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Q.
An emf of $20\, V$ is applied at time $t = 0$ to a circuit containing in series $10\, mH$ inductor and $5 \,\Omega$ resistor. The ratio of the currents at time $t = \infty$ and at $t = 40$ s is close to : (Take $e^2 = 7.389$)
$\text { We know, } i = i _{0}\left(1- e ^{- t / \tau}\right) $
$\tau=\frac{ L }{ R }=\frac{0.01}{5}=0.002 $
$i = i _{0}\left(1- e ^{\frac{- t }{0.002}}\right)= i _{0}\left(1- e ^{-500 t }\right) $
$\text { at } t =40 s $
$i _{40}= i _{0}\left(1- e ^{-500 \times 40}\right) $
$=\frac{20}{5}\left(1- e ^{-20000}\right) $
$i _{40}=4\left(1- e ^{-20000}\right) \quad\left[\because i _{0}=\frac{20 v }{5 \omega}\right] $
$\text { at } t =\infty$
$i _{\infty}= i _{0}\left(1- e ^{- t / \tau}\right)= i _{0}\left(1- e ^{-\infty}\right)$
$i _{\infty}=\frac{20}{5}(1-0)=4$
$\Rightarrow \frac{ i _{\infty}}{ i _{40}}=\frac{4}{4\left(1- e ^{-20000}\right)}=\frac{1}{\left(1- e ^{-20000}\right)}$
$e ^{-20000} \simeq 0 \& 1>\left(1- e ^{-20000}\right)$
$\text { So the most probable answer is } 1.06$