Consider a simple series RC circuit at steady state (capacitor is full).

I've been told that once the capacitor is full we can literally "cut" the circuit because no current can flow.

That argument makes sense, but because the capacitor is full, doesn't that also mean it has a potential difference across it? In other words, the capacitor is a new battery? Then we should apply Kirchhoff Law in the circuit if we want to find the current? (even though I just said it was zero).

What is going on? What is this contradiction?

There is no current in the circuit yet clearly the potential difference in the capacitor is NOT zero.


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For part (c), why do I have to use $I = I_0 e^{-t/ \tau}$? Clearly at steady state, the current of the right circuit is just the potential difference across the capacitor (which now can be treated as a battery) and the resistance. As pointed out before I can only apply Ohm's Law to resistors and batteries, but at steady state the capacitor can be thought of as a battery. So the contradiction lies here.

On the left circuit the current is simply $10V = I_1(50k \Omega)$, and on the right it should be $10V = I_2(100k \Omega)$ only because this is steady state

EDIT2 Let me clarify my question again. I am also confused why $I = I_0 e^{-t/ \tau}$ is used? Why does the capacitor discharge? Yes it's considered an isolated circuit once the switch is shut, but does that mean even though part of the circuit is still hooked up to the battery, that the current doesn't continue to build to prevent discharge?


2 Answers 2


You are right, there is a potential difference across capacitor according to the expression

$$\Delta V = \frac{Q}{C},$$

where $Q$ is charge on the capacitor and $C$ is capacity of the capacitor. And yes, capacitor can be thought of as being a tiny rechargeable battery (see below).


  • When you close the switch you are no longer in steady state.
  • Ideal battery is taken to have infinitive charge $Q = \infty$ and constant voltage $\Delta V = \text{const}$ . So $\Delta V = \frac{Q}{C}$ is not valid for ideal batteries.
  • Capacitor on the other hand can change voltage according to $\Delta V = \frac{Q}{C}$.
  • $\begingroup$ That's my point, so why do we often "short out" the capacitor during steady-state? It contradicts V = IR, Ohm's Law and Kirchoff's Junction rule (that is, the shorting part) $\endgroup$
    – Lemon
    May 20, 2012 at 20:56
  • $\begingroup$ You cannot short out the capacitor during steady state. You just say that current through it is zero! And Ohm's law is by no means valid either for capacitors or for inductors, only for resistors. $\endgroup$
    – Pygmalion
    May 20, 2012 at 20:59
  • $\begingroup$ Let me edit my question. Also, current can't go through a capacitor in the first place. $\endgroup$
    – Lemon
    May 20, 2012 at 21:04
  • $\begingroup$ @jak see the edit in my answer $\endgroup$
    – Pygmalion
    May 20, 2012 at 22:02

Cutting out the capacitor of a circuit in a steady-state situation is a valuable trick to simplify a more complex system.

While you are correct, that there is a potential difference between the capacitor, this is also true for other parts of your circuit. In a 'complete' model you need capacitors between every bit of a circuit that is on a different voltage level. This is called stray capacitance. Modeling on this level of detail is hard and often only necessary when you use very high frequencies.

So you need to apply Kirchhoff's law if there is current flowing, i.e. you are not in a steady-state situation.

Just charge a big capacitor with a DC power supply through a resistor and measure the voltage across the capacitor. It will steadily rise to the output voltage of the power supply. Then cut off one of the legs of the capacitor. Your voltmeter will not show any difference. That is why you can neglect it.


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