1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    CEL

    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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Overdamped RLC circuit
  1. RLC circuit. (Replies: 1)

  2. RLC circuit (Replies: 3)

Loading...