Animated Solution for Physics - Electromagnetic Induction: The current ($i$) at time $t=0$ and $t=\infty$ respectively for the given circuit is

When dealing with DC circuits containing inductors, the key to unlocking the problem lies in understanding the inductor's behavior at two extreme time boundaries: the exact moment the circuit is closed ($t=0$), and after a long time has passed ($t=\infty$).

An inductor fundamentally opposes any change in the current flowing through it, governed by Faraday's law of induction, $V_L=-L\frac{di}{dt}$. Before the switch is closed, the current is zero. Therefore, at $t=0$, the inductor refuses to let the current jump instantly, effectively acting as an open circuit (infinite resistance).

Conversely, after a long time ($t=\infty$), the current reaches a steady, constant value. Since the current is no longer changing, the derivative $\frac{di}{dt}$ becomes zero, meaning there is no voltage drop across the ideal inductor. It behaves exactly like a perfect conducting wire, or a short circuit.

Let's apply this logic to our specific circuit. At $t=0$, we replace the inductor with an open circuit. This breaks the bottom-most wire, meaning no current can flow through that path.

The current $i$ supplied by the battery $E$ travels up to the top node and splits. One path goes left, flowing through the top-left $5\Omega$ resistor and then down through the bottom-left $1\Omega$ resistor. Because the inductor branch is open, these two resistors are strictly in series. The resistance of this left branch is:

Similarly, the other path goes right, flowing through the top-right $5\Omega$ resistor and the bottom-right $4\Omega$ resistor in series. The resistance of this right branch is:

These two branches are connected in parallel across the battery. The equivalent resistance $R_{eq}$ of the entire circuit at $t=0$ is:

Using Ohm's law, the current $i$ at $t=0$ is:

Now, let's fast forward to $t=\infty$. The inductor has reached a steady state and now acts as a short circuit. This is where the topology of the circuit gets interesting.

The short circuit effectively connects the bottom-left node and the bottom-right node together with a zero-resistance wire. Because the left and right vertical wires are now at the exact same electrical potential, the circuit folds onto itself.

The two top $5\Omega$ resistors are now connected between the top battery node and this common outer potential. This means they are in parallel with each other. Their equivalent resistance is:

By the exact same logic, the bottom $1\Omega$ and $4\Omega$ resistors are also in parallel with each other. Their equivalent resistance is:

The total equivalent resistance of the circuit is now the series combination of these two parallel blocks:

Finally, applying Ohm's law one last time, the steady-state current $i$ is:

Comparing our two calculated currents, $\frac{5E}{18}$ and $\frac{10E}{33}$, we find that they perfectly match option (d).