QJEEAdv17Ph7909: JEE Advanced 2017 Physics - Class 12 JEEAdv17Ph7909 Solution: A source of constant voltage V is connected to a resistance R…

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 :

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

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