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Veritasium "How Electricity Actually Works"
electrodacus:
--- Quote from: hamster_nz on May 03, 2022, 10:52:14 pm ---
--- Quote from: electrodacus on May 03, 2022, 10:16:57 pm ---Also your do not provide any feedback on where my math is wrong. Math that shows the amount of energy in a capacitor.
--- End quote ---
I have done that many times.... but one more time...
Can we agree that the total charge on the top half of the system is:
Qtotal = Cc1 Vc1 + Cc2 Vc2
So here is the initial state of the system:
Qtotal = 1F * 3V + 1F * 0V = 3 Coulomb.
I hope both agree that no charge can cross either capacitor regardless of the position of the switch. So the total charges on the top and bottom halves must remain the same before and after the switch is closed. If not, please let me know what mechanism is taking electrons out of the top half and moving them to the bottom half, as I am sure we both agree that no positively charged atoms are moving around either.
The switch is closed, and allowing for the incorrect assumption that everything settles down to a stable state when using ideal components:
Qtotal = Cc1 Vc1 + Cc2 Vc2
Qtotal = 1F * 1.5V + 1F * 1.5 = 3 Coulomb.
This is the same as you would have if you consider the two 1F capacitors now a single 2F capacitor:
Qtotal = 2F * 1.5V = 3 Coulomb.
We have the same amount of charge, but are left with only half the voltage over each capacitor. The incorrect assumption that the system becomes stable has let us down.
A simpler example that shows how flawed your position is
What about a single charged ideal capacitor (no ESR), ideal wires (zero ohms) and a single ideal switch (zero ohms when closed, perfect isolation when open), that shorts the capacitor. If, as you say, energy must be conserved, what happens if the capacitor has been charged to 3V and the switch is closed?
What happens with "close to ideal wires", that have zero ohms resistance, but a small amount of inductance?
The inductance is important, as it gives a solution where the total energy in the system is conserved.
--- End quote ---
As I mentioned you are confusing charge with energy.
The important law is called Conservation of energy and not conservation of charge.
Please do a energy balance not a charge balance and you will understand what I'm saying.
When you do the transfer from one capacitor to the other you will be losing half of the energy as heat.
Once you understand the difference between charge and energy it should be easier to also understand what I'm saying.
3 Coulomb is like saying 0.833mAh so you say nothing about energy.
TimFox:
If one ignores the inductance in the circuit, and just treats it as having a finite resistance R in series with the two equal capacitors:
The final distribution of charge (1/2 on each capacitor) is independent of R, but the time required to equilibrate decreases as R decreases.
Since the final distribution is independent of R, it will still be true in the limit as R --> 0.
Also, the total energy after equilibrium in the two-capacitor system will also be 1/2 the original total energy in only one capacitor, independent of R and therefore in the limit as R --> 0.
hamster_nz:
--- Quote from: electrodacus on May 03, 2022, 11:03:35 pm ---
--- Quote from: hamster_nz on May 03, 2022, 10:52:14 pm ---A simpler example that shows how flawed your position is
What about a single charged ideal capacitor (no ESR), ideal wires (zero ohms) and a single ideal switch (zero ohms when closed, perfect isolation when open), that shorts the capacitor. If, as you say, energy must be conserved, what happens if the capacitor has been charged to 3V and the switch is closed?
What happens with "close to ideal wires", that have zero ohms resistance, but a small amount of inductance?
The inductance is important, as it gives a solution where the total energy in the system is conserved.
--- End quote ---
As I mentioned you are confusing charge with energy.
The important law is called Conservation of energy and not conservation of charge.
Please do a energy balance not a charge balance and you will understand what I'm saying.
When you do the transfer from one capacitor to the other you will be losing half of the energy as heat.
Once you understand the difference between charge and energy it should be easier to also understand what I'm saying.
3 Coulomb is like saying 0.833mAh so you say nothing about energy.
--- End quote ---
But voltage is also clearly defined, through the capacitance formula of Q = C V, and you can calculate energy directly from charge and capacitance U = 1/2 Q^2 / C
And the answer in the "one cap, one switch" situation?
electrodacus:
--- Quote from: TimFox on May 03, 2022, 11:03:55 pm ---If one ignores the inductance in the circuit, and just treats it as having a finite resistance R in series with the two capacitors:
The final distribution of charge (1/2 on each capacitor) is independent of R, but the time required to equilibrate decreases as R decreases.
Since the final distribution is independent of R, it will still be true in the limit as R --> 0.
--- End quote ---
Yes 1/2 of the charge but 1/4 of energy. Half of the total energy was dissipate as heat on the series resistance no matter the value of the series resistance.
Please understand that discussion is about transfer of energy.
TimFox:
--- Quote from: electrodacus on May 03, 2022, 11:06:58 pm ---
--- Quote from: TimFox on May 03, 2022, 11:03:55 pm ---If one ignores the inductance in the circuit, and just treats it as having a finite resistance R in series with the two capacitors:
The final distribution of charge (1/2 on each capacitor) is independent of R, but the time required to equilibrate decreases as R decreases.
Since the final distribution is independent of R, it will still be true in the limit as R --> 0.
--- End quote ---
Yes 1/2 of the charge but 1/4 of energy. Half of the total energy was dissipate as heat on the series resistance no matter the value of the series resistance.
Please understand that discussion is about transfer of energy.
--- End quote ---
1/4 of the energy in each capacitor, for a total energy in the two-capacitor system half the original energy.
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