Capacitor current problem

Discussion in 'Homework Help' started by peter_morley, Mar 27, 2011.

  1. peter_morley

    Thread Starter Member

    Mar 12, 2011
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    I'm having trouble understanding how to find A1 and A2 in the attached problem. I did a bunch of work and got stuck. please help

    Peter
     
  2. Georacer

    Moderator

    Nov 25, 2009
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    1,266
    The derivative of
    A_1 t e^{-1500 t} is actually
    A_1 e^{-1500 t} + A_1 t (-1500) e^{-1500 t}
    so, you should correct that.

    Also, for times before zero, the voltage is constant and as a result, the current of the capacitor is zero, not 90mA as you wrote.

    Since the voltage on a capacitor is a continuous quantity, you can safely say that
    A_1 \cdot 0 \cdot e^{-1500 \cdot 0}+A_2 \cdot e^{1500 \cdot 0}=A_2=25V.

    I don't know, however what you could do for A_1.
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    This is a rather cryptic problem.

    A typical (if not unique) circuit which would satisfy the general solution would be a capacitor with an initial voltage [25V] discharging into a series R+L network. The effective circuit would be critically damped.

    Under these conditions one would obtain the general solution for the capacitor voltage

    Vc(t)=25(1+at)e^(-at)

    We are given a=1500

    Hence

    Vc(t)=25e^(-1500t)+25x1500te^(-1500t)

    So A2=25 and A1=25x1500=37,500.
     
    Last edited: Mar 28, 2011
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    This solution supports Georacer's comments about the initial capacitor current being zero. If there were a current at zero, the general solution for the cap voltage may not be satisfied - I've not checked whether this is in fact the case. It's probably beyond the original scope of the problem anyway.
     
    Last edited: Mar 28, 2011
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