Veritasium "How Electricity Actually Works"

[Naej]

Really, where then?

Now you're trolling.

You gave no example of an alternative breaking any of this . Coincidence?

If you couldn't recognize it, this means you didn't go very far in your understanding of the problem.

Right so there are plenty of problems with S=JV but you are forbidden to discuss them.

Sure so in the 20th century and 21st physicists never heard of antennae. 

Who knows? What is important to understand is that the alternatives never prospered.

Prosperity is knowledge. Or science I don't know. Or is it scientology?
And influencers like Derek are needed to keep physicists in the right direction. It's definitely how science works.

And it doesn't bother you at all that energy is supposedly flowing through the region A yet there is no circuit you can put in region A in the diagram that can extract energy from it?
(Hopefully the answer is no)

No, with no voltage potential over region A, there is no way you can extract energy from the field in just that region.

So what really matters is the potential, while Poynting's energy flow give no useful information.

Are you bothered???
Do you think that if every electron could communicate with far away electrons, humans could make machines hacking this property to communicate messages?

It did bother me just a little - Every charge being constantly aware of every other charge in the universe does not have the feel of being fundamental to the universe.

An electric field can still be used to communicate messages. "Ripples in the fields" is preferable to "very small pushes and shoves over great distances".

Every night, you can check that electrons in your eye are "aware" of what's going on with electrons in far away stars.

A wire is an inductor. With 2 you have an inductor and half a capacitor. And it works even for infinitesimal wires. 

Fair call. But I still feel that a lumped model of a continuous thing is (very useful) approximation.

Yes it's quite bizarre: the lumped model approximation has no reason to be a good approximation with light-speed processes, yet it is an extremely precise one with transmission lines.
I guess the main reason is they hardly behave as antennae, due to their counter-currents.

[TimFox]

(PS:  when I distinguished active and passive circuits, I mean that an active circuit has an external power source.  So a diode in the dark is passive, but it has losses.) 
What happens if you leave S2 on for a long time in your simulation?

What do you think it will happen ?
The energy will flow back and forth between the two capacitors and inductor (RLC resonant circuit) until half of the initial energy will be dissipated as heat so you end up with half the voltage.
 

Yes, with two equal capacitors and any passive two-terminal network between them, that is what I expect.  Each capacitor has 1/2 the original voltage, and therefore 1/2 the original charge and 1/4 the original total energy in both capacitors, until the leakage resistance not included in the model eventually discharges both of them.  I'm busy this weekend, but I will do a very simple experiment next week with polypropylene capacitors and a 10 megohm voltmeter.
If I use two 10 uF capacitors, and then one 10 uF (original charge) and a second 20 uF (originally discharged), I will have discharge time constants of 200 sec (both caps) and 300 sec (both caps), so I will lose 1% of the charge due to the parasitic voltmeter in 2 sec or 3 sec, respectively, after I throw the switch, first disconnecting C1 from the battery, and then connecting C1 to C2  (DPDT CO, L&N huge honking switch).

[electrodacus]

Yes, with two equal capacitors and any passive two-terminal network between them, that is what I expect.  Each capacitor has 1/2 the original voltage, and therefore 1/2 the original charge and 1/4 the original total energy in both capacitors, until the leakage resistance not included in the model eventually discharges both of them.  I'm busy this weekend, but I will do a very simple experiment next week with polypropylene capacitors and a 10 megohm voltmeter.
If I use two 10 uF capacitors, and then one 10 uF (original charge) and a second 20 uF (originally discharged), I will have discharge time constants of 200 sec (both caps) and 300 sec (both caps), so I will lose 1% of the charge due to the parasitic voltmeter in 2 sec or 3 sec, respectively, after I throw the switch, first disconnecting C1 from the battery, and then connecting C1 to C2  (DPDT CO, L&N huge honking switch).

Not sure I understand what you want to test.
Will you parallel two 10uF capacitors so that you get a 20uF and use that 20uF capacitor as the discharged capacitor and then use a 10uF charged capacitor from witch you will charge the 20uF one ?
If that is the case then you will see that voltage will drop down to 33% so 1V if your charged capacitor is at 3V initially.
So a lot more energy will be lost as heat compared to the two identical capacitor test.
You start with 45uWs of energy 0.5 * 10uF * 3²
And you end up with 15uWs of energy 0.5 * 30uF * 1²
The other 30uWs ends up as heat in the conductors.

