A $16 μ F$ capacitor is charged to a 20 volt potential. Battery is then disconnected and inductor of inductance 40 mH is connected across the capacitor, so that LC oscillations are step-up. Maximum current in the coil is .

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2 A

0.4 A

0.2 A

0.8 A

answer is A.

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Detailed Solution

$\frac{1}{2} {CV}^{2} = \frac{1}{2} {L i}_{max}^{-2} \Rightarrow i_{max} = V \sqrt{\frac{C}{L}} i_{max} = 20 \sqrt[]{\frac{16 \times 10^{-6}}{4 \times 10^{- 2}}} = 0.4 A$

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