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Consider two capacitors connected to each other, where one is charged (5.0V) and the other has no charge stored:

Capacitors1

When they are connected, current flows to even out the charge and the resulting voltages can easily be determined from the ratio of the two capacitances. My question is: when a resistor is added between them (diagram below), does that affect the conservation of charge and result in lower final voltages?

Capacitors2

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  • \$\begingroup\$ Well... what do you think happens when current flows through a resistor? \$\endgroup\$ – Finbarr Mar 11 '17 at 10:06
  • \$\begingroup\$ Voltages? Surely you mean VOLTAGE. You do know they both end up at the same level riiiight? \$\endgroup\$ – Trevor_G Mar 11 '17 at 15:07
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when a resistor is added between them (diagram below), does that affect the conservation of charge and result in lower final voltages?

Nope.

Think about it this way: your circuit is split into two halves by the capacitor dielectrics. On your schematic, you can draw a vertical line in the middle, going through the capacitors.

Conservation of charge means that charge on both sides of this line is constant.

Since Q=CV, then C1*V1 + C2*V2 will be constant, and that will allow you to calculate the capacitor voltages once steady state is reached.

The resistor does not influence the end result. A higher value will simply need more time to equalize voltage in your caps.

Now, the real amusing thing here is if you calculate the energy. Since E=1/2 CV^2 you'll probably notice that the energy at the end is lower than the energy at the beginning, even if the resistor is zero...

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The resistor makes the charge redistribution slower, but finally the voltage is the same in both cases. The charge remains constant in both cases. In the first case probably the energy waste is more via radiation and less via heating the wires. In the second case the most is dissipated in the resistor, but the total energy loss is the same in both cases.

You can calculate the final voltage VT = Q/15uF where Q is the total original charge; Q = 10uF * 5V

The loss is the energy difference between 0.5 * 10uF* (5V)^2 and

0.5 * 15uF *(VT)^2

In pure theory with no losses in anywhere, the circuit gets never stable. But in the practice the oscillation fades away very soon due the losses and my formulas are based on that.

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  • \$\begingroup\$ How can the voltages be the same in both cases? In the second, there is energy transfer as the resistor dissipates power. In the first there is no power dissipation. ."'m,, \$\endgroup\$ – Chu Mar 11 '17 at 17:09
  • \$\begingroup\$ @Chu when 2 capacitors with different voltages are connected together, there appears a current that balances the voltages. In the first case the current spike is huge in the practice, too. Theoretically it's infinite (=U/ 0 Ohms) That causes a radiowave that dissipates energy out of the circuit as much as the resistor generates heat in the second case. Imagine into case 1 added some superconductive shielding that prevents the radiowave to escape => the system oscillates infinitely, no final balance would be found. \$\endgroup\$ – user287001 Mar 11 '17 at 17:20
  • \$\begingroup\$ Your answers are pure fantasy. Radio wave, continuous oscillation, no balance ... \$\endgroup\$ – Chu Mar 11 '17 at 17:44
  • \$\begingroup\$ @Chu zero ohm wiring in the question is the start of the fantasy. But the radio wave is detectable from a practical circuit. Put the version 1 together (=connect an empty and a charged capacitor ) near an AM radio receiver and hear the click. \$\endgroup\$ – user287001 Mar 11 '17 at 17:51
  • \$\begingroup\$ @Chu Maybe you should learn some history. Mathematician James C Maxwell ruminated at his writing table all the easy to derive contradictions in the electricity. You seem to ruminating one of those contradictions now. Finally he find a solution: Add a short ad hoc term to the basic equations and voila' - no more contradictions. That term gave the birth to the theory of radiowaves. Radiowaves start from where the current or the voltage changes fast. In our example the connection starts a high current spike \$\endgroup\$ – user287001 Mar 12 '17 at 16:48
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To add a little to the other answers, although intuitively it seems that the voltages should be lower in the case with the resistor (because the resistor will dissipate energy as heat, and lower electrical energy should result in lower capacitor voltages, and in turn lower charge), actually the energy loss is the same and not a function of the resistor. I.e. if the resistor was reduced from 1k the result would be the same, even toward zero resistance.

I think this page should help convince you of that. It deals with energy and not charge conservation, but I think if you agree with the energy loss not being affected by resistance you can then accept the charge conservation result.

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