Q 12-06-115JEE MainJEE Main 2020 (8 Jan, Shift 2)Medium
A battery of emf $\varepsilon$ is connected in series with an inductor $L$, a resistance $R$ and a switch S. The switch is closed at $t = 0$. The total charge that flows from the battery, between $t = 0$ and $t = t_c$ ($t_c$ is the time constant of the circuit) is:
Answer: (A) $\dfrac{\varepsilon L}{eR^2}$
The current grows as $i = \dfrac{\varepsilon}{R}\left(1 - e^{-t/\tau}\right)$ with $\tau = t_c = L/R$.
$$q = \int_0^{\tau} i\,dt = \frac{\varepsilon}{R}\left[t + \tau e^{-t/\tau}\right]_0^{\tau} = \frac{\varepsilon}{R}\left(\tau + \frac{\tau}{e} - \tau\right) = \frac{\varepsilon\tau}{eR}$$
$$q = \frac{\varepsilon L}{eR^2}$$
Solution by Sreeraj P, M.Sc Physics