Capacitor charge - LRC series circuit

america8371
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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.

Homework Statement


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


Homework 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)


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\pm40i

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 there's more to this exercise than just solving the diff. eq.
 
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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 let's you find A and B. And then you need to determine the value of t such that qp(t)= 0.
 
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