General > General Technical Chat
Veritasium "How Electricity Actually Works"
T3sl4co1l:
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
TimFox:
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.
electrodacus:
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.
pepsi:
@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.
electrodacus:
--- Quote from: pepsi on May 01, 2022, 04:15:48 am ---@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.
--- End quote ---
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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