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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Electromagnetic Induction
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An inductor $(L = 100\: mH)$, a resistor $(R = 100\Omega)$ and a battery $(E = 100 \: V)$ are initially connected in series as shown in the figure. After a long time the battery is disconnected after short circuiting the points A and B. The current in the circuit $1\: ms$ after the circuit is


$\begin {array} {1 1} (A)\;1A & \quad (B)\;\large\frac{1}{e}A \\ (C)\;eA & \quad (D)\;0.1A \end {array}$


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$I=I_0\: e^{\large\frac{-Rt}{L}}$$= e^{-1}A$
Ans : (B)
answered Feb 16, 2014 by thanvigandhi_1
edited Mar 14, 2014 by thanvigandhi_1

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