Question

As shown in the figure, a battery of emf $\epsilon$ 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, be tween $t = 0$ and $t = t_c$ ($t_c$ is the time constant of the circuit) is :

• A. $\frac{\epsilon L}{R^{2}}\left(1-\frac{1}{e}\right)$
• B. $\frac{\epsilon L}{R^{2}}$
• C. $\frac{\epsilon R}{eL^{2}}$
• D. $\frac{\epsilon L}{eR^{2}}$

Answer 1

Answer: (D)

$i=i_{0}\left(1-e^{-Rt/L}\right)=i_{0}\left(1-e^{-t/T_{C}}\right)
q=$$\int\limits^{T_C}_{{0}}i\,dt$
$=\int\limits^{T_C}_{{0}}$$\frac{\varepsilon}{R}\left(1-e^{-t/T_{C}}\right)$
$=\frac{\varepsilon }{R}\left(t-\frac{e^{-t/T_{C}}}{-1/T_{C}}\right)|^{T_C}_0$
$=\frac{e}{R}\left(T_{C}-T_{C}e^{-1}\right)-\frac{e}{R}\left(0+T_{C}\right)$
$q=\frac{e}{R}\times T_{C}\,e^{-1}$
$=\frac{\varepsilon}{R}\times\frac{L}{R} \, \frac{1}{e}$
$=\frac{\varepsilon L}{eR^{2}}$