# Homework Help: Discharging capacitor, the differential equation

1. Feb 21, 2012

### Sefrez

First applying Kirchhoff's loop rule with an ideal emf device with potential difference V, a capacitor with capacitance C, and a resistor with resistance R, you get:
V - q/C - iR = 0 or V = q/C + dq/dt*R

For charging a capacitor. This makes sense. But for a discharging capacitor in the same circuit but without the emf device, you have the differential equation: q/C + dq/dt*R = 0

But this makes no sense to me. The current is in the other direction when the capacitor is being discharged. The higher potential side marks the direction of that current. That being said, applying Kirchhoff's loop rule, you end up with -iR + q/C = 0 which is not the same.

Am I applying the rule incorrectly?

2. Feb 21, 2012

### kushan

if i am not wrong dq/dt is negative in case of discharging

3. Feb 21, 2012

### Sefrez

Oh, that might be correct. It has to be stated as negative in the differential to indicate decreasing of charge. My book through me off. It simply used the excuse that the battery is no longer present and thus V = q/C + dq/dt*R becomes 0 = q/C + dq/dt*R. When really there is more going on. Thanks. :)

4. Feb 21, 2012

### ehild

The homogeneous part of the equation is the same, but solving the equation with the initial condition q=Q0 at t=0, you get negative current and negative dq/dt. Kushan is right, the current flows in the opposite direction during discharge and the charge of the capacitor decreases which makes dq/dt negative.

ehild

5. Feb 21, 2012

### Sefrez

Yeah, I see. Thanks to both of you!