# Capacitor in circuit

1. Feb 18, 2008

### parkskier

1. The problem statement, all variables and given/known data
V = 115 V, R = 30 , and the switch has been closed for a very long time.

Diagram:

2. Relevant equations

Q=CV
V=IR

3. The attempt at a solution

I thought it would be a good idea to use the Loop Laws I get the following equations (If Loop 1 is the larger loop with the battery, and Loop 2 is the smaller one):

L1: 115-60I1-30I3=0
L2: 30I3-10I2-Q/2.0microF=0
I1=I2+I3

Where I1 is the current through the battery and 60 ohm resistor; I2 is the current through the 10 ohm resistor and capacitor; I3 is the current through the middle resistor.

However, with this there are 3 equations and 4 variables, so something is wrong, or I'm approaching it wrong. Thanks for any help.

2. Feb 18, 2008

### Defennder

First of all you should note that if the switch has been closed for an infinite period of time and you're using a DC source, then the current flowing through the right loop containing the 10 ohm resistor and the capacitor is 0A. You need to solve it for the case where t<0 and t>0 separately.

3. Feb 18, 2008

### parkskier

Sorry, I forgot one part of the problem. I'm supposed to solve for what the charge on the capacitor is after the switch has been closed for a very long time...and then solve for how much time it takes for the capacitor to lose 10% of it's charge when the switch is opened.