A capacitor of capacity 2μF is charged to a potential difference of 12 V.It is then connected across an inductor of inductance 6μH. What is the current (in A) in the circuit at a time when the potential difference across the capacitor is 6.0 V?

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answer is 6.

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Detailed Solution

From energy conservation : 12CV02=12CV2+12LI2⇒12×2×10-6×122=12×2×10-6×62+12×6×10-6I2⇒I=6 A

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