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# An inductor of inductance $L = 400\: mH$ and resistors of resistances $R1 = 2\Omega$ and $R2 = 2\Omega$ are connected to a battery of $emf\: 12V$ as shown in the figure. The internal resistance of the battery is negligible. The switch $K$ is closed at $t= 0$. The potential drop across $L$ as a function of time is

$\begin {array} {1 1} (A)\;6e^{-5t}V & \quad (B)\;\bigg( \large\frac{12}{t} \bigg) e^{-3t}V \\ (C)\;6 \bigg(1-e^{\large\frac{-t}{0.2}} \bigg)V & \quad (D)\;12e^{-5t}V \end {array}$

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## 1 Answer

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$I_1 = \large\frac{F}{R_1}$ $\large\frac{12}{2}=6A$
$E=L \large\frac{dI_2}{dt}$ $+ R_2 \times I_2$
$I_2 = I_0 \bigg ( 1- e^{\large\frac{-t}{t_c}} \bigg)$ $\Rightarrow I_0 = \large\frac{E}{R_2}$ $\large\frac{12}{2}$ $= 6A$
$t_c =\large\frac{L}{R}$ $= \large\frac{400 \times 10^{-3}} {2}$ $0.2$
$I_2 = 6\bigg(1-e^{\large\frac{-t}{0.2}} \bigg)$
Potential drop across $L =E - R_2I_2 = 12 -2 \times 6 ( 1-e^{-bt}) = 12e^{-5t}$
Ans : (D)
answered Feb 13, 2014
edited Mar 14, 2014

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