GATE 2026 BM – Question 25
The shown circuit has a 10 V source, a 10 μH inductor and two 100 Ω branches. It is initially at DC steady state with S open. Immediately after S closes, find VA in volts.
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Correct answer: (C) 5
Explanation
Initially the inductor is a short at DC, and current is 10/100=0.1 A through the connected resistor. Inductor current cannot change instantaneously. Closing S puts the two 100 Ω resistors in parallel, equivalent to fifty ohms. Thus VA(0+)=0.1×50=5 V, C.