[TimFox]

A simple demonstration at maybe 2% accuracy:
Connect voltmeter across C1, and discharge C2
Charge C1 = 10 uF to 9 V (battery), measure the voltage immediately after opening the battery-C1 switch (must be faster than a few seconds with the 100 sec time constant of 10 uF & 10 megohm).
Close C1 to C2 switch and measure voltage as soon as it settles, paying attention to the 200 sec discharge time constant.
Do again with C2 = 20 uF to demonstrate unequal capacitor case.
It should agree with your calculation, to show that total charge rather than total energy in the open system is conserved.

[electrodacus]

A simple demonstration at maybe 2% accuracy:
Connect voltmeter across C1, and discharge C2
Charge C1 = 10 uF to 9 V (battery), measure the voltage immediately after opening the battery-C1 switch (must be faster than a few seconds with the 100 sec time constant of 10 uF & 10 megohm).
Close C1 to C2 switch and measure voltage as soon as it settles, paying attention to the 200 sec discharge time constant.
Do again with C2 = 20 uF to demonstrate unequal capacitor case.
It should agree with your calculation, to show that total charge rather than total energy in the open system is conserved.

Total energy is always conserved in an isolated system.
So in this case 1)
charged 10uF charging a discharged 10uF half of the energy will remain in the two combined capacitors and half of the energy will be dissipated as heat.

In case 2)
charged 10uF charging a discharged 20uF one third of the energy will remain stored in the capacitors while two thirds will be dissipated as heat so again energy is conserved.

There is no rule about charge being conserved it just happen due to symmetry as there is resistance everywhere including the capacitor plates.
If you add a discharged inductor in the circuit you are not adding either charge or energy and you are just adding another energy storage device as an intermediary.

The point is that energy is always conserved in an isolated system and using this equation is easy to see where energy ended up (in the conductors) from witch should be easy to conclude what path the energy has traveled.
If energy transfer was not done through conductors then there will be no reason to have all the energy loss (exactly all) inside the conductors.

The symmetry is broken when you add an inductor even if you are actually not bringing any charge from outside and system is still isolated. Inductor is basically a fairly long wire.

Edit:
I'm thinking of ways you can break the symmetry but unless one of the capacitors has superconductor plates I do not see how it can be done other than by adding an inductor as intermediary energy storage device.
Parallel capacitors will always have this symmetry as they have resistance (they are like a transmission line).
Adding an inductor will just break this symmetry reducing the loss as heat in wire.
I will try to think of a mechanical analog.
It will not make sense to do the experiment as you probably agree with me on what the result of that test will be.

[TimFox]

Yes, “isolated system” is the key point.
If resistor heat dissipation can escape from the bench top, then the system is not isolated.

[electrodacus]

Yes, “isolated system” is the key point.
If resistor heat dissipation can escape from the bench top, then the system is not isolated.

You can drop the circuit in to an insulated box then it is isolated It will have some small leakage but still isolated.
I hope you do not doubt that all that missing energy is not ending up as heat due to resistance.

It is basically the equivalent of charging a 1.5V cell from a 3V supply through a linear regulator or resistor so half the energy is wasted as heat in the resistor or linear regulator.
The total charge Q will of course always the the same in this scenarios and a lot of energy will be lost as heat.
Adding an extra energy storage device that is first charged from the source and then discharged on the load can increase this efficiency significantly and in that case the Q will no longer be the same at the start and the end of the experiment.
Conservation of charge is no more even if you did not added charge from any external source you just used the available energy more efficiently.
So energy conservation is a law that can not be violated where charge conservation is more of a coincidence when resistive losses are involved and there is no mechanism to reduce those losses like adding a inductor as intermediary energy storage.


The important part of all this is that there is no mystery in the identical parallel capacitors and all energy is accounted for and another important part is that energy travels through wires as that is where all this heat loss originate.

