How to Calculate Energy Dissipated by a Resistor with Changing Current

[mstrobel]

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

 

[Tide]

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.

 

[mstrobel]

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?

 

[Tide]

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