GATE 2021 EE – Question 39
In the circuit, switch 'S' is in the closed position for a very long time. If the switch is opened at time $t=0$, then $i_L(t)$ in amperes, for $t\ge0$ is

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Correct answer: (C) $8+2e^{-10t}$
Explanation
With S closed the $4\ \Omega$-30 V branch is shorted, so $i_L(0^-)=\frac{10}{1}=10$ A. After opening, the two sources add in series: $\frac{10+30}{4+1}=8$ A final current. The time constant is $\frac{L}{R}=\frac{0.5}{5}=0.1$ s, so $i_L=8+(10-8)e^{-10t}=8+2e^{-10t}$.