A source of constant voltage V is connected to a resistance R and...

As ${L}_{1}$ and ${L}_{2}$ are in parallel so equivalent

inductance, $L = \dfrac{{L}_{1}{L}_{2}}{{L}_{1} + {L}_{2}}$

Here, ${L}_{1}$ and ${L}_{2}$ are the inductance (shown in circuit)

The current through the $RL$ series circuit at anytime is given as 

$I = \dfrac{V}{R}\left(1 - e\dfrac{- Rt}{L}\right)$

At $t = 0$,

$I = \dfrac{V}{R}\left(1 - {e}^{0}\right) = 0$

After a longtime i.e. at $t = \infty$

$I = \dfrac{V}{R}\left(1 - {e}^{- \infty}\right)$

$⇒ I = \dfrac{V}{R}$

Let currents through ${L}_{1}$ and ${L}_{2}$ are ${I}_{1}$ and ${I}_{2}$ respectively. As ${L}_{1}$ and ${L}_{2}$ are joined in parallel

Voltage across ${L}_{1} =$ Voltage across ${L}_{2}$

$⇒ {V}_{{L}_{1}} = {V}_{{L}_{2}}$

$⇒ {L}_{1}\dfrac{d {I}_{1}}{d t} = {L}_{2}\dfrac{d {I}_{2}}{d t}$

$⇒ {L}_{1}{I}_{1} = {L}_{2}{I}_{2}$

Thus, $\dfrac{{I}_{1}}{{I}_{2}} = \dfrac{{L}_{2}}{{L}_{1}}$

By current division rule,

Current ${I}_{1}$ across ${L}_{1}$ is 

${I}_{1} = \dfrac{{L}_{2}}{{L}_{1} + {L}_{2}} \times I$

$⇒ {I}_{1} = \dfrac{V}{R}\left(\dfrac{{L}_{2}}{{L}_{1} + {L}_{2}}\right)$

Current ${I}_{2}$ across ${L}_{2}$ is

${I}_{2} = \dfrac{{L}_{1}}{{L}_{1} + {l}_{2}} \times I$

${I}_{2} = \dfrac{V}{R}\left(\dfrac{{L}_{1}}{{L}_{1} + {L}_{2}}\right)$