An RL Circuit with Switches at T=0 Switch S1 Opens
An RL circuit with switches is shown in the figure. Switch s1 remains closed and s2 remains open for a long time (steady-state). At t=0 switch s1 opens and s2 closes. Find V_1(t) for t > 0.
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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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