# Capacitor charge - LRC series circuit

1. Nov 25, 2007

### america8371

In differential equations I was assigned a few circuit exercises. I'm having trouble understanding them, but I'm pretty sure that if can understand the first one then I can get the rest of them. The first exercise asks for time of capacitor charge and the others ask for things like capacitor maximum charge, steady state current, current/charge equations, etc.

1. The problem statement, all variables and given/known data
Find the charge on the capacitor in an LRC series circuit at t=0.01s when L=0.05h, R=2$$\Omega$$, C=0.01f, E(t)=0V, q(0)=5C, and i(0)=0A. Determine the first time at which the charge on the capacitor is equal to zero. Please give answer to four decimal places.

a. t = 0.5095s
b. t = 0.9595s
c. t = 0.0509s
d. t = 0.5959s
e. t = 0.9059s

2. Relevant equations
I'm not too sure but i think the correct diff. eq. to use is

L $$\frac{d^{2}q}{dt^{2}}$$+R $$\frac{dq}{dt}$$+ $$\frac{1}{C}$$ q = e(t)

3. The attempt at a solution
I attempted to solve it by method of undetermined coefficients.

q''+40q'+2000q=0

r$$^{2}$$+40r+2000=0 r=-20$$\pm$$40i

q$$_{p}$$(t) = Ae$$^{-20t}$$cos 40t + Be$$^{-20t}$$sin 40t

etc.

It feels like I'm going in the wrong direction because it seem like theres more to this exercise than just solving the diff. eq.

2. Nov 25, 2007

### HallsofIvy

Staff Emeritus
Well, yes, there is more! But you are going in exactly the right direction. You just haven't yet answered the question! You are told that q(0)=5C, and i(0)= q'(0)= 0A. That lets you find A and B. And then you need to determine the value of t such that qp(t)= 0.