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Energy Dissipated by Resistor |
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| Nov8-04, 07:57 PM | #1 |
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Energy Dissipated by Resistor
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 |
| Nov8-04, 08:27 PM | #2 |
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Recognitions:
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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.
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| Nov8-04, 08:52 PM | #3 |
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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? |
| Nov8-04, 10:20 PM | #4 |
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Recognitions:
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Energy Dissipated by Resistor
One problem is that the current through the resistor is not a constant.
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