Question

The figure below shows a battery of emf $\epsilon $ 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. $\frac{\epsilon L}{R^{2}}\left(1 - \frac{1}{e}\right)$
• B. $\frac{\epsilon L}{2 e R^{2}}$
• C. $\frac{\epsilon L}{4 e R^{2}}$
• D. $\frac{\epsilon L}{e R^{2}}$

Answer 1

Answer: (D)

$q=\int\limits _{0}^{\tau}\frac{\epsilon }{R}\left(\right.1-e^{- t / \tau}\left.\right)dt$
$q=\frac{\epsilon }{R}\left[t - \frac{e^{- t / \tau}}{\frac{- 1}{\tau}}\right]_{0}^{\tau}=\frac{\epsilon }{R}\left[\tau + \tau e^{- 1} - \tau\right]$
$q=\frac{\epsilon }{R}\times \frac{1}{e}\times \frac{L}{R}=\frac{\epsilon L}{R^{2} e}$