A capacitor of capacitance 500 μF is charged completely using a dc supply of 100 V. It is now connected to an inductor of inductance 50 mH to form an LC circuit. The maximum current in LC circuit will be ______ A

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answer is 10.

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Detailed Solution

Energy stored in capacitor
$\begin{array}{l} = \frac{1}{2} {CV}^{2} = \frac{1}{2} 500 \times 10^{- 6} \times 10^{4} \\ = \frac{5}{2} J \end{array}$
Current will be maximum when whole energy of capacitor becomes energy of inductor.
$\begin{array}{l} \frac{1}{2} {LI}^{2} = \frac{5}{2} \\ I = \sqrt{\frac{5}{L}} = \sqrt{\frac{5}{50 \times 10^{- 3}}} = 10 A \end{array}$

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