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Veritasium "How Electricity Actually Works"

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TimFox:
Also, check what the charge and voltage ratios as a function of the series resistance and what the limits as R-->0 are.

electrodacus:

--- Quote from: SandyCox on May 13, 2022, 06:25:48 pm ---Check if your solution satisfies the law of preservation of charge.

--- End quote ---

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.



electrodacus:

--- Quote from: TimFox on May 13, 2022, 06:40:01 pm ---Also, check what the charge and voltage ratios as a function of the series resistance and what the limits as R-->0 are.

--- End quote ---

Any resistance different from zero will result in the same exact final state for two identical parallel capacitors and that will be that half of the energy will be lost as heat in conductors due to resistance to current flow.
You basically have a resistor divider. The value of the resistance will only influence the time it takes for the energy to be transferred and wasted as heat.
To test this just add a 1 Ohm resistor between the two capacitors and then add a 1KOhm resistor and you will see that same amount of energy will be wasted as heat and you will have the same half of the original voltage in the two identical capacitors.

TimFox:
Exactly.
Since the final result is independent of the finite resistance in the circuit, it will also be the result in the limit as R-->0.

Although the famous conservation of energy law is true, it is not always relevant to a given situation unless you can include the exact losses due to, for example, resistor heat loss and EM radiation.
In first-year physics, we learned to distinguish between conservation of momentum (true) and conservation of kinetic energy (not always true) in simple mechanical calculations.

electrodacus:

--- Quote from: TimFox on May 13, 2022, 07:07:33 pm ---Exactly.
Since the final result is independent of the finite resistance in the circuit, it will also be the result in the limit as R-->0.

--- End quote ---

Can you rephrase that ? Id not not understand what you mean.
As long as resistance is higher than absolute zero half of the energy will be lost as heat in that resistance so you end up with just half the energy in this particular systemic case of two identical capacitors.
If capacitors are not identical then percentage of loss as heat will be different than half.
All energy is accounted for in any setup. Q on the other hand that is Q = C*V will not be the same in any other experiment other than the two identical capacitor case with resistance higher than zero. This is due to symmetry so it is just a coincidence and not a rulle.

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