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

(A) $\frac{\mathrm{EL}}{7.3 \mathrm{R}^{2}}$

(C) $\frac{7.3 \mathrm{EL}}{\mathrm{R}^{2}}$

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Electromagnetic Induction

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 = \frac{L}{R}

is :

(A) $\frac{EL}{7.3 R ^{2}}$
(B) $\frac{E L}{2.7 R ^{2}}$ ✓ Correct
(C) $\frac{7.3 EL}{R ^{2}}$
(D) $\frac{2.7 E L}{R ^{2}}$

Answer: (B)

Solution

The current builds up as

I (t) = \frac{E}{R} (1 - e^{- R t / L})

, and the charge is its time integral. Integrating from

0

\frac{L}{R}

q = \frac{E}{R} [t + \frac{L}{R} e^{- R t / L}]_{0}^{L / R} = \frac{E}{R} (\frac{L}{R} + \frac{L}{R} e^{- 1} - \frac{L}{R}) = \frac{E L}{R ^{2}} e^{- 1}

. With

e \approx 2.7

, this charge is

\frac{E L}{2.7 R ^{2}}

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option

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