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Homework Help: LC-circuit problem

  1. Nov 7, 2011 #1
    An LC-circuit has L = 64mH, C = 121nF.
    At the time t=0, the charge on the condensator is 10μC and the current in the inductor is 0.3A. What is the current in the inductor as a function of time?

    With:[itex]q = Q_mcos(\omega_0t+\phi)[/itex] we get that: [itex]I = Q_m\omega_0cos(\omega_0t+\phi + \pi/2)[/itex] (because I = dq/dt), where [itex]\omega_0 = 1/(√LC)[/itex]

    I thought that the initial conditions would mean that I have to solve:
    [itex]q(0) = 10*10^{-6} =Q_mcos(\phi) [/itex] and [itex]I(0) = 0.3 = Q_m\omega_0cos(\phi + \pi/2)[/itex]. But these equations have no solution(!)
  2. jcsd
  3. Nov 7, 2011 #2

    rude man

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    Here's what I do with a problem like this:

    1. write the differential equation without regard to initial conditions.
    2. s-transform the equation term-by-term, paying attention to
    f'(t) <--> sF(s) - f(0+) and f''(t) <--> s2F(s) - sf(0+) - f'(0+).

    The you just invert the ensuing transfer function back to the time domain, and you get all the i.c.'s included.

    (If you haven't had the Laplace transform yet I don't know what to tell you.)
  4. Nov 8, 2011 #3
    No, I don't know Laplace:uhh:
  5. Nov 8, 2011 #4

    rude man

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    OK, then you have to use standard differential equations for the network, and solve in the traditional diff e manner including initial conditions.

    I didn't follow you argument but I would write

    i = -CdV/dt (i is + if flowing out of C)
    V = Ldi/dt (i is + if flowing into L)

    So i = -C(d/dt)Ldi/dt = -LCd2i/dt2 or

    LCd2i/dt2 + i = 0

    I.C. 1: i(0) = 0.3A
    I.C. 2: di/dt(0+) = V0/L but V0 = Q0/C = 1e-5/C so di/dt(0+) = Q0/LC

    So now just solve the 2nd order diff eq. with those two initial conditions.
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