JEE Main 2024 · Physics · Electromagnetic Induction and Alternating Current · medium

A capacitor of 10 muF charged to 10 V is connected across an ideal inductor of 0.1 H at t=0. At the instant when the charge on the capacitor has…

Numerical answer

A capacitor of

10\ \mu\mathrm{F}

charged to

10\ \mathrm{V}

is connected across an ideal inductor of

0.1\ \mathrm{H}

t=0

. At the instant when the charge on the capacitor has fallen to

60\%

of its initial value, the current in the circuit (in mA) is

Show answer and solution

Answer: 80

Initial charge

q_0=CV=10^{-4}\ \mathrm{C}

. Angular frequency

\omega=\dfrac1{\sqrt{LC}}=\dfrac1{\sqrt{0.1\times10^{-5}}}=1000\ \mathrm{rad\,s^{-1}}

. Energy conservation:

\dfrac{q^2}{2C}+\tfrac12Li^2=\dfrac{q_0^2}{2C}

, so

i=\omega\sqrt{q_0^2-q^2}=\omega q_0\sqrt{1-0.36}=1000\times10^{-4}\times0.8=0.08\ \mathrm{A}=80\ \mathrm{mA}

. Maximum current would be

100\ \mathrm{mA}

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