JEE Mains · Physics · STD 12 -7. Alternating current
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 \(.........A\).
- A \(10\)
- B \(1\)
- C \(0\)
- D \(100\)
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. \(\frac{1}{2} LI ^{2}=\frac{5}{2}\)…
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