On the long transmission line you have both inductive and capacitive storage but the capacitive storage is responsible for the initial energy flow through lamp as the lamp is in series with the two capacitors being charged so lamp becomes part of the total resistance of the circuit so part of the losses will be on the lamp (significant part when you use a 1KOhm resistor as a lamp) and thick copper pipes as transmission line.

[TimFox]

Yes, you can do calorimeter on the insulated box and measure the thermal portion of the system energy.

[electrodacus]

Yes, you can do calorimeter on the insulated box and measure the thermal portion of the system energy.

But you think that is needed ? Is clear missing energy is ending as heat in the wires/conductors.

[TimFox]

Spell check error: “calorimetry”. 
No, I don’t think it’s necessary.  Just an example of how energy conservation is not the most relevant calculation method in certain situations.

[electrodacus]

Spell check error: “calorimetry”. 
No, I don’t think it’s necessary.  Just an example of how energy conservation is not the most relevant calculation method in certain situations.

The relevant part is that energy conservation can not be violated.
If you were to measure energy going out of the source in Derek's experiment and energy received by the lamp/resistor you will see the situation in this graph


That shows that much more power exits the source than gets to the lamp and that is explained by the fact that energy is stored in the line capacitance.
To make things even more clear the second graph shows what happens if the switch is only closed circuit for 30ns so less than half it is required for the electron wave to travel the transmission line.

 

In both graphs the green is the power from source so the integral of that is the energy and the purple line is the lamp/resistor power.
The difference that is missing ends as heat on the line but most of the energy gets to lamp after the time needed for the electron wave to get there.

[Sredni]

A selection of references for the two-capacitor problem (one of which has already been given in this very thread)


The two-capacitor problem with radiation
Timothy B. Boykin, Dennis Hite, and Nagendra Singh
Department of Electrical and Computer Engineering,
The University of Alabama in Huntsville, Huntsville, Alabama 35899
~Received 12 June 2001; accepted 8 November 2001; Published Online: 11 March 2002
American Association of Physics Teachers. @DOI: 10.1119/1.1435344
available upon request to the author from this ResearchGate page:
https://www.researchgate.net/publication/243492397_The_two-capacitor_problem_with_radiation


Capacitors can radiate: Some consequences of the two-capacitor problem with radiation
14 May 2003
T.C. CHOY
arXiv:physics/0305062v1 [physics.class-ph]
https://www.researchgate.net/publication/2168891_Capacitors_can_radiate_-_some_consequences_of_the_two-capacitor_problemwith_radiation
freely available at ResearchGate (direct link)
https://www.researchgate.net/profile/Tuck-Choy/publication/2168891_Capacitors_can_radiate_-_some_consequences_of_the_two-capacitor_problemwith_radiation/links/5935528daca272fc5556a317/Capacitors-can-radiate-some-consequences-of-the-two-capacitor-problem-with-radiation.pdf


Radiative effects and the missing energy paradox in the two capacitor problem
Gilberto A Urzua, Omar Jimenez, Fernando Maass and Alvaro Restuccia
Departamento de Fisica, Universidad de Antofagasta, Casilla 170, Antofagasta, Chile
E-mail: gilberto.urzua@uantof.cl, omar.jimenez@uantof.cl, fernando.maass@uantof.cl, alvaro.restuccia@uantof.cl
Journal of Physics: Conference Series 720 (2016) 012054
doi:10.1088/1742-6596/720/1/012054
https://www.researchgate.net/publication/303980311_Radiative_effects_and_the_missing_energy_paradox_in_the_ideal_two_capacitors_problem
freely available at ResearchGate (direct link)
https://www.researchgate.net/publication/303980311_Radiative_effects_and_the_missing_energy_paradox_in_the_ideal_two_capacitors_problem/fulltext/57a4bae308ae455e8539f85d/Radiative-effects-and-the-missing-energy-paradox-in-the-ideal-two-capacitors-problem.pdf


