Let i₁ be the current through L₁, i₂ be the current through L₂ and i be the current through R.

The inductors $ L_1 $ and $ L_2 $ are connected in parallel. Therefore, the self-induced emf in $ L_1 $ and $ L_2 $ are equal, which can be expressed as:

$ \varepsilon_1 = -L_1 \frac{d i_1}{dt} = \varepsilon_2 = -L_2 \frac{d i_2}{dt} $

Upon integrating this equation, we get :

$ L_1 i_1 = L_2 i_2 $ ............(1)

This indicates that the ratio of the currents through $ L_1 $ and $ L_2 $ is constant.

Using Kirchhoff's current law at junction A, we have :

$ i = i_1 + i_2 $ .............(2)

At the moment the switch is closed (t = 0), the inductors act as open circuits (since the inductive reactance $ X_L = \omega L $ approaches infinity as $ \omega \to \infty $). Therefore, initially :

$ i = i_1 + i_2 = 0 $

and the current through the resistor is :

$ i = i_1 + i_2 = 0 $

After a long time, the inductive reactance of the inductors becomes zero. Applying Kirchhoff's loop law, the current through the resistor $ R $ is :

$ i = \frac{V}{R} $

Substituting $ i = \frac{V}{R} $ into equation (2) and solving equations (1) and (2), we find :

$ i_1 = \frac{V}{R} \frac{L_2}{L_1 + L_2} $

$ i_2 = \frac{V}{R} \frac{L_1}{L_1 + L_2} $