# Energy dissipated in resistor

1. The problem statement, all variables and given/known data

In the circuit shown in the figure, both switches operate together; that is, they either open or close at the same time. The switches are closed a long time before opening at t=0. (Figure 1) How many microjoules of energy have been dissipated in the 12 kΩ resistor 22ms after the switches open? How long does it take to dissipate 24% of the initially stored energy?

2. Relevant equations

P=i(t)v(t)
E=0.5*C*v^2
i = C * dv/dt
KVL

3. The attempt at a solution

First I found the energy in the capacitor:

E = 0.5*C*v^2 = 0.5 * (10/3)*(10^-6)*(120)^2 = 0.024 J

I then wrote a KVL equation around the center loop (only closed loop remaining with switches open) using passive sign convention to get:

-V_C + V_R = 0, V_R = V_C
Using i=C*dv/dt, V_C = (1/C) * ∫i dt
V_R = C*dv_C/dt
(1/C)*∫i dt = iR
(1/C)*i = R*di/dt
i = K * e^(t/(RC))

I’m not exactly sure where to go from here, obviously I need to solve for K but I’m not sure what I can use for the initial condition. After solving for that, I assume you can plug that value into p=i^2*r, and w=$\int$$\stackrel{t2}{t1}$p dt.

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