# Two Capacitors with a Resistor

Discussion in 'Homework Help' started by ErnieWork, Sep 28, 2011.

1. ### ErnieWork Thread Starter New Member

Sep 28, 2011
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0
I am given two circuits, each with two capacitors and a switch. Capacitor 1 is initially charged. I am asked to find the final charge in each capacitor after the system reaches equilibrium.

In P1 I set the final voltage of each capacitor equal to each other and I get

Q1/C1 = Q2/C2 Q1 + Q2 = Q0

subsituting Q2 = Q0 - Q1 and solving, i get

Q1 = (2/5)*Q0 and Q2 = (3/5)*Q0

But when the resistor is included, I cannot figure out what will happen. There are two things that do not add up to me:

-The voltage between the terminals of both capacitors will be equal again
-The total charge Q0 will be the same.

-The potential energy of the system will be lower than it was at t=0 due to the resistor dissipating some of the energy.

It seems like I should use the same method as p1 and set the voltages equal to each other, but I know that in part p1 the potential energy is conserved but in p2 some energy is dissipated.

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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The final voltage and total charge (conservation of charge) will be the same in both cases. As you rightly point out, there is a net energy loss which again is the same in either case. The problem is accounting for the energy loss in the first case where there is no resistor. In the second case the energy is lost as heat in the resistor. In the first case there is no dissipative resistance element - so where did that energy go? This an idealized situation, because in a practical circuit there will always be a means of dissipating energy in the wiring & component parasitic resistances.

As a conceptual exercise it is possible to account for the energy loss in the idealized situation with zero resistance, by allowing for the generation of an electromagnetic impulse wave propagating away from the circuit. But this is a rather (fanciful?) fancy way of explaining a situation that can't exist in practice. Or maybe not - one could search for an electromagnetic impulse wave near a practical circuit (with an intentionally low a resistance as possible) with a very high rate of change in current and somehow do an energy balance. Not an experiment I'm interested in however.

Last edited: Sep 28, 2011
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3. ### ErnieWork Thread Starter New Member

Sep 28, 2011
2
0
Ahh I see, I realize that now that I look at the potential energy equation. Are all circuits with time varying current releasing EM radiation then?

Thank you for clearing this up.