In the circuit shown in Fig. , switch $S$ is closed at time $t=0$. Calculate current $i_1$ and $i_2$ through inductances $L_1$ and $L_2$ respectively at time $t$.

When switch $S$ is closed, battery forces a current in the circuit which increases gradually. Due to increase in current emf is induced in both the solenoids. Since two solenoids are in parallel with each other, therefore emf in the two solenoids is always equal to each other, through solenoids $L_1$ and $L_2$ be $i_1$ and $i_2$ respectively, then magnitude of induced emf in the solenoids will be 

$\left|e_1\right|=L_1\frac{d{i}_1}{dt}$ and $\left|e_2\right|=L_2\frac{d{i}_2}{dt}$ respectively.

But, $\left|e_1\right|=\left|e_2\right|$

$\therefore L_1\frac{d{i}_1}{dt}=L_2\frac{d{i}_2}{dt}$

$d{i}_2=\frac{L_1}{L_2}d{i}_1$

Integrating      $i_2=\frac{L_1}{L_2}{i}_1\dots \dots \dots \dots .\left(1\right)$

According to Lenz’s law, induced emfs will oppose increase of current in respective solenoids. Hence at time $t$, current and polarities of induced emfs in the circuit will be as shown in Fig

Applying Kirchhoff’s voltage law on outer mesh

$+\left|e_1\right|=(i_1+i_2)R-E=0$

$\therefore L_1\frac{d{i}_1}{dt}+\left(i_1+\frac{L_1{i}_1}{L_2}\right)R-E=0$

Or, $L_1\frac{d{i}_1}{dt}=\backslash \mathrm{farc}E{L}_2-(L_1+L_2)i_{1}R{L}_2$

Or, $\frac{d{i}_1}{E{L}_2-(L_1+L_2)i_{1}R}=\frac{dt}{L_1{L}_2}\dots \dots \dots \dots ..\left(2\right)$

But at time $t=0,i_1=0$

Integrating equation $\left(2\right),{\int }_0^{i_1}\frac{d{i}_1}{E{L}_2-(L_1+L_2)i_{1}R}={\int }_0^t\frac{dt}{L_1{L}_2}$

$i_1=\frac{E{L}_2}{(L_1+L_2)R}\left[1-e^{-}\frac{Rt(L_1+L_2)}{L_1{L}_2}\right]$ Ans.

Substituting this value in equation $\left(1\right),$

$i_2=\frac{E{L}_1}{(L_1+L_2)R}\left[1-e^{-}\frac{Rt(L_1+L_2)}{L_1{L}_2}\right]$ Ans.