An RL Circuit with Switches at T=0 Switch S1 Opens

Answer is B. Because

During steady state, S2 is open, means; the last inductor is cut off from the circuit. So, what we have is, 6 Ohm, 4 ohm and 1 H. And since, it's the DC current which is flowing, inductor acts a short circuit at steady state.

Hence we have 5A current source in parallel with 6 and 4 ohm resistors which are in parallel with each other. So, the voltage developed at node V1, when V1 is close is,

5A * [(6*4)/(6+4)] = 12 V.

But, at t=0, V1 opens and V2 closes. So we have inductor in the circuit. But the 6 ohm resistor and the 5A source are cut off from the circuits. Effectively we have an RL tank circuit. with time constant T = 1/[R/L] =1/[4/1] = 1/4

Where,

L= 1H ; since we are asked to calculate V2 at t=0. So only the first inductor comes into picture and the other inductor is shorted at that point of time.

So, the voltage V2, at t=0 is

-12 exp (-t/T) = -12 exp (-4t)

-Ve sign is due to the current flowing out of the node V2.

B is your ans.

Hope this helped.

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