Veritasium "How Electricity Actually Works"

[rfeecs]

I thought it might be good to start a new thread for Veritasium's new video, the follow up to the "Big Misconception"  video:



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. 

[rfeecs]

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.

[HuronKing]

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.

[rfeecs]

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.

[snarkysparky]

What pushes the charges inside the wire if there is no electric field inside the wire and the electrons don't push themselves.

[rfeecs]

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.

[electrodacus]

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.

[T3sl4co1l]

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...


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

[electrodacus]

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.

[hamster_nz]

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

[electrodacus]

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.

[hamster_nz]

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

[SandyCox]

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.

[hamster_nz]

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 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?

[hamster_nz]

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.

EDIT: Updated video to have two meters.

[EPAIII]

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.

[TimFox]

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.

[Brumby]

I can do whatever I want

We've noticed.

[electrodacus]

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.

[T3sl4co1l]

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

[electrodacus]

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.

[E63S4Me]

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.

[T3sl4co1l]

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

[electrodacus]

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.

[T3sl4co1l]

Then you will agree the final voltages will be sqrt(2)/2, not 0.5, or 1 (initial)?

Please provide waveforms showing this.

Tim