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Homework Help: Overdamped RLC circuit

  1. Mar 30, 2008 #1
    [SOLVED] Overdamped RLC circuit

    1. The problem statement, all variables and given/known data

    In a parallel RLC circuit determine [tex]i_R(t)[/tex].
    R = 20 mohms
    L = 2mH
    C = 50 mF

    v(0+) = 0 (capacitor)
    i(0-) = 2mA (inductor)

    My question is what is [tex]i_R(0^+)[/tex]? According to my final answer, it should be 0. However, went I graph it with PSPICE, it looks like it starts out somewhere at -186mA. I know that i_R can change instantaneously but the graph that pspice makes, makes it look like it will never be 0. I was under the impression that if the voltage of the capacitor is 0 then iR(t) will be 0 regardless of the current in the inductor. Is this correct?

    2. Relevant equations

    [tex]\alpha = 500 Hz[/tex]
    [tex]\omega_0 = 100 Hz[/tex]
    [tex]i_R(t) = Ae^{-10.10t} + Be^{-989.9t}[/tex]

    3. The attempt at a solution

    Using these two equations
    A + B = 0
    [tex]i_c(0+) + i_L(0+) + i_R(0+) = 0[/tex]
    [tex]RC di_R/dt = -2 mA[/tex]
    [tex]i_R(t) = -2e^{-10.10t} + 2e^{-989.9t} mA[/tex]
    Last edited: Mar 30, 2008
  2. jcsd
  3. Mar 30, 2008 #2


    User Avatar

    Yes, you are correct. Since [tex]V_R(0^+) = V_C(0^+) = 0[/tex], [tex]i_R(0^+)[/tex] should also be zero.
    You should write your differential equation using [tex]i_L[/tex] as the independent variable.
    You get [tex]i_L(t) = Ae^{-10.10t} + Be^{-989.9t}[/tex], with [tex]i_L(0) = 2mA[/tex] and [tex] \frac{di_L}{dt}(0) = 0[/tex]
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