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Q.

As shown in the figure, a battery of emf $\in$ is connected to an inductor L and resistance R in series. 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 :

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a

$\frac{\in L}{R^{2}} (1 - \frac{1}{e})$

b

$\frac{\in R}{e L^{2}}$

c

$\frac{\in L}{R^{2}}$

d

$\frac{\in L}{e R^{2}}$

answer is D.

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Detailed Solution

$\begin{array}{l} q =_{0}^{t_{c}} i d t \\ \Rightarrow q =_{0}^{t_{c}} \frac{\in}{R} (1 - e^{\frac{- t}{t_{c}}}) d t \\ \Rightarrow q = \frac{\in}{R} (t_{c} + t_{c} (e^{- 1} - 1)) \\ \Rightarrow q = \frac{\in}{R} \times \frac{t_{c}}{e} \\ \Rightarrow q = \frac{\in}{R} \times \frac{L}{e R} \\ \Rightarrow q = \frac{\in L}{R^{2} e} \end{array}$

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