An ideal capacitor of capacitance 0.2 μ F is charged to a potential difference of 10 ~V . The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance 0.5 mH . The

Correct answer: A. 0.17 ~A Given: Capacitance, C=0.2 μ F=0.2 × 10^ -6 F Inductance L=0.5 mH=0.5 × 10^ -3 H Current I = ? Using energy conservation (1)/(2) C V²= (1)/(2) C V_1^2+ (1)/(2) L I² (1)/(2) × 0.2 × 10^ -6 × 10^2+0=(1)/(2) × 0.2 × 10^ -6 × 5^2+(1)/(2) × 0.5 × 10^ -3 I^2 I= 3 × 10^ -1 ~A=0.17 ~A

JEE Main 201815 AprilMorning ShiftOnline

Exam: JEE Main
Year: 2018
Date: 15 April
Shift: Morning Shift
Mode: Online
Subject: Physics
Chapter: Electromagnetic Induction

JEE Main 2018 · 15 April · Morning Shift · Online · Physics · Electromagnetic Induction

An ideal capacitor of capacitance 0.2 μ F is charged to a potential...

An ideal capacitor of capacitance

0.2 \mu \mathrm{F}

is charged to a potential difference of

10 \mathrm{~V}

. The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance

0.5 \mathrm{mH}

. The current at a time when the potential difference across the capacitor is

5 \mathrm{~V}

, is:

Options

A $0.17 \mathrm{~A}$
B $0.15 \mathrm{~A}$
C $0.34 \mathrm{~A}$
D $0.25 \mathrm{~A}$

Correct answer

A. $0.17 \mathrm{~A}$

Step-by-step solution

Given: Capacitance,

C=0.2 \mu \mathrm{F}=0.2 \times 10^{-6}\mathrm{F}

Inductance

\mathrm{L}=0.5 \mathrm{mH}=0.5 \times 10^{-3} \mathrm{H}

Current

\mathrm{I}

= ? Using energy conservation

\frac{1}{2}

C V²=

\frac{1}{2}

C V_1^2+

\frac{1}{2}

L I²

\frac{1}{2} \times 0.2 \times 10^{-6} \times 10^2+0=\frac{1}{2} \times 0.2 \times 10^{-6} \times 5^2+\frac{1}{2} \times 0.5 \times 10^{-3} I^2\therefore \quad I=\sqrt{3} \times 10^{-1} \mathrm{~A}=0.17 \mathrm{~A}

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