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Question

A capacitor of capacity 2 μF is charged to a potential difference of 12 V. It is then connected across an inductor of inductance 0.6 mH. The current in the circuit at a time when the P.D. across the capacitor is 6 V is (in ampere).

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Solution

ω=1LC=10512
q=q0cosωt
2×106×6=2×106×12cosωt
cosωt=12ωt=π3
i=dqdt=q0ωsinωt
i=0.6 A

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