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.QuoteIf 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.QuoteThe 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.
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
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.
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
I mean, how would you know? Given the above information, can you solve for the resistance (if any) in my circuit?
Tim
Actually, power peaks for R = Zo. If it were true that smaller resistance heats faster -- what would superconductors do? Just explode?
Tim
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.
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.
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.
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.
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.
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.
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:QuoteThe 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 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.
The transient at connection [of a DC source to a transformer] is not DC.
Only a tiniest initial amount is will be transferred directly over the 1m gap.
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?
QuoteThe 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.
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 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.
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.
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.