The two-capacitor problem revisited: a mechanical harmonic oscillator model approach
Keeyung Lee
Department of Physics, Inha University, Incheon, 402-751, Korea
Received 17 July 2008, in final form 4 September 2008; Published 6 November 2008
EUROPEAN JOURNAL OF PHYSICS
Eur. J. Phys. 30 (2009) 69–74    doi:10.1088/0143-0807/30/1/007
Online at stacks.iop.org/EJP/30/69
Preprint available from ArXiv: https://arxiv.org/abs/1210.4155


Entropic Considerations of the Two-Capacitor Problem
January 19, 2012
V.O.M. Lara, A. P. Lima, and A. Costa
Instituto de Fısica - Universidade Federal Fluminense
https://www.researchgate.net/publication/51990322_Entropic_considerations_on_the_Two-Capacitor_Problem
freely available at ResearchGate (direct link)
https://www.researchgate.net/profile/Ap-Lima/publication/51990322_Entropic_considerations_on_the_Two-Capacitor_Problem/links/55e9aa6208ae21d099c302fb/Entropic-considerations-on-the-Two-Capacitor-Problem.pdf


The Paradox of Two Charged Capacitors -- A New Perspective
August 2013
Authors: Ashok K. Singal
Indian Space Research Organization arXiv
https://www.researchgate.net/publication/256762725_The_Paradox_of_Two_Charged_Capacitors_--_A_New_Perspective
(direct link)
https://www.researchgate.net/profile/Ashok-Singal/publication/256762725_The_Paradox_of_Two_Charged_Capacitors_--_A_New_Perspective/links/559e349008ae76bed0bb6d46/The-Paradox-of-Two-Charged-Capacitors--A-New-Perspective.pdf


The idea I have come to so far:
You start with energy E, you end up with energy E/2 split between the two caps and E/2 lost in some way (doesn't actually matter which one and in which measure). Charge is conserved, voltage is the same.
The transformation is irreversible, therefore some energy MUST be lost and entropy must increase.
Turns out that energy it can be lost in different ways, depending on the actual circuit:  if there is appreciable resistance (a handful of microohms might suffice), it goes basically all into heat. If there is nearly no resistance or exactly zero resistance it goes into radiation (even without inductance in the loop) either by magnetic dipole or by electric dipole. If there is appreciable inductance it can go back and forth rapidly enough to lose energy by radiation of the LC oscillator (but it's usually the other ways that predominate).

[electrodacus]

A selection of references for the two-capacitor problem (one of which has already been given in this very thread)


The idea I have come to so far:
You start with energy E, you end up with energy E/2 split between the two caps and E/2 lost in some way (doesn't actually matter which one and in which measure). Charge is conserved, voltage is the same.
The transformation is irreversible, therefore some energy MUST be lost and entropy must increase.
Turns out that energy it can be lost in different ways, depending on the actual circuit:  if there is appreciable resistance (a handful of microohms might suffice), it goes basically all into heat. If there is nearly no resistance or exactly zero resistance it goes into radiation (even without inductance in the loop) either by magnetic dipole or by electric dipole. If there is appreciable inductance it can go back and forth rapidly enough to lose energy by radiation of the LC oscillator (but it's usually the other ways that predominate).

While I have not read any of those papers the fact that there are so many is ridiculous.
If you read any of those paper can you tell me if any of them actually did the test ?
Even if they have a challenge understanding the theory the experimental result will show them exactly what happens and that is the missing energy is all found in the conductors proportional with their resistance.

I guess none of them had the budget to do the superconductor experiment as they will have found out all energy remains stored and it is not radiated as some of the titles of those papers you link may suggest.

Even my simple simulation shows that losses in the case of two identical capacitors can be reduced from 50% to something like 12% or less by using an inductor and a diode to transfer the energy.

I can even provide you with a simple way to test that all loss is thermal loss not radiated (unless you call infrared photons radiated energy).
Experiment will be fairly simple:

Take to large capacitors either large electrolytic or super capacitors (just because is easier to measure the energy lost as heat not that is needed as it can be calculated).

I will give this example
3V 1F for the charged capacitor
identical 1F discharged capacitor.
Take this for example https://www.mouser.com/datasheet/2/40/AVX_SCC_3_0V-1128335.pdf
SCCR12E105PRB it has a max DC ESR of 860mOhm I will rounded up to 1Ohm.
Use a 8Ohm series resistor just that total resistance in the circuit is a round 10Ohm the 1Ohm from the two capacitors than the 8Ohm you add.

