An emf of 20V is applied at time t = 0 to a circuit containing in series 10 mH inductor and 5$\Omega$ resistor. The ratio of currents at time  $t=\infty$ and t = 40 s is close to: (Take $e^2$ = 7.389).

$\begin{array}{l}Growth of current in RL circuit\\ \text{}i=i_0\left(1-e^{-Rt/L}\right) \text{}\\ At t = \infty , i_1 = i_0\\ At t =40s, i_2=i_0\times \left(1-e^{-\frac{5\times 40}{10\times {10}^{-3}}}\right)=i_0\times \left(1-e^{-2\times {10}^4}\right)\\ Hence, ratio = \frac{i_1}{i_2}=\frac{1}{1-e^{-2\times {10}^4}} \\ \Rightarrow \frac{i_1}{i_2}=1 \simeq 1.06 \left[Since, e^{-2\times {10}^4}is very small\right] \end{array}$