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Series RLC

  1. Mar 31, 2008 #1
    1. The problem statement, all variables and given/known data
    I'm having some problems solving this RLC circuit, if anyone could help.

    R = 2 ohms
    C = 2/3 F
    L = 1/2 H

    The top picture is when t < 0
    The bottom picture is when t > 0
    Find [tex]v_o(t)[/tex] (notice V_0 is the defined +- voltage over the resistor)
    [​IMG]

    2. Relevant equations



    3. The attempt at a solution

    Well I found that:
    [tex]i_L(0^+) = 2 A[/tex]
    [tex]v_c(0^+) = 0[/tex]
    [tex]v_o(0^+) = -4 V [/tex]
    Which I believe are correct,
    I found the general equation to be [tex]v_0(t) = Ae^{-t} + Be^{-3t}[/tex]
    (My main problem is finding A and B, they should be 2 and -6 but I just can't get that)

    A + B = -4
    [tex]dv(0^+)/dt = -A + -3B[/tex]
    This is where I get a little sketchy, but since Cdv/dt = i_c then I was thinking that since [tex]i_L(0^+) = i_C(0^+)[/tex] for series circuits I could just say
    [tex]-A + -3B = 2/C[/tex]
    However when I solve those two equations I get A = -4.5, B = 1/2
    Does anyone know what I did wrong?
     
  2. jcsd
  3. Apr 2, 2008 #2

    CEL

    User Avatar

    You should always solve a series circuit for [tex]v_c(t)[/tex] and a parallel circuit for [tex]i_L(t)[/tex].
    Once you have [tex]v_c(t)[/tex] , you can differentiate it in order to get [tex]\frac{dv_C}{dt}[/tex]. Since the current in a series circuit is the same for all elements, you have [tex]v_c(0)[/tex] and [tex]\frac{dv_C}{dt}(0)[/tex].
    Knowing [tex]v_c(t)[/tex] you can calculate [tex]v_o(t)[/tex].
     
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