Veritasium "How Electricity Actually Works"

[electrodacus]

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.

[hamster_nz]

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

[pepsi]

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.

[electrodacus]

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.

[hamster_nz]

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.

[electrodacus]

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.

[rfeecs]

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

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.

[hamster_nz]

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.

[electrodacus]

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 ?

[hamster_nz]

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.

[electrodacus]

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).

[Naej]

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?

[hamster_nz]

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

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.

[eti]

He should leave this alone. Lol.

[electrodacus]

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.

[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]

@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.

[EEVblog]

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. 

I mostly stopped reading the orignal thread.

[EEVblog]

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....

[Naej]

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.

[electrodacus]

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.

[rfeecs]

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.