Q 12-06-143JEE MainJEE Main 2019 (12 Apr, Shift 2)Medium
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 $t = \dfrac LR$ is:
Answer: (D) $\dfrac{EL}{2.7R^2}$
$i = \dfrac ER\left(1 - e^{-t/\tau}\right)$ with $\tau = \dfrac LR$.
$$q = \int_0^\tau i\,dt = \frac ER\left[\tau - \tau\left(1 - e^{-1}\right)\right] = \frac ER\cdot\frac{\tau}{e} = \frac{EL}{eR^2} \approx \frac{EL}{2.7R^2}$$
Solution by Sreeraj P, M.Sc Physics