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Math Help - RLC Circuit Question

  1. #1
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    RLC Circuit Question

    I have the following question to work out:-

    A defibrillator consists of a open circuit containing a capacitor of 40 \muF 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.
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  2. #2
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    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.
    Thanks from topsquark
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