# Does resistance affect a capacitor charging from another capacitor?

Consider two capacitors connected to each other, where one is charged (5.0V) and the other has no charge stored: 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? • Well... what do you think happens when current flows through a resistor? Mar 11, 2017 at 10:06
• Voltages? Surely you mean VOLTAGE. You do know they both end up at the same level riiiight? Mar 11, 2017 at 15:07

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...

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.

• 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,,
– Chu
Mar 11, 2017 at 17:09
• @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. Mar 11, 2017 at 17:20