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

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$I= I_0(1-e^{\large\frac{-Ro}{L}})$

$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.

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