Electromagnetic Induction

JEE Main level, single correct, medium

112 of 153

Question 112 of 153

An ideal capacitor of capacitance 0.2μF0.2 \mu \mathrm{F} is charged to a potential difference of 10 V10 \mathrm{~V}. The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance 0.5mH0.5 \mathrm{mH}. The current at a time when the potential difference across the capacitor is 5 V5 \mathrm{~V}, is:

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Answer: (A) 0.17 A0.17 \mathrm{~A}

Given: Capacitance, C=0.2μF=0.2×10−6C=0.2 \mu \mathrm{F}=0.2 \times 10^{-6} F\mathrm{F}
Inductance L=0.5mH=0.5×10−3H\mathrm{L}=0.5 \mathrm{mH}=0.5 \times 10^{-3} \mathrm{H}
Current I\mathrm{I} = ?
Using energy conservation
12CV2=12CV12+12LI2\frac{1}{2} C V^2=\frac{1}{2} C V_1^2+\frac{1}{2} L I^2
12×0.2×10−6×102+0\frac{1}{2} \times 0.2 \times 10^{-6} \times 10^2+0
=12×0.2×10−6×52+12×0.5×10−3I2=\frac{1}{2} \times 0.2 \times 10^{-6} \times 5^2+\frac{1}{2} \times 0.5 \times 10^{-3} I^2
∴I=3×10−1 A=0.17 A\therefore \quad I=\sqrt{3} \times 10^{-1} \mathrm{~A}=0.17 \mathrm{~A}

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