Now just use a multimeter with data logging to measure the voltage drop across the 8Ohm resistor.
You know for sure that voltage drop divided by resistance will give you the power dissipated as heat on the resistor so no need to measure any temperature and you can have very accurate measurement.
Make 5 or 10 measurements per second depending on how capable your logging multimeter is as experiment will likely need to run for about one minute to properly charge the discharged capacitor from the charged one.
Then integrate that power measurement over time to get energy and divide by 0.8 to get the entire energy dissipated as heat including the heat dissipated on the capacitor's plates that 1Ohm DC ESR.
You will get exactly all the missing energy so half of 4.5Ws that you started as as heat and the other half is split between the two capacitors as stored energy.
With accurate equipment and careful measurement you will find that not even a faction of a percent of energy is unaccounted for.

But for anyone that understand physics this experiment will not even be needed.
And yes it is sad to see all those universities and research institutions wasting resources and making fools of themselves.

[Sredni]

While I have not read any of those papers the fact that there are so many is ridiculous.

Not necessarily: the first one made the analysis for a magnetic dipole radiator, obtaining under certain simplifications a nonlinear differential equation of third order that gives an exponentially decaying solution.
IIRC, the second one generalized the results by making the analysis more general, obtaining a fifth order differential equation and finding a new type of solution that goes to zero in a finite time ('sudden death'). The third one considers radiation by electric dipole, as well.

The others consider the problem from a more general point of view (increase in entropy and loss of energy). But get the same conclusion about the necessity to lose half the initial energy. So, they are not necessarily in contrast with one another, but rather explore different ways to see the same problem.

Incidentally, there is a video by youtuber SimplyPut that reaches the same conclusions about entropy in a rather intuitive way. The guy is a bus driver, but in my opinion he got the problem right.

If you read any of those paper can you tell me if any of them actually did the test ?

I guess they had a little problem in finding ideal capacitor and getting a grant to set up the superconducting wires. So, no, I don't think you can test the ideal problem in this world. But the first paper has a very interesting graph that gives you how much of the energy goes into heat and how much into radiation, based on the value of resistance in the wires.

Even if they have a challenge understanding the theory the experimental result will show them exactly what happens and that is the missing energy is all found in the conductors proportional with their resistance.

Did you perform the experiment with an ideal capacitor and superconducting wires?

[electrodacus]

Did you perform the experiment with an ideal capacitor and superconducting wires?

No ideal capacitor or superconducting wires are needed.  To prove that all energy that is not in capacitors at the end of the test is found as heat on the resistive elements (basically all conductors in the circuit).
There will be no energy left that is not accounted for either as energy stored in capacitor or as heat so no energy is "radiated away" other than as heat.

Hope you understand that supercapacitors have nothing to do with superconductors as I mentioned supercapacitors for the test just because they have higher capacity so discharge is slower and you can use a multimeter but you can use any other type of capacitors and an oscilloscope tho you lose the vertical resolution compared to a multimeter so accuracy may be a bit lower.
If you did not heard of supercapacitors before you can check the spec that I listed for the example.

If you understand the mechanisms you do not need to do the test to know what the results will be. This is my main hobby (energy generation and energy storage) so it will be boring to do a test for witch I know exactly the result.
If you measure a voltage drop across a resistor you know that all that energy calculated form that ended up as heat.is not like an LED where some was radiated as visible photons there will be infrared radiated from the resistor but that is not the type of radiation discussed in those papers.
I feel frustrated not being able to properly explain what looks like such a simple problem.

[T3sl4co1l]

You can just admit you don't know how superconductors behave, like, it's okay to not know things sometimes??  I know it's awfully late in the thread to do so [sunk cost fallacy], but late is still better than never?

The whole reason superconductors were brought up is you claim a conservation of energy, when none exists in any possible conceivable experiment that can be constructed, using such materials and no active circuitry (e.g. DC-DC).

