Alternating Current

JEE Main level, single correct, medium

134 of 190

Question 134 of 190

A 2020 H\text{H} inductor coil is connected to a 1010 Ω\Omega 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:

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Answer: (B) 2ln⁡22\ln 2

Rate of dissipation of energy in resistor =i2R={i}^{2}R
Rate of energy stored in inductor =ddt(12Li2)=Lididt=\dfrac{d}{dt}\left(\dfrac{1}{2}L{i}^{2}\right)=Li\dfrac{di}{dt}
i2R=Lididt{i}^{2}R=Li\dfrac{di}{dt}
didt=iRL…(i)\dfrac{di}{dt}=i\dfrac{R}{L}\ldots \left(i\right)
In L−RL-R circuit:
i=i0(1−e−tτ)(∴τ=LR=2)i={i}_{0}\left(1-{e}^{-\dfrac{t}{\tau }}\right)(∴\tau =\dfrac{L}{R}=2)
didt=i0τe−t/τ\dfrac{di}{dt}=\dfrac{{i}_{0}}{\tau }{e}^{-t/\tau }
From equation (i),
i0τe−t/τ=i0(1−e−tτ)RL\dfrac{{i}_{0}}{\tau }{e}^{-t/\tau }={i}_{0}\left(1-{e}^{-\dfrac{t}{\tau }}\right)\dfrac{R}{L}
e−t/τ=1−e−t/τ{e}^{-t/\tau }=1-{e}^{-t/\tau }
e−tτ=12⇒t=τln⁡2{e}^{-\dfrac{t}{\tau }}=\dfrac{1}{2}\Rightarrow t=\tau \ln 2
=2ln⁡2=2\ln 2

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