GATE 2025 IN – Question 63
In the shown circuit, the switch is opened at $t=0$. The current $i(t)$ at $t=2$ ms is ______ mA (rounded to two decimal places).
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Correct answer: 4.60 to 5.00
Explanation
Before opening, the switch shorts the current source and the settled inductor current is zero.
After opening, the ideal source forces 10 mA into the parallel combination of 120 Ω and the series branch containing 80 Ω plus the inductor.
The inductor sees $R_{th}=120+40+40=200\ \Omega$, giving $\tau=L/R_{th}=0.2/200=1$ ms.
The final inductor current is $10(120/200)=6$ mA.
At 2 ms, $i_L=6(1-e^{-2})$ mA.
The marked 120 Ω current is $i=10-i_L=4+6e^{-2}=\boxed{4.81\ \mathrm{mA}}$.