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 \(.........A\).
Pick an option to check your answer.
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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