Capacitors

JEE Main level, single correct, medium

154 of 180

Question 154 of 180

A capacitor of capacitance 100μF100\mu F is charged to a potential of 12V12V and connected to a 6.4mH6.4mH inductor to produce oscillations. The maximum current in the circuit would be :

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Answer: (B) 1.5A1.5A

By energy conservation, it follows that
12CV2=12LImax⁡2...(1)\dfrac{1}{2}C{V}^{2}=\dfrac{1}{2}L{I}_{\max }^{2}...\left(1\right)
Equation (1) implies that
Imax⁡=CLV=100×10−66.4×10−3×12=128=1.5A\begin{matrix}{I}_{\max } & = & \sqrt{\dfrac{C}{L}}V \\ & = & \sqrt{\dfrac{100\times {10}^{-6}}{6.4\times {10}^{-3}}}\times 12 \\ & = & \dfrac{12}{8} \\ & = & 1.5A\end{matrix}

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