In the circuit shown in Figure. The steady state currents I₁ and I₂ in the coils after the switch S is closed are given by

$\begin{array}{l} {\mathrm{L}}_1\frac{{\mathrm{dI}}_1}{\mathrm{dt}}={\mathrm{L}}_2\frac{{\mathrm{dI}}_2}{\mathrm{dt}}\\ \Rightarrow {\mathrm{L}}_1{\mathrm{I}}_1={\mathrm{L}}_2{\mathrm{I}}_2\\ \Rightarrow \frac{{\mathrm{I}}_1}{{\mathrm{I}}_2}=\frac{{\mathrm{L}}_2}{{\mathrm{L}}_1} .......\left(1\right)\end{array}$

Steady state current passing through the resistor is 

$\begin{array}{l} \mathrm{I}={\mathrm{I}}_1+{\mathrm{I}}_2=\frac{\mathrm{E}}{\mathrm{R}}\\ \Rightarrow {\mathrm{I}}_1+\left(\frac{{\mathrm{L}}_1}{{\mathrm{L}}_2}\right){\mathrm{I}}_1=\frac{\mathrm{E}}{\mathrm{R}}\\ \Rightarrow {\mathrm{I}}_1=\frac{{\mathrm{EL}}_2}{\mathrm{R}\left({\mathrm{L}}_1+{\mathrm{L}}_2\right)}\text{ and }{\mathrm{I}}_2=\frac{{\mathrm{EL}}_1}{\mathrm{R}\left({\mathrm{L}}_1+{\mathrm{L}}_2\right)}\end{array}$