JEE Main 2018 · Physics · Electromagnetic Induction and Alternating Current · medium

A coil of self-inductance 0.5 H and resistance 10 Ω is connected across a 20 V battery of negligible internal resistance at t=0. The time at which…

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A coil of self-inductance

0.5\ \mathrm{H}

and resistance

10\ \Omega

is connected across a

20\ \mathrm{V}

battery of negligible internal resistance at

t=0

. The time at which the energy stored in the coil is one quarter of its final (maximum) value is

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Answer: (A) $34.7\ \mathrm{ms}$

The current grows as

I=I_0(1-e^{-t/\tau})

with

I_0=\dfrac{20}{10}=2\ \mathrm{A}

and

\tau=\dfrac LR=\dfrac{0.5}{10}=50\ \mathrm{ms}

. Stored energy is

\tfrac12LI^2

, so one quarter of the maximum means

I=\tfrac12I_0

(not

\tfrac14I_0

). Then

1-e^{-t/\tau}=\tfrac12

, i.e.

e^{-t/\tau}=\tfrac12

and

t=\tau\ln2=50\times0.693=34.7\ \mathrm{ms}

. Taking the current to be one quarter of its maximum would give

\tau\ln\tfrac43=14.4\ \mathrm{ms}

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