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In the circuit shown, a capacitor of capacitance 100 uF is initially charged to 200 V. Switch S1 is closed for some time and then opened. Next, switch S2 is closed for the minimum time required so that the 400 uF capacitor gets charged to 100 V, and then S2 is opened. The minimum time for which S2 must be closed is expressed as (pi / n) seconds. Find the value of n.

1. 1
2. 2
3. 3
4. 4

Correct answer: 4

Solution

The S2 sub-circuit forms an LC oscillator. The charge on 400 uF varies sinusoidally with the LC period. The minimum time for it to reach 100 V corresponds to a quarter-period (pi/2) or related fraction, giving n = 4.

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