# RL circuit problem

#### Mr. Johnson

1. The problem statement, all variables and given/known data
[PLAIN]http://img14.imageshack.us/img14/4355/screenshot20110807at249.png [Broken]

2. Relevant equations

v(t) = Vmax * e^(-Rt/L)
i(t) = imax * e^(-Rt/L)
P = V^2/R
w = integral of power w/ respect to time
w(0) = .5Li^2(0) = energy stored in inductor

3. The attempt at a solution

Well for #3, for the estimates of I and R, all I did was put the v(t) = Vmax*e^(-Rt/L) in my calculator w/ Vmax as 60 and L as 40mH. I played around with different R values and found that if I put R =1, the graph in my calculator is identical to the one as above.

To solve for I (current source), I used V = IR => I = 60 Amps.

For #4, I just used the energy equation w(0) = .5Li^2(0)
= .5*40e^-3 * 60^2 = 72 Joules

For #5, I integrated the power w/ respect to time to get the energy. But first I needed to solve for the power equation. P = V^2(t)/R where v(t) = -60e^(-t/40e^-3) and R = 1.

So then the energy equation was then the integral of 3600e^(-50t) dt from 0 to .02 milli seconds, which equals .716412 Joules.

The percentage = .07164/72 * 100 = .1%

I don't feel good about these answers and not sure if I did them right. Any help would be greatly appreciated.

Thank you.

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#### gneill

Mentor
You should be able to find a reasonably good value for the time constant (rather than mucking about with trial and error curve fitting) by taking a data point from the provided graph. I see that the voltage decays through the 5V level at t = 0.1ms...

With the time constant in hand, a reasonably accurate value for the resistor Rx can be found, and then the initial current... etc.

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