been touched on?
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Has "how does noise voltage work then?" been touched on?
Can you be more specific ? Are you talking about an isolated circuit ?
Derek's real world circuit was not isolated from external influence but that part (contribution from that) was ignored. Was fairly small so easy to ignore.
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Noise voltage across a bare resistor.
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Noise voltage across a bare resistor.
The simplest model to analyze thermal noise in a resistor is to connect a capacitor in parallel with the resistor (at finite temperature) and measure the random voltage across the circuit with an ideal voltmeter.
The capacitor determines the bandwidth in the ideal circuit.
Drs. Johnson and Nyquist submitted their papers to Physical Review at the same time--it is rumored that Johnson shoved his under the door of the office at Columbia--(grad student rumor).
The two papers were published in the same issue. See https://web.stanford.edu/~edwin98/johnson-shot-noise-paper.pdf
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Noise voltage across a bare resistor.
The simplest model to analyze thermal noise in a resistor is to connect a capacitor in parallel with the resistor (at finite temperature) and measure the random voltage across the circuit with an ideal voltmeter.
The capacitor determines the bandwidth in the ideal circuit.
Drs. Johnson and Nyquist submitted their papers to Physical Review at the same time--it is rumored that Johnson shoved his under the door of the office at Columbia--(grad student rumor).
The two papers were published in the same issue. See https://web.stanford.edu/~edwin98/johnson-shot-noise-paper.pdf
Exactly. A function of the bandwidth of the measurement and the absolute temperature of the resistance. So. What's the explanation that isn't elections bouncing around?
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Noise voltage across a bare resistor.
The simplest model to analyze thermal noise in a resistor is to connect a capacitor in parallel with the resistor (at finite temperature) and measure the random voltage across the circuit with an ideal voltmeter.
The capacitor determines the bandwidth in the ideal circuit.
Drs. Johnson and Nyquist submitted their papers to Physical Review at the same time--it is rumored that Johnson shoved his under the door of the office at Columbia--(grad student rumor).
The two papers were published in the same issue. See https://web.stanford.edu/~edwin98/johnson-shot-noise-paper.pdf
Exactly. A function of the bandwidth of the measurement and the absolute temperature of the resistance. So. What's the explanation that isn't elections bouncing around?
"Elections"--nice Freudian slip!
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:-DD
Yeah whoops.
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Some time back Dave pointed out that for most practical engineers it doesn’t matter. I'd agree with that. But it may or may not matter, it depends on the situation. It matters very much to ic designers and semiconductor physicists.
No it never matter.
Yes, that which we call Ohm's Law is a result of the scattering of electrons by lots of things in a conductor, so that they do not achieve a high velocity from the voltage gradient along the wire.
An example calculation in https://www.macmillanlearning.com/studentresources/college/physics/tiplermodernphysics6e/classial_concept_review/chapter_10_ccr_10_mean_free_path.pdf (https://www.macmillanlearning.com/studentresources/college/physics/tiplermodernphysics6e/classial_concept_review/chapter_10_ccr_10_mean_free_path.pdf) gives a mean free path between scattering events for electrons in copper as 0.39 nm.
This is completely incorrect.
https://homepages.rpi.edu/~galld/publications/PDF-files/Gall-116.pdf (https://homepages.rpi.edu/~galld/publications/PDF-files/Gall-116.pdf)
Indicates 39.9 nm.
He assumed that electrons scatters with atoms, while they scatter with phonons.
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Some time back Dave pointed out that for most practical engineers it doesn’t matter. I'd agree with that. But it may or may not matter, it depends on the situation. It matters very much to ic designers and semiconductor physicists.
No it never matter.
Yes, that which we call Ohm's Law is a result of the scattering of electrons by lots of things in a conductor, so that they do not achieve a high velocity from the voltage gradient along the wire.