You've had many "outs" presented to you throughout this thread:
- Energy goes into AC mode
- "Oops, I meant RMS voltage"
- Admit the trivial circuit (two capacitances and ""ideal"" wires) cannot be constructed
- Admit the trivial circuit (two capacitances and real wires, superconducting or otherwise) behaves as expected

After the first few pages, the main curiosity in this thread, I think, revolves around the peculiar psychology causing this reluctance; the technical topics are all well settled, after all.

Tim

[SandyCox]

Check if your solution satisfies the law of preservation of charge.

You do not understand what conservation of charge means and how it is applied.
The clue is in the fact that adding another charged particle to a capacitor that is at 1V and same capacitor that is at 2V requires different amounts of energy.

What you are trying to do is not the law of conservation of charge.

Just integrate the current through the capacitors with respect to time to calculate the charge in each capacitor.

Think of it this way: You are in a sealed room with an empty box on the floor and a box full of tennis balls high op on a shelf. You move half of the balls from the full box to the empty box. In the process some of the gravitational potential energy of the balls is converted to another form (probably heat). The point is that the number of tennis balls stays the same. So we have the law of conservation of tennis balls. The law of conservation of charge is the same concept. It is about the number of charged particles, not their electric potential energy!

[electrodacus]

You can just admit you don't know how superconductors behave, like, it's okay to not know things sometimes??  I know it's awfully late in the thread to do so [sunk cost fallacy], but late is still better than never?

The whole reason superconductors were brought up is you claim a conservation of energy, when none exists in any possible conceivable experiment that can be constructed, using such materials and no active circuitry (e.g. DC-DC).

You've had many "outs" presented to you throughout this thread:
- Energy goes into AC mode
- "Oops, I meant RMS voltage"
- Admit the trivial circuit (two capacitances and ""ideal"" wires) cannot be constructed
- Admit the trivial circuit (two capacitances and real wires, superconducting or otherwise) behaves as expected

After the first few pages, the main curiosity in this thread, I think, revolves around the peculiar psychology causing this reluctance; the technical topics are all well settled, after all.

Tim

I never did experiments with superconductors.
- The properties of superconductors are zero resistance to current flow (not super low resistance but zero).
- Conservation of energy in an isolated system is always true no matter if superconductor materials are used or not.

What are you calling an active circuit ? If the circuit is powered by the only source of energy at the start of the experiment meaning the charged capacitor in this case it is completely irrelevant what components you are using as long as the system stay isolated.
The inductor and simple diode is all that is needed as I have shown in spice simulation. Neither the diode nor the inductor have any stored energy at the beginning of the experiment.
If I can demonstrate that significantly less energy is lost by just adding the diode and inductor and sowed that conservation of charge is no longer a thing then what makes you so shure that can not be shown with superconductors where no conductor in circuit will have any resistance.

I also hope you understand the concept of an isolated system.

[electrodacus]

What you are trying to do is not the law of conservation of charge.

Just integrate the current through the capacitors with respect to time to calculate the charge in each capacitor.

Think of it this way: You are in a sealed room with an empty box on the floor and a box full of tennis balls high op on a shelf. You move half of the balls from the full box to the empty box. In the process some of the gravitational potential energy of the balls is converted to another form (probably heat). The point is that the number of tennis balls stays the same. So we have the law of conservation of tennis balls. The law of conservation of charge is the same concept. It is about the number of charged particles, not their electric potential energy!

In a circuit with resistance you have charged capacitor in series with resistor in series with discharged capacitor.
Since all this elements are in series current will be the same through the loop.
So at the start you have 3V in charged capacitor say 3Ohm total circuit resistance an 0V on discharged capacitor.
Thus almost all energy is dissipated as heat on the resistor as you have 3V / 3Ohm = 1A * 3V = 3W dissipated as heat on the resistor.
A bit latter you have 2V on the charged capacitor and 1V on the discharged capacitor as current exiting the charged capacitor can only be equal with the one entering the discharged capacitor thus the special case of charge being conserved and half of the energy lost as heat.
Now you will have (2V - 1V ) / 3Ohms = 0.333A * 1V = 0.333W loss on the resistor as heat.

