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In the circuit shown here, the point $C$ is kept connected to point $A$ till the current flowing through the circuit becomes constant. Afterward, suddenly, point $C$ is disconnected from point $A$ and connected to point $B$ at time $t = 0.$ Ratio of the voltage across resistance and the inductor at $t = \large\frac{L}{R}$ will be equal to

$\begin{array}{1 1}(A)\;-1 \\ (B)\;\frac{1-e}{e} \\(C)\;\frac{e}{1-e} \\(D)\;1 \end{array} $

1 Answer

Applying Kirchoff's law of voltage in closed loop, $-V_R - V_C = 0$
$\implies \frac{V_R}{V_C} = -1$
Hence A is the correct answer.
answered Apr 7, 2014 by meena.p
edited Nov 25, 2017 by priyanka.c

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