An example calculation in https://www.macmillanlearning.com/studentresources/college/physics/tiplermodernphysics6e/classial_concept_review/chapter_10_ccr_10_mean_free_path.pdf (https://www.macmillanlearning.com/studentresources/college/physics/tiplermodernphysics6e/classial_concept_review/chapter_10_ccr_10_mean_free_path.pdf) gives a mean free path between scattering events for electrons in copper as 0.39 nm.
This is completely incorrect.
https://homepages.rpi.edu/~galld/publications/PDF-files/Gall-116.pdf (https://homepages.rpi.edu/~galld/publications/PDF-files/Gall-116.pdf)
Indicates 39.9 nm.
He assumed that electrons scatters with atoms, while they scatter with phonons.
Yes, there are lots of things inside a conductor to scatter electrons.
In grad school, I learned about phonons, lattice imperfections, and other defects as contributions to limiting the conductivity.
Note that, in general, alloys (which include impurities) have much lower conductivity than pure metals under normal conditions.
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But electrons don't matter anyway? ;D
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"Bremsstrahlung... what? Why are you all hiding behind lead shields? I'm surely not in the path of any Poynting vectors"
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"Bremsstrahlung... what? Why are you all hiding behind lead shields? I'm surely not in the path of any Poynting vectors"
Again, in grad school we had a visiting professor from Darmstadt. He was giving a series of lectures on electron interactions. Although his English was very good, he was a little nervous about his lecture and asked an American student who was fluent in German to sit in the front row, in case he had a language problem. At one point, he asked "Wie sagt man Bremsstrahlung auf Englisch?, to which the student replied "Bremsstrahlung".
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But electrons don't matter anyway? ;D
Poynting vector does not matter. Electrons, obviously, do.
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Some time back Dave pointed out that for most practical engineers it doesn’t matter. I'd agree with that. But it may or may not matter, it depends on the situation. It matters very much to ic designers and semiconductor physicists.
No it never matter.
Yes, that which we call Ohm's Law is a result of the scattering of electrons by lots of things in a conductor, so that they do not achieve a high velocity from the voltage gradient along the wire.
An example calculation in https://www.macmillanlearning.com/studentresources/college/physics/tiplermodernphysics6e/classial_concept_review/chapter_10_ccr_10_mean_free_path.pdf (https://www.macmillanlearning.com/studentresources/college/physics/tiplermodernphysics6e/classial_concept_review/chapter_10_ccr_10_mean_free_path.pdf) gives a mean free path between scattering events for electrons in copper as 0.39 nm.
This is completely incorrect.
https://homepages.rpi.edu/~galld/publications/PDF-files/Gall-116.pdf (https://homepages.rpi.edu/~galld/publications/PDF-files/Gall-116.pdf)
Indicates 39.9 nm.
He assumed that electrons scatters with atoms, while they scatter with phonons.
Yes, there are lots of things inside a conductor to scatter electrons.
In grad school, I learned about phonons, lattice imperfections, and other defects as contributions to limiting the conductivity.
Note that, in general, alloys (which include impurities) have much lower conductivity than pure metals under normal conditions.
Yes, but alloys in wires do not affect the speed of the elektons (elekticity) on the surface of the wire.
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Some time back Dave pointed out that for most practical engineers it doesn’t matter. I'd agree with that. But it may or may not matter, it depends on the situation. It matters very much to ic designers and semiconductor physicists.
No it never matter.
Yes, that which we call Ohm's Law is a result of the scattering of electrons by lots of things in a conductor, so that they do not achieve a high velocity from the voltage gradient along the wire.
An example calculation in https://www.macmillanlearning.com/studentresources/college/physics/tiplermodernphysics6e/classial_concept_review/chapter_10_ccr_10_mean_free_path.pdf (https://www.macmillanlearning.com/studentresources/college/physics/tiplermodernphysics6e/classial_concept_review/chapter_10_ccr_10_mean_free_path.pdf) gives a mean free path between scattering events for electrons in copper as 0.39 nm.
This is completely incorrect.
https://homepages.rpi.edu/~galld/publications/PDF-files/Gall-116.pdf (https://homepages.rpi.edu/~galld/publications/PDF-files/Gall-116.pdf)
Indicates 39.9 nm.
