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A capacitor of capacitor $2 μF$ is charged to a potential difference of 12V. It is then connected across an inductor of inductance $6 μH$ . The current in the circuit when the potential across the capacitor 6V:

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0.6 A

19 A

3.6 A

2.4 A

answer is D.

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Detailed Solution

$\begin{array}{l} u = \frac{1}{2} {Cv}^{2} + \frac{1}{2} {LI}^{2} \\ \frac{1}{2} C {(12)}^{2} - \frac{1}{2} C {(6)}^{2} = \frac{1}{2} L I^{2} \\ I = 6 A \end{array}$

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