# RLC Circuit Question

• Apr 17th 2013, 05:19 AM
tammyl
RLC Circuit Question
I have the following question to work out:-

A defibrillator consists of a open circuit containing a capacitor of 40 $\mu$F and an inductor of 0.05H. A patient has a resistance of 60 $\Omega$ when the defibrillator is discharged through him. Find the current during discharge if the capacitor is charged initially to 5000V.

Solving the second order linear D.E. for RLC circuits from:

$L\frac{d^2Q}{dt^2} + R\frac{dQ}{dt} + \frac{1}{C}Q = E$

I got the following general solution for Q (where L = 0.05, R = 60, C = $40 x 10^{-6}$ and E = 5000) -

$Q(t) = c_{1} e^{-600t} cos 100 \sqrt{14}t + c_{2} e^{-600t} sin 100 \sqrt{14}t + 0.02$

I am confused as how to get $c_{1}$ and $c_{2}$. I used the intial conditions of t=0 and Q=0 to get $c_{1}$ = -0.02 but I am unsure as how to get $c_{2}$.

Also how do I get the current? Would it be also at t=0?

Any suggesstions would be greatly appreciated.
• Apr 17th 2013, 06:35 AM
russo
Re: RLC Circuit Question
First of all I would (just for convenience) rewrite Q(t) as $Q(t) = e^{-600t}\left[c_1 cos\left(100 \sqrt{14} \, t\right) +c_2 sin\left(100 \sqrt{14} \, t\right)\right]\, \, (1)$.

Secondly, $i(t) = \frac{dQ(t)}{dt}$ where i is current.

Also, have you assumed that Q(0) = 0?, because, as I understand, since $C = \frac{Q}{\Delta V}\, \, (2)$ and $\Delta V = 5000\, Volts$ and C is constant, there are charges involved from the beggining (the capacitor is storing that charge eventually).

The problem also states that the capacitor is discharging so the charges are moving (which means that $i(0) \neq 0$).

As for the constants, I would get Q from (2) and replace it in (1). There we have our first equation. For the second one I would replace (1) in the differential equation and that seems to be solvable.

Finally, the current will be a time function.

As an engineering student, I'd approach it that way. Hope it helps.