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## Homework Statement

http://img167.imageshack.us/img167/7747/picture2mg7.png [Broken]

## Homework Equations

iL(t) = current through inductor

**Using The Step by Step Method:**

Assume the solution iL(t) = K1 + K2*e^(-t/tau)

where K1 = iL(infinity) and K2 = iL(0+) - iL(infinity)

so, Assume the solution has the form:

iL(t) = iL(infinity) + [iL(0+) - iL(infinity]*e^(-t/tau)

Now, what I did was I calculated iL(0-) first... I replaced the inductor with a short circuit because we've assumed the SW has been held for a while.

So, we know that

**V1(t) = 6 V**... I applied

*MESH ANALYSIS*to the middle loop and to the right loop (both going clockwise) to find iL(0-).

**Middle_Loop = 2A**

**Right_Loop**= 2( iL(0-) - 2 ) + 12*iL(0-) = 30V... I get

**iL(0-) = 32/14 amps going CW**

Is that right? If so, isn't it safe to assume that just after the switch is opened, for that one instant in time, the inductor will act like a current source since the current through an inductor CANNOT change instanteously.

So, I believe that you have to rewrite the circuit with the switch open and put a temporary current source in place of the inductor. After that, what do I do though? That's where I'm lost.

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