# Finding the Steady State Charges

Here the problem states to find the steady state charges on the condensers.<<<<

According to me the charges on second at steady must be zero else there would be some current in the resistance in it's parallel(which can't be true because then $C$ too will have current in it and it will not be steady state) . Is this argument correct or there would be something else in this case?

How does the charge initially that started accumulating on the $2C$ capacitor goes away at steady state?

Initially the capacitor does not pose resistance so, maximum charges pass through the capacitor instead of the resistance.But finally the charge on it is zero,please explain what happens in mean time.

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This question saved my day.Thank you! – ZOZ Oct 31 '15 at 18:31

Your argument is correct that there if there is no current in the resistor which is in parallel with the $2C$ condenser, then the charge on the $2C$ condenser must be $0$. This, as you probably already know, is because the two elements are in parallel and so they must have the same potential across them. However, you should give a clear argument as to why the current through the resistor is zero in the first place.

Having done this you ought to be able to find the potential drop across the series resistor. After this you should be able to find the potential drop, and charge on, the remaining condenser. I will let you fill in the details yourself.

The short answer to how the capacitor discharges is that it discharges through the resistor it is in parallel with.

It will probably be useful to view the problem this way:

You can consider the power supply and the resistor together to be one compound object: an power supply with internal resistance. Any real power supply will have such an internal resistance.

You can consider the capacitor and resistor in parallel as a leaky capacitor that will discharge if left for long enough. Any real battery behaves like this. The other capacitor, then, is like an ideal battery.

Phrasing the problem this way, the situation is this: you are charging a perfect battery and a leaky battery in series with a non-ideal power supply. At first, both batteries start charging, but as the leaky battery leaks (through its internal resistor), the voltage drop concentrates more and more on the ideal battery, and the voltage drop on the leaky battery goes to zero.

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Sorry, the series resistor is the one on the bottom right. – NowIGetToLearnWhatAHeadIs Apr 30 '13 at 13:25
I was saying you should be able to find the potential drop across the resistor, and by your previous comment you just did. You found it to be zero. – NowIGetToLearnWhatAHeadIs Apr 30 '13 at 13:38
See my additional question. – ABC May 27 '13 at 4:04