A 20 Henry inductor coil is connected to a 10 ohm resistance in series as shown in figure. The time at which rate of dissipation of energy (joule's heat) across resistance is equal to the rate at which magnetic energy is stored in the inductor is :

Electromagnetic Induction PYQ — JEE Main 2019 · 8 Apr · Shift 1 · Q63 with Solution

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Correct answer: C

For a series $RL$ circuit connected to a DC source, the current grows as

i(t)=I_0\left(1-e^{-t/\tau}\right)

where

\tau=\frac{L}{R}

Given:

L=20\,\text{H}, \qquad R=10\,\Omega

So,

\tau=\frac{20}{10}=2\,\text{s}

Hence,

i(t)=I_0\left(1-e^{-t/2}\right)

Power dissipated in the resistor is

P_R=i^2R

So,

P_R=R I_0^2\left(1-e^{-t/2}\right)^2

Magnetic energy stored in the inductor is

U=\frac12 Li^2

Therefore,

\frac{dU}{dt}=Li\frac{di}{dt}

Now,

i=I_0\left(1-e^{-t/2}\right)

Differentiate:

\frac{di}{dt}=I_0\cdot \frac{1}{2}e^{-t/2}

Thus,

\frac{dU}{dt}=L\cdot I_0\left(1-e^{-t/2}\right)\cdot I_0\frac12 e^{-t/2}

Using $L=20$ ,

\frac{dU}{dt}=20\cdot I_0^2\cdot \frac12 \left(1-e^{-t/2}\right)e^{-t/2}

\frac{dU}{dt}=10I_0^2\left(1-e^{-t/2}\right)e^{-t/2}

We need

P_R=\frac{dU}{dt}

So,

10I_0^2\left(1-e^{-t/2}\right)^2=10I_0^2\left(1-e^{-t/2}\right)e^{-t/2}

Cancel common factors $10I_0^2$ and $\left(1-e^{-t/2}\right)$ :

1-e^{-t/2}=e^{-t/2}

Therefore,

1=2e^{-t/2}

e^{-t/2}=\frac12

Taking natural logarithm,

-\frac{t}{2}=\ln\left(\frac12\right)=-\ln 2

\frac{t}{2}=\ln 2

t=2\ln 2

t=2\ln 2

So the correct option is:

C: $2\ln 2$