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Thread: RLC Circuit Question

  1. #1
    Junior Member
    Oct 2012

    RLC Circuit Question

    I have the following question to work out:-

    A defibrillator consists of a open circuit containing a capacitor of 40 $\displaystyle \mu$F and an inductor of 0.05H. A patient has a resistance of 60$\displaystyle \Omega$ 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:

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

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

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

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

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

    Any suggesstions would be greatly appreciated.
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  2. #2
    Newbie russo's Avatar
    Jan 2013
    Mar del Plata

    Re: RLC Circuit Question

    First of all I would (just for convenience) rewrite Q(t) as $\displaystyle Q(t) = e^{-600t}\left[c_1 cos\left(100 \sqrt{14} \, t\right) +c_2 sin\left(100 \sqrt{14} \, t\right)\right]\, \, (1)$.

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

    Also, have you assumed that Q(0) = 0?, because, as I understand, since $\displaystyle C = \frac{Q}{\Delta V}\, \, (2)$ and $\displaystyle \Delta V = 5000\, Volts$ 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 $\displaystyle i(0) \neq 0$).

    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.
    Thanks from topsquark
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