# Homework Help: Energy Dissipated by Resistor

1. Nov 8, 2004

### 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 [Broken]

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 by a moderator: May 1, 2017
2. Nov 8, 2004

### 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.

3. Nov 8, 2004

### 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?

4. Nov 8, 2004

### Tide

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