In the figure shown, a circuit contains two identical resistors with resistance R = 5 ohms and an in

Q: Why do we ignore the 2 mH value of the inductor?

In steady state, the rate of change of current is zero. Since $V = L \frac{dI}{dt}$, the voltage drop is zero regardless of the value of $L$. The 2 mH only matters during the transient phase.

Q: What if the question asked for current at $t=0$?

At $t=0$, the inductor opposes any change in current, acting as an open circuit. Current would only flow through the rightmost resistor, making $I = \frac{15}{5} = 3\text{ A}$.

Q: Do the two resistors share the current equally?

Yes! Since they are in parallel and have identical resistance ($5\Omega$), the $6\text{ A}$ total current splits equally, with $3\text{ A}$ flowing through each branch.

Q: Why is the inductor considered a short circuit and not an open circuit?

An inductor is basically a coil of wire. Once the magnetic field stops changing (steady state), it's just a regular wire with negligible resistance, hence a short circuit.

Q: Would the answer change if the battery was AC instead of DC?

Absolutely! With an AC source, the current is constantly changing, so the inductor would continuously provide inductive reactance ($X_L = \omega L$), and it would never act as a simple short circuit.

Question

In the figure shown, a circuit contains two identical resistors with resistance $R = 5\Omega$ and an inductance with $L = 2	ext{ mH}$. An ideal battery of $15	ext{ V}$ is connected in the circuit. What will be the current through the battery long after the switch is closed?

Options:

(A) 6 A

(B) 3 A

(C) 5.5 A

(D) 7.5 A

Sigma Percentile · Accepted Answer

### The Anatomy of the Circuit

Imagine you are an electron standing at the positive terminal of a $15	ext{ V}$ battery. The switch is thrown, and suddenly, a path opens up before you. But this isn't just a straight highway; it's a branching river.

The circuit splits into two parallel paths. One path is a simple, straightforward $5\Omega$ resistor. The other path, however, contains a twist: an inductor of $2	ext{ mH}$ sitting in series with another $5\Omega$ resistor.

### The Magic of "Long After"

The most critical phrase in this entire problem is **"long after the switch is closed"**. In the world of physics, this is a secret code for the **steady state**.

When a switch is first closed, the circuit experiences a sudden jolt. The current tries to rush in, but the inductor—a coil of wire—hates change. It fights back, creating a back-EMF that opposes the sudden surge of current. This is the transient state.

But what happens if we wait? Long after the switch is closed, the current stops changing. It reaches a constant, steady flow.

### Simplifying the Maze

Mathematically, the voltage across an inductor is given by $V_L = L \frac{dI}{dt}$. In the steady state, the current $I$ is constant, which means its rate of change $\frac{dI}{dt}$ is exactly zero.

If $\frac{dI}{dt} = 0$, then $V_L = 0$. The inductor loses all its resistance to the flow. It stops fighting and simply becomes a perfect, zero-resistance wire! It acts as a **short circuit**.

Suddenly, our complex circuit becomes beautifully simple. The inductor vanishes from our calculations, leaving us with two identical $5\Omega$ resistors connected perfectly in parallel across the $15	ext{ V}$ battery.

### The Final Calculation

Now, we just need to find the equivalent resistance of these two parallel branches. Using the parallel resistance formula:

$$R_{eq} = \frac{R_1 R_2}{R_1 + R_2}$$

Substitute our values:

$$R_{eq} = \frac{5 	imes 5}{5 + 5} = \frac{25}{10} = 2.5 \Omega$$

With the total equivalent resistance in hand, we apply Ohm's Law to find the total current drawn from the battery:

$$I = \frac{V}{R_{eq}}$$

$$I = \frac{15}{2.5} = 6	ext{ A}$$

And there we have it! The steady-state current flowing through the battery is exactly $6	ext{ A}$.

**A Quick Thought Experiment:** What if the question asked for the current *immediately* after the switch was closed? At $t=0$, the inductor acts as an open circuit, completely blocking the middle branch. The current would only flow through the rightmost $5\Omega$ resistor, giving $I = \frac{15}{5} = 3	ext{ A}$. Always watch the clock in inductor problems!

Select Answer:

Visualized Solution

Initial Circuit Setup

The Steady State Condition

Inductor in Steady State

Simplified Circuit

Equivalent Resistance $R_{e q}$

Total Current $I$

Final Answer

Transient State at $t = 0$

The Sigma Insight: Self and Mutual Inductance

The Anatomy of the Circuit

The Magic of "Long After"

Simplifying the Maze

The Final Calculation

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