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Energy Dissipated by Resistor

  1. Nov 8, 2004 #1
    I have the solution to this problem, but I can't figure out how to solve it for the life of me. I'd appreciate any assistance:

    Image: http://www.cc.gatech.edu/~strobel/problem.gif

    The switch in the figure above has been in position 'a' for a very long time. It is suddenly flipped to position 'b' for a period of 1.25 ms, then back to 'a'. How much energy is dissipated by the 50 Ohm resistor?

    The solution is 23 mJ.

    Thanks,
    Mike
     
    Last edited: Nov 8, 2004
  2. jcsd
  3. Nov 8, 2004 #2

    Tide

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    First find out how much charge has accumulated on the capacitor then solve the equation for an LC circuit given that charge as your inititial condition.
     
  4. Nov 8, 2004 #3
    That seemed like the logical place to start, but it seems I messed up somewhere after that. Here's my attempt to solve it:

    At position ‘a’:
    Q = C*EMF = (20*10^-6 F)(50 V) = 0.001 C

    At position ‘b’:
    Q_0 = 0.001 C
    Q = Q_0*e^(-t/(R*C)) = 0.001*e^((-0.00125)/(50 * 20*10^-6)) = 0.000287
    dQ = 0.000287 – 0.001 = -0.000713
    I = -dQ/dt = 0.000713/0.00125 = 0.570 A

    P_R = I*V_R = (I^2)(R) = (0.570^2)(50) = 16.245 J

    Quite a bit off... any idea where I went wrong?
     
  5. Nov 8, 2004 #4

    Tide

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    One problem is that the current through the resistor is not a constant.
     
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