He assumed that electrons scatters with atoms, while they scatter with phonons.
Yes, there are lots of things inside a conductor to scatter electrons.
In grad school, I learned about phonons, lattice imperfections, and other defects as contributions to limiting the conductivity.
Note that, in general, alloys (which include impurities) have much lower conductivity than pure metals under normal conditions.
Yes, but alloys in wires do not affect the speed of the electons (electricity) on the surface of the wire.
So tell me: why does the alloy composition affect the conductivity of current down the wire?
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Some time back Dave pointed out that for most practical engineers it doesn’t matter. I'd agree with that. But it may or may not matter, it depends on the situation. It matters very much to ic designers and semiconductor physicists.
No it never matter.
Yes, that which we call Ohm's Law is a result of the scattering of electrons by lots of things in a conductor, so that they do not achieve a high velocity from the voltage gradient along the wire.
An example calculation in https://www.macmillanlearning.com/studentresources/college/physics/tiplermodernphysics6e/classial_concept_review/chapter_10_ccr_10_mean_free_path.pdf (https://www.macmillanlearning.com/studentresources/college/physics/tiplermodernphysics6e/classial_concept_review/chapter_10_ccr_10_mean_free_path.pdf) gives a mean free path between scattering events for electrons in copper as 0.39 nm.
This is completely incorrect.
https://homepages.rpi.edu/~galld/publications/PDF-files/Gall-116.pdf (https://homepages.rpi.edu/~galld/publications/PDF-files/Gall-116.pdf)
Indicates 39.9 nm.
He assumed that electrons scatters with atoms, while they scatter with phonons.
Yes, there are lots of things inside a conductor to scatter electrons.
In grad school, I learned about phonons, lattice imperfections, and other defects as contributions to limiting the conductivity.
Note that, in general, alloys (which include impurities) have much lower conductivity than pure metals under normal conditions.
Yes, but alloys in wires do not affect the speed of the electons (electricity) on the surface of the wire.
So tell me: why does the alloy composition affect the conductivity of current down the wire?
Yes, good question, it is so good (& obvious) that it must have kumup earlier here, i forget.
Elektons have a nett E×H field (in the nearfield & in the farfield), whereas photons don’t (at least not in the far field).
The elekton's E×H acts on elektrons, particularly on free electrons, in the wire.
The free electrons are moved, & the movement is resisted by the Cu.
The resistance ends up heating the Cu.
Every action has a reaction. Hence u would think that conductivity must affect the speed of electricity. In other words my elektons must be slowed moreso by bad conductors. This subject too must have kumup before. Praps they are slowed moreso. But praps that slowing is only say 1%, which would show, but would not show if u were not looking for it, or if u were ignoring it, or if it twernt important in the particular case under test.
For insulated wires, & on PCBs, there is a major slowing due to the insulation (insulation usually covers the whole surface), & due to the board material (in effect insulated on a half of the surface i think), hence a minor speed effect due to conductivity might not show up.
Such tests (speed of electricity) comparing metals & alloys etc would be simple -- have they ever been done? Still thinking.
Ooops. No. The speed of electricity should be slower the better the conductor.
A good conductor acts more strongly on elektons. A poor conductor (plastic) has so little action on elektons that the elektons will not jump onto its surface, to an elekton the plastic duznt exist (unless it is in the plastic insulation of a wire)(ie whilst the elekton is propagating/flowing along the wire).
Elektons act on other elektons, hence the surface area of a wire has a saturation value, which depends on the conduction/resistance of the wire, & depends on the voltage of the source, & depends on the rate of production/supply of elektons. However, this interaction is a chicken & egg thing i suppose (as usual).
Here i need to remind myself that elekton elekticity is just one of the 3 kinds -- the others being (2) the movement of free electrons on the surface, & (3) the movement of free elektrons inside the wire.
Re that there saturation value – i wonder if it is affected by insulation. Praps knot. Still thinking.