In the circuit shown below, $i_1$ and $i_2$ are the steady-state values of the current through $L_1$ and $L_2$ respectively, then $i_1$ is

Since the potential difference across the inductors is same, we get

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

$L_{1}d{i}_1=L_{2}d{i}_2$

$L_1∆i_1=L_2∆i_2$

$L_1{i}_1=L_2{i}_2$ (Since the initial value of current through both the inductors was zero) 

$\frac{i_1}{i_2}=\frac{L_2}{L_1}$

The steady current $i$ passing through $R$ is

$\mathit{\text{i}}\mathit{=}\frac{\mathit{\text{E}}}{\mathit{\text{R}}}$

$i_1=\left(\frac{L_2}{L_1+L_2}\right)i=\frac{E{L}_2}{R\left(L_1+L_2\right)}$