If you do the integration either as a calculation or as a measurement you will see that no energy is missing and half of initial energy is still in the two capacitors while the other half was lost as heat on the resistor.  There is no energy unaccounted for so no paradox.

This is as basic of electrical knowledge as it gets.  If you want to claim that there is energy lost some other way other that as heat on the circuit resistance then you are welcome to prove that either with math or with an experiment.

[TimFox]

Again, in that simple ideal circuit with a finite-value resistor connecting the second (initially uncharged) capacitor to the first (charged) capacitor, and no additional lossy elements (such as capacitor ESR or leakage) the total energy lost by the circuit itself (into heat) is independent of the resistance value, and my earlier statement holds, that it is the same value in the limit as R-->0.  Of course, the time required to transfer the charge is directly proportional to that resistor value and the limit as R-->0 of transfer time (to a specific settling fractional value such as 1 ppm, or 14 time constants) is zero.  Here, "limit" is the normal mathematical meaning of the term.

[electrodacus]

Again, in that simple ideal circuit with a finite-value resistor connecting the second (initially uncharged) capacitor to the first (charged) capacitor, and no additional lossy elements (such as capacitor ESR or leakage) the total energy lost by the circuit itself (into heat) is independent of the resistance value, and my earlier statement holds, that it is the same value in the limit as R-->0.  Of course, the time required to transfer the charge is directly proportional to that resistor value and the limit as R-->0 of transfer time (to a specific settling fractional value such as 1 ppm, or 14 time constants) is zero.  Here, "limit" is the normal mathematical meaning of the term.

Capacitor DC ESR is already included in that resistance. So there is no additional loss. Energy that is not found stored in the two capacitors at the end of the test will be found as heat in the resistive elements and that is super obvious includes the DC ESR witch is the resistance of the capacitor plates which are no different from wires.
If you admit that there is no energy unaccounted for either as stored in capacitors or as heat in the conductors than you admit that there is no paradox and there is no magical energy that is not explained by this two.

[TimFox]

Again, in that simple ideal circuit with a finite-value resistor connecting the second (initially uncharged) capacitor to the first (charged) capacitor, and no additional lossy elements (such as capacitor ESR or leakage) the total energy lost by the circuit itself (into heat) is independent of the resistance value, and my earlier statement holds, that it is the same value in the limit as R-->0.  Of course, the time required to transfer the charge is directly proportional to that resistor value and the limit as R-->0 of transfer time (to a specific settling fractional value such as 1 ppm, or 14 time constants) is zero.  Here, "limit" is the normal mathematical meaning of the term.

Capacitor DC ESR is already included in that resistance. So there is no additional loss. Energy that is not found stored in the two capacitors at the end of the test will be found as heat in the resistive elements and that is super obvious includes the DC ESR witch is the resistance of the capacitor plates which are no different from wires.
If you admit that there is no energy unaccounted for either as stored in capacitors or as heat in the conductors than you admit that there is no paradox and there is no magical energy that is not explained by this two.

Yes, all the energy is accounted for when you include the "loss" away from the circuit as heat.  If it were important, one could do calorimetry to track that loss.  Note that Spice simulations do not handle heat or EM radiation, only voltages and currents in the discrete components  of the Spice model.  My comments about the limits just are to emphasize that the "lost" energy is independent of the total circuit (series) resistance, and that the electrical components themselves are not an isolated system in the thermodynamic meaning of "isolated".  I never considered the two-capacitor system to be paradoxical.

At a trade show, I had an interesting conversation with an engineer from SBE, who make very low-loss polypropylene capacitors.  One of their series, used in high-voltage pulse systems, has such low loss that they indeed used calorimetry to measure it.

[electrodacus]

Yes, all the energy is accounted for when you include the "loss" away from the circuit as heat.  If it were important, one could do calorimetry to track that loss.  Note that Spice simulations do not handle heat or EM radiation, only voltages and currents in the discrete components  of the Spice model.  My comments about the limits just are to emphasize that the "lost" energy is independent of the total circuit (series) resistance, and that the electrical components themselves are not an isolated system in the thermodynamic meaning of "isolated".  I never considered the two-capacitor system to be paradoxical.

