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JEE Main 2021 · Physics · Magnetic Effects of Current · hard

An inductor L=0.2 H is connected in series with R=10 Ω and an ideal battery at t=0. At the instant when the rate at which the battery's energy is…

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An inductor L=0.2 HL=0.2\ \mathrm{H} is connected in series with R=10 ΩR=10\ \Omega and an ideal battery at t=0t=0. At the instant when the rate at which the battery's energy is converted to heat in RR equals the rate at which energy is stored in the inductor, the time elapsed is approximately
  1. A. 6.9 ms6.9\ \mathrm{ms}
  2. B. 20 ms20\ \mathrm{ms}
  3. C. 27.7 ms27.7\ \mathrm{ms}
  4. D. 13.9 ms13.9\ \mathrm{ms}
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Answer: (D) 13.9 ms13.9\ \mathrm{ms}

Heat rate =I2R=I^2R; storage rate =LI dIdt=I(E−IR)=LI\,\dfrac{dI}{dt}=I(\mathcal E-IR). Equating gives IR=E−IR⇒I=E2R=I02IR=\mathcal E-IR\Rightarrow I=\dfrac{\mathcal E}{2R}=\dfrac{I_0}{2}. With I=I0(1−e−t/τ)I=I_0(1-e^{-t/\tau}), e−t/τ=12e^{-t/\tau}=\tfrac12, so t=τln⁡2=0.210(0.693)=13.9 mst=\tau\ln2=\dfrac{0.2}{10}(0.693)=13.9\ \mathrm{ms}.
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