Question

The figure below shows a battery of emf $ϵ$ connected to an inductor $L$ and a resistance $R$ in series. If the switch is closed at $t=0$ , then the total charge that flows from the battery in one time constant of the circuit is

• A. $\frac{ϵL}{R^2}\left(1-\frac{1}{e}\right)$
• B. $\frac{ϵL}{2e{R}^2}$
• C. $\frac{ϵL}{4e{R}^2}$
• D. $\frac{ϵL}{e{R}^2}$

Answer 1

Answer: (D)

$q=\underset{0}{\overset{\tau }{\int }}\frac{ϵ}{R}(1-e^{-t/\tau })dt$
$q=\frac{ϵ}{R}{\left[t-\frac{e^{-t/\tau }}{\frac{-1}{\tau }}\right]}_0^{\tau }=\frac{ϵ}{R}\left[\tau +\tau e^{-1}-\tau \right]$
$q=\frac{ϵ}{R}\times \frac{1}{e}\times \frac{L}{R}=\frac{ϵL}{R^{2}e}$