JEE Advanced 2020 Paper 2 · Physics · Electromagnetic Induction and Alternating Current · medium

An inductor of inductance 2.0 H in series with a 8.0 Ω resistor is connected at t=0 across an ideal 24 V battery. The time (in seconds) at which the…

Numerical answer

An inductor of inductance

2.0\ \mathrm{H}

in series with a

8.0\ \Omega

resistor is connected at

t=0

across an ideal

24\ \mathrm{V}

battery. The time (in seconds) at which the energy stored in the inductor reaches half of its final (steady-state) value is. Give the answer correct to two decimal places.

Show answer and solution

Answer: 0.31

The current grows as

I(t)=I_\infty\left(1-e^{-t/\tau}\right)

with

\tau=L/R=2/8=0.25\ \mathrm{s}

and

I_\infty=24/8=3\ \mathrm{A}

. The stored energy is

U=\tfrac12LI^2\propto I^2

, so

U=\tfrac12U_\infty

requires

I=I_\infty/\sqrt2

. Therefore

1-e^{-t/\tau}=\dfrac1{\sqrt2}

, so

e^{-t/\tau}=1-0.7071=0.2929

and

t=-\tau\ln(0.2929)=0.25\times1.2279=0.3070\ \mathrm{s}

, i.e.

0.31\ \mathrm{s}

. The frequent error is to wait until the current is half its final value,

t=\tau\ln2=0.17\ \mathrm{s}

, which makes the energy only one quarter of the final energy. The quantities

24\ \mathrm{V}

and

8\ \Omega

enter only through

I_\infty

and

\tau

. As a check, at

t=0.307\ \mathrm{s}

the current is

3(0.7071)=2.12\ \mathrm{A}

, giving energy

\tfrac12(2)(2.12)^2=4.5\ \mathrm{J}

, which is half of the final

\tfrac12(2)(3)^2=9\ \mathrm{J}

Report a mistake

← Previous Electromagnetic Induction and Alternating Current question Next Electromagnetic Induction and Alternating Current question →

Original question written in the JEE Advanced 2020 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake