Parallel Inductors Connected to Constant Voltage Source

To determine the correct statements, we analyze the behavior of the given circuit containing a DC voltage source $V$, a resistor $R$, and two ideal inductors $L_1$ and $L_2$ connected in parallel with no mutual inductance ($M = 0$).

1. Relation Between Currents in $L_1$ and $L_2$ for $t > 0$

Let $i_1(t)$ be the current through $L_1$, $i_2(t)$ be the current through $L_2$, and $i(t)$ be the total current passing through the resistor $R$. By Kirchhoff's Current Law: $i(t) = i_1(t) + i_2(t)$

Since $L_1$ and $L_2$ are connected in parallel, the potential difference across each inductor at any instant $t$ must be equal: $V_{parallel}(t) = L_1 \frac{di_1}{dt} = L_2 \frac{di_2}{dt}$

Integrating both sides with respect to time from $t = 0$ (at which $i_1(0) = 0$ and $i_2(0) = 0$ because inductors oppose instantaneous changes in current) to any time $t > 0$: $\int_{0}^{i_1(t)} L_1 \, di_1 = \int_{0}^{i_2(t)} L_2 \, di_2$

$L_1 i_1(t) = L_2 i_2(t)$

Rearranging this gives the ratio of the currents through $L_1$ and $L_2$: $\frac{i_1(t)}{i_2(t)} = \frac{L_2}{L_1} = \text{constant}$

This proves that the ratio of the currents through $L_1$ and $L_2$ is independent of time and fixed for all $t > 0$. Hence, Statement C is correct.

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2. Current as a Function of Time $t$

From $L_1 i_1(t) = L_2 i_2(t)$, we can express $i_1(t)$ and $i_2(t)$ in terms of the total current $i(t)$: $i(t) = i_1(t) + \frac{L_1}{L_2} i_1(t) = i_1(t) \left( \frac{L_1 + L_2}{L_2} \right)$

Therefore: $i_1(t) = i(t) \frac{L_2}{L_1 + L_2}$ $i_2(t) = i(t) \frac{L_1}{L_1 + L_2}$

The equivalent inductance $L_{eq}$ of two inductors connected in parallel without mutual inductance is given by: $L_{eq} = \frac{L_1 L_2}{L_1 + L_2}$

Applying Kirchhoff's Voltage Law around the circuit loop: $V - i(t) R - L_{eq} \frac{di}{dt} = 0$

Solving this standard first-order linear differential equation with the initial condition $i(0) = 0$: $i(t) = \frac{V}{R} \left( 1 - e^{-\frac{R}{L_{eq}} t} \right)$

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3. Current at $t = 0$

At $t = 0$: $i(0) = \frac{V}{R} (1 - e^0) = 0$

Since $i(0) = 0$, the current through the resistor $R$ at $t = 0$ is $0$, not $\frac{V}{R}$. Hence, Statement D is incorrect.

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4. Currents After a Long Time Interval ($t \to \infty$)

As $t \to \infty$, the exponential term $e^{-\frac{R}{L_{eq}} t} \to 0$, giving the steady-state total current through the resistor: $i(\infty) = \frac{V}{R}$

Substituting $i(\infty)$ into the expressions for $i_1$ and $i_2$: $i_1(\infty) = \frac{V}{R} \frac{L_2}{L_1 + L_2}$ $i_2(\infty) = \frac{V}{R} \frac{L_1}{L_1 + L_2}$

Hence, Statement A is correct and Statement B is correct.

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Conclusion

The correct statements are A, B, and C.