An emf of 15   V is applied to a circuit containing 5   H inductance and 10   Ω…

Given: The inductance $L=5 \mathrm{H}$, resistance $R=10 \mathrm{\Omega }$ and emf applied $V=15 \mathrm{V}$.

The time constant for $LR$ constant is given by,

${\tau }_L=\frac{L}{R}=\frac{1}{2} \mathrm{s}$

The current in the $LR$ circuit is given by,

$i=i_0\left(1-e^{-t/{\tau }_L}\right)$; where $i_0=\frac{V}{R}$ is the maximum current at $t=\infty$.

The ratio of the maximum current to current will be,

$\therefore \frac{i_0}{i}=\frac{1}{1-e^{-t/{\tau }_L}}$

Substituting, $t=1 \mathrm{s}$ and ${\tau }_L=\frac{1}{2} \mathrm{s}$

We get, $\frac{i_0}{i}=\frac{e^2}{e^2-1}$.