Veritasium "How Electricity Actually Works"

[Sredni]

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. 

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)?

[rfeecs]

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.

[bsfeechannel]

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.

[TimFox]

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]

[T3sl4co1l]

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

[Naej]

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. 

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.

[electrodacus]

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.

[EEVblog]

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?" 

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.

[timenutgoblin]

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?



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

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.

[electrodacus]

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?



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 * V_i

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 * V_i
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 V_i / 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.

[timenutgoblin]

If those two capacitors are identical (same capacity) the voltage after switch is closed will be 0.707 * V_i

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.

[electrodacus]

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 * V_i.
Voltage is not energy.

[SiliconWizard]

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.

[electrodacus]

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.

[hamster_nz]

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 * V_i.
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 * V_i.

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."

[electrodacus]

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 * V_i.

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.


V_i = 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.

[hamster_nz]

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 V_i 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 V_i. 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

[electrodacus]

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 V_i 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.

[electrodacus]

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 V_i 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 V_i. 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 V_i   but you do not need a DC-DC converter if there is no resistance to get the same result.

[hamster_nz]

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 V_i to V_i ) 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 V_i.

(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).

[electrodacus]

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.

[cj]

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

[T3sl4co1l]

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

[electrodacus]

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 * V_i) ?

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.

[T3sl4co1l]

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