GATE 2014 EC (EC2) – Question 17
The ideal switch shown has been open for a long time. It closes at t = 0. Find the magnitude of current through the 4 kΩ resistor at t = 0⁺ (mA).
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Correct answer: 1.2 to 1.3
Explanation
Before switching, DC steady state makes the capacitor open and inductor short. The series resistance is5k+4k+1k=10kΩ, giving1mA and capacitor voltage10−1mA×5k=5V. At0+ that capacitor voltage cannot jump; the switch grounds the other end of4kΩ. Hence its current is5/4000=**1.25mA**.