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JEE Main 2019 (12 Apr, Shift 2) · Inductors & DC Transients (Mixed)

Inductors & DC Transients (Mixed) question from JEE Main 2019 (12 Apr, Shift 2)

Q20JEE Main 2019 · 12 APR · SHIFT 2Difficult

Inductors & DC Transients (Mixed)

Consider the LR circuit shown in the figure. If the switch S is closed at t = 0 then the amount of charge that passes through the battery between t = 0 and t = LR\frac{L}{R} is:

Figure

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Answer: (b)

  1. i=ER(1−e−t/τ)i = \frac{E}{R}(1 - e^{-t/\tau}), τ=LR\tau = \frac{L}{R}; charge q=∫0τi dtq = \int_0^{\tau} i\,dt.
  2. q=ER[t+τe−t/τ]0τ=ER(τ+τe−1−τ)=EτeRq = \frac{E}{R}\left[t + \tau e^{-t/\tau}\right]_0^{\tau} = \frac{E}{R}\left(\tau + \tau e^{-1} - \tau\right) = \frac{E\tau}{eR}
  3. q=ELeR2=EL2.7R2q = \frac{EL}{eR^2} = \frac{EL}{2.7R^2}
  4. So the answer is (b).
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