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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