1. Nov 1, 2009

### jemitu

RC Circuit discharge question - test prep for tomorrow

1. The problem statement, all variables and given/known data

A charged 6$$\mu$$F capacitor is discharged across a 2$$\Omega$$ resistor. How long does it take to lose half its stored energy?

2. Relevant equations
....so many:

Q = CV
V = IR
W = I2R
U = 1/2 QV
V = V0e-t/RC

3. The attempt at a solution

I'm not really sure where to start on this one. I'm trying to find a way to resolve Q and V where I only have one unknown, given C and R, but I'm having trouble. This is prep for an exam tomorrow, so any help is greatly appreciated.

Last edited: Nov 1, 2009
2. Nov 1, 2009

### rl.bhat

Hi jemitu, welcome to PF.
Energy stored in the capacitor is given by
E = 1/2*C*Vo^2.
In the expression C remains constant.
To change E to E/2, what should be the value of V in terms of Vo?

3. Nov 1, 2009

### jemitu

Thanks for responding so quickly, I'm getting a little overwhelmed here.

Given the energy equation you get Vo = sqrt(2E/C).
plugging in to V(t), V= sqrt(2E/C) e^(-t/RC)

But when you do the math, you don't have the original E so you can't solve it. The question is multiple choice with values, not expressions, hence my severe confusion. I think I need a little more hand-holding.

4. Nov 1, 2009

### rl.bhat

V0 = sqrt(2E/C)
When E changes to E/2
V(t) = sqrt(2*E/2*1/C) = 1/Sqrt(2)*sqrt(2E/C) = V0/sqrt(2)
Substitute this in the equation and solve for t