Step-by-Step Solution:

The problem asks to analyze the currents in an RL circuit just after a switch is closed and after a long time.

1. Analysis at t = 0 (immediately after closing the switch S)

• An inductor resists an instantaneous change in the current flowing through it. Before the switch is closed (at $t = 0^-$), the current through both inductors $L_1$ and $L_2$ is zero.

• Therefore, immediately after the switch is closed (at $t = 0^+$), the currents through the inductors must still be zero. $I_{L1}(0) = 0$ $I_{L2}(0) = 0$

• The current through the resistor R, denoted as $I_R$, is the sum of the currents through the parallel branches containing $L_1$ and $L_2$ (by Kirchhoff's Current Law). $I_R(t) = I_{L1}(t) + I_{L2}(t)$

• At t=0, the current through the resistor is: $I_R(0) = I_{L1}(0) + I_{L2}(0) = 0 + 0 = 0$

• Evaluation of Option D: Option D states that at t = 0, the current through the resistance R is V/R. Our analysis shows that $I_R(0) = 0$. Thus, Option D is incorrect.

2. Analysis for any time t > 0

• The two inductors $L_1$ and $L_2$ are connected in parallel. This means the voltage across them is the same at all times t > 0. Let's call this voltage $V_L(t)$.

• The voltage across an inductor is given by $V_L = L (di/dt)$. Therefore: $V_L(t) = L_1 \frac{dI_{L1}}{dt}$ $V_L(t) = L_2 \frac{dI_{L2}}{dt}$

• Equating the two expressions for $V_L(t)$: $L_1 \frac{dI_{L1}}{dt} = L_2 \frac{dI_{L2}}{dt}$

• We can integrate this equation with respect to time from t=0 to any time t: $\int_0^t L_1 \frac{dI_{L1}}{dt'} dt' = \int_0^t L_2 \frac{dI_{L2}}{dt'} dt'$ $L_1 \int_{I_{L1}(0)}^{I_{L1}(t)} dI_{L1} = L_2 \int_{I_{L2}(0)}^{I_{L2}(t)} dI_{L2}$

• Using the initial conditions $I_{L1}(0) = 0$ and $I_{L2}(0) = 0$: $L_1 [I_{L1}(t) - 0] = L_2 [I_{L2}(t) - 0]$ $L_1 I_{L1}(t) = L_2 I_{L2}(t)$

• This relation holds for all times t > 0. From this, we can find the ratio of the currents: $\frac{I_{L1}(t)}{I_{L2}(t)} = \frac{L_2}{L_1}$

• Evaluation of Option C: Since $L_1$ and $L_2$ are constants, the ratio $L_2/L_1$ is also a constant. This means the ratio of the currents through $L_1$ and $L_2$ is fixed at all times t > 0. Thus, Option C is correct.

3. Analysis after a long time (t → ∞, steady state)

• In a DC circuit, after a long time, the currents become constant (steady state). In this condition, an ideal inductor behaves like a short circuit (a wire with zero resistance), because the voltage across it, $V_L = L(di/dt)$, becomes zero as di/dt = 0.

• So, the parallel combination of $L_1$ and $L_2$ acts as a short circuit, meaning the voltage across this parallel branch is zero.

• Applying Kirchhoff's Voltage Law to the main loop: $V - I_R(\infty) R - V_L(\infty) = 0$ Since $V_L(\infty) = 0$, we have: $V - I_R(\infty) R = 0$ $I_R(\infty) = \frac{V}{R}$

• This total steady-state current $I_R(\infty)$ splits between the two inductors: $I_R(\infty) = I_{L1}(\infty) + I_{L2}(\infty) = \frac{V}{R}$ (Equation 1)

• The relationship $L_1 I_{L1}(t) = L_2 I_{L2}(t)$ derived in step 2 is valid for all t, including $t \to \infty$: $L_1 I_{L1}(\infty) = L_2 I_{L2}(\infty)$ (Equation 2)

• Now we solve the system of two equations for $I_{L1}(\infty)$ and $I_{L2}(\infty)$. From Equation 2: $I_{L2}(\infty) = \frac{L_1}{L_2} I_{L1}(\infty)$

• Substitute this into Equation 1: $I_{L1}(\infty) + \frac{L_1}{L_2} I_{L1}(\infty) = \frac{V}{R}$ $I_{L1}(\infty) \left(1 + \frac{L_1}{L_2}\right) = \frac{V}{R}$ $I_{L1}(\infty) \left(\frac{L_2 + L_1}{L_2}\right) = \frac{V}{R}$ $I_{L1}(\infty) = \frac{V}{R} \frac{L_2}{L_1 + L_2}$

• Evaluation of Option A: This matches the expression given in Option A. Thus, Option A is correct.

• Now, let's find $I_{L2}(\infty)$: $I_{L2}(\infty) = I_R(\infty) - I_{L1}(\infty) = \frac{V}{R} - \frac{V}{R} \frac{L_2}{L_1 + L_2}$ $I_{L2}(\infty) = \frac{V}{R} \left(1 - \frac{L_2}{L_1 + L_2}\right) = \frac{V}{R} \left(\frac{L_1 + L_2 - L_2}{L_1 + L_2}\right)$ $I_{L2}(\infty) = \frac{V}{R} \frac{L_1}{L_1 + L_2}$

• Evaluation of Option B: This matches the expression given in Option B. Thus, Option B is correct.

Conclusion:

Based on the analysis, options A, B, and C are correct, while option D is incorrect.