# RC-circuit for charged capacitor

1. Jan 13, 2010

### Apteronotus

1. The problem statement, all variables and given/known data
I have a simple circuit consisting of a charged capacitor and a resistor connected in series to a battery.
Suppose that initially the potential across the capacitor is greater than that of the battery. (ie. $$\frac{q}{C}>V$$).

What happens to the capacitor when the switch is closed and the circuit completed?

2. Relevant equations
Kirchhoff's 2nd:
$$V+IR-\frac{q}{C}=0$$
Solving for q:
$$q(t)=CV+(q_0-CV)e^{(t-t_0)/RC}$$
where $$q_0$$ is the initial charge on the capacitor at time $$t_0$$

3. The attempt at a solution
I'm guessing that when the switch is closed we should expect the capacitor to discharge to a certain degree and after some time have the same potential as the battery.

But my equation $$q(t)=CV+(q_0-CV)e^{(t-t_0)/RC}$$ does not reflect this.

As $$t\rightarrow\infty$$ the charge on the capaictor $$q(t)\rightarrow (q_0-CV)e^{(t-t_0)/RC}\rightarrow\infty$$

What am I doing wrong?

2. Jan 13, 2010

### vela

Staff Emeritus
You have the wrong sign on the IR term.

3. Jan 13, 2010

### Apteronotus

Are you sure? My reasoning is that if the potential on the capacitor is larger than on the battery then
1. electrons will flow from the battery to the capacitor
so
2. current is flowing in counter-clockwise direction - from the capacitor to the battery.

Lastly, if a resistor is traversed in the direction opposite the current, the potential difference across the resistor is +IR.

4. Jan 13, 2010

### vela

Staff Emeritus
The problem you're running into is that $I_r = -I_c$. You're assuming $I=dq/dt$, which is negative if the capacitor is discharging, and using it as a positive quantity when calculating the voltage drop across the resistor.

Trying to get the signs right by reasoning which way the current flows is a mistake waiting to happen. It's much easier to just assume a direction for the current. Then for capacitors and resistors, the voltage "drops" going from where the current enters to where it leaves. For the battery, the opposite convention holds because the battery is a source, not a sink.

So for this circuit, assume the current flows clockwise through the circuit. Starting at the negative terminal of the battery and going clockwise, first, the potential go up by V; then it drops by IR across the resistor; and then it drops by q/C across the capacitor. If the current actually flows in the other direction, I becomes negative, and the direction of the voltage drop across the resistor automatically flips. By following this convention, you're also guaranteed that I=dq/dt.

5. Jan 13, 2010

### Apteronotus

Vela thank you for your explanation.

So we always treat a battery as positive and a resistor and capacitor as negative?

I guess my confusion lies in the fact that I expect the capacitor to act as a current source since it potential is larger than that of the battery.