Question

In an $L-R$ circuit switch is closed at $t=0$ then find the charge which passes through the battery till $t=\frac{L}{R}$. ($e$ is Euler number)

Answer 1

$\frac{VL}{eR^2}$

Answer 2

The current in an LR circuit when a DC voltage $V$ is applied at $t=0$ is given by $I(t) = \frac{V}{R}(1 - e^{-Rt/L})$. The charge $Q(t)$ is the integral of the current: $Q(t) = \int_0^t I(t') dt' = \frac{V}{R} \left( t + \frac{L}{R} e^{-Rt/L} - \frac{L}{R} \right)$. At $t = \frac{L}{R}$, the charge is $Q(\frac{L}{R}) = \frac{V}{R} \left( \frac{L}{R} + \frac{L}{R} e^{-1} - \frac{L}{R} \right) = \frac{VL}{eR^2}$.