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Veritasium "How Electricity Actually Works"
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electrodacus:

--- Quote from: TimFox on May 14, 2022, 04:25:09 pm ---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.

--- End quote ---

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:

--- Quote from: electrodacus on May 14, 2022, 04:30:28 pm ---
--- Quote from: TimFox on May 14, 2022, 04:25:09 pm ---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.

--- End quote ---

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.

--- End quote ---

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:

--- Quote from: TimFox on May 14, 2022, 05:11:30 pm ---
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.

--- End quote ---

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:

--- Quote from: TimFox on May 14, 2022, 06:00:07 pm ---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.

--- End quote ---

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.
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