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JEE Mains · Physics · STD 12 -Alternating current
  • JEE Mains 2022
  • Medium
  • Numerical

A capacitor of capacitance \(500\,\mu F\) 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 & Solution

Correct Answer

(A) \(10\)

Step-by-step Solution

Detailed explanation

Energy stored in capacitor

\(=\frac{1}{2} CV ^{2}=\frac{1}{2} 500 \times 10^{-6} \times 10^{4}\)

\(=\frac{5}{2}\,J\)

Current will be maximum when whole energy of capacitor becomes energy of inductor.

\(\frac{1}{2} LI ^{2}=\frac{5}{2}\)

\(I =\sqrt{\frac{5}{ L }}=\sqrt{\frac{5}{50 \times 10^{-3}}}=10\,A\)

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