At a trade show, I had an interesting conversation with an engineer from SBE, who make very low-loss polypropylene capacitors.  One of their series, used in high-voltage pulse systems, has such low loss that they indeed used calorimetry to measure it.

The important part is where the loss originated in order to know what medium energy traveled through.
Since all the loss originated in wires (not outside the wires) that means energy traveled through wires.
Yes energy in the form of infrared photons will be radiated from the wires to outside the wire with circuit in a vacuum but that is after the energy was already transported.
Of course the loss is independent in this particular case of the resistance as we are talking about energy that is power integrated over time.
With a lower resistance you have higher power loss but for a shorter time duration and with a higher resistance you have lower power loss but over a longer period of time.
Glad to hear that you do not consider the two capacitor problem a paradox.
That loss is in dielectric like a parallel resistance exists with any capacitor as we do not have an ideal insulator like we have an ideal conductor.



The main point of discussion is if energy travels through wire or not.
Electrical energy is electrical power integrated over time and electrical power is electrical potential multiplied with electrical current.
Since electrical current is the rate at which the charge is moving and charge in this case is an electron and they will flow through wires unless energy is so high that air becomes a conductor (not the case in Derek's experiment).
Thus from the above the conclusion will be that energy travels through wire.   

[TimFox]

I don't see that we disagree about energy traveling through a wire and being dissipated as it travels.
With respect to capacitor resistance, a physical capacitor (either with a dielectric, or some other insulators to hold the plates in fixed locations, such as a vacuum capacitor) will have a series resistance and a parallel resistance.
When engineers discuss "ESR" of a real capacitor, they must take into account its frequency dependence.
At a single frequency, the series and parallel capacitances can be combined into a single ESR (or into a equivalent parallel resistance), by elementary circuit theory.
A physical resistance in series with the capacitor (plates and wires) will dissipate energy as heat, but conserves the charge.
A parallel (leakage) resistance will also dissipate energy as heat, but will eventually discharge the capacitors, which does not conserve the charge.

[electrodacus]

I don't see that we disagree about energy traveling through a wire and being dissipated as it travels.
With respect to capacitor resistance, a physical capacitor (either with a dielectric, or some other insulators to hold the plates in fixed locations, such as a vacuum capacitor) will have a series resistance and a parallel resistance.
When engineers discuss "ESR" of a real capacitor, they must take into account its frequency dependence.
At a single frequency, the series and parallel capacitances can be combined into a single ESR (or into a equivalent parallel resistance), by elementary circuit theory.
A physical resistance in series with the capacitor (plates and wires) will dissipate energy as heat, but conserves the charge.
A parallel (leakage) resistance will also dissipate energy as heat, but will eventually discharge the capacitors, which does not conserve the charge.

For this examples where a fairly large capacitor is discharged over a few seconds in to another capacitor the ESR of the capacitor will basically be the DC ESR witch is dependent on the DC resistance of the plates.
The parallel leakage resistance is so high for typical capacitors that will be within the measurement error of such a setup so it will not be relevant.

When Derek's says that "energy doesn't travel in wires" he is not referring to leakage he is implying that all energy travels outside the wires.
His proof is that energy arrives at the lamp in the time it takes light to travel 1m (distance between the source and the lamp/resistor) instead of the much longer transmission line.

That argument is completely irrelevant because he ignores the fact that the transmission line has capacitance. He completely ignored the line capacitance in the first video and while he acknowledged that in second video he ignored the effect of that even if that capacitance is what explains fully that energy through the lamp in first 65ns.

This can be seen as a charged parallel plate capacitor (open circuit) with large gap between the plates where you insert another plate of negligible thickness. This will transform that single capacitor in to two capacitors in series. That plate will see charges separate on each side of the plate so electrons move from one face of the plate to the other. When the plate is removed the charges will get back to neutral.
Now if you instead of one plate insert two plates connected together with a lamp then when the electrons move from one plate to the other they will pass through the lamp and the same in reverse will happen when you get those plates out of the electric field.
The energy in the charged capacitor will remain the same so energy was not provided by that but by the person that inserted those plates.