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JEE Mains · Physics · STD 12 -Alternating current
  • JEE Mains 2024
  • Hard
  • MCQ

A capacitor of capacitance \(100\  \mu \mathrm{F}\) is charged to a potential of \(12  \mathrm{~V}\) and connected to a \(6.4 \ \mathrm{mH}\) inductor to produce oscillations. The maximum current in the circuit would be :

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Answer & Solution

Correct Answer

(B)  \(1.5 \mathrm{~A}\)

Step-by-step Solution

Detailed explanation

By energy conservation

\( \frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \mathrm{LI}_{\max }^2 \)

\( \mathrm{I}_{\max }=\sqrt{\frac{\mathrm{C}}{\mathrm{L}}} \mathrm{V} \)

\( =\sqrt{\frac{100 \times 10^{-6}}{6.4 \times 10^{-3}}} \times 12\)

\( =\frac{12}{8}=\frac{3}{2}=1.5 \mathrm{~A}\)

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