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Veritasium "How Electricity Actually Works"

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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:

--- Quote from: TimFox on May 14, 2022, 01:18:50 am ---Yes, “isolated system” is the key point.
If resistor heat dissipation can escape from the bench top, then the system is not isolated.

--- End quote ---

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:

--- Quote from: TimFox on May 14, 2022, 01:53:47 am ---Yes, you can do calorimeter on the insulated box and measure the thermal portion of the system energy.

--- End quote ---

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.

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