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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class12  >>  Electromagnetic Induction

A emf of 10V is applied in a circuit containing inductance L and resistance R. $L=5H$ and $R= 10 \;\Omega$ . The ratio of current at $t =\infty$ and $t= 1\;s$ in

$(a)\;\frac{1}{e^2-1} \\ (b)\;\frac{e^2}{e^2-1} \\(c)\;\frac{1}{1-e^2} \\(d)\;e^2 $

1 Answer

$I= I_0(1-e^{\large\frac{-Ro}{L}})$
$I_o=\large\frac{E}{R}=\frac{10}{10}$$=1$
at $t = \infty$
$I_1= I_0 (1-e^{- \infty})$
$\quad=1$
at $t=1$
$I_2=1(1-e^{-10/5})$
$\quad= (1-\large\frac{1}{e^2})$
$\large\frac{I_1}{I_2} =\large\frac{1}{\bigg(1- \Large\frac{1}{e^2}\bigg)}$
$\qquad= \large\frac{e^2}{e^2-1}$
Hence b is the correct answer.
answered Mar 24, 2014 by meena.p
 

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