Q.A light bulb and an open coil inductor are connected to an ac source through a key as shown in Fig. 7.9.

The key idea is that inserting an iron rod increases the inductor's inductance $L$, which raises its inductive reactance $X_L = \omega L$.

1. The bulb's brightness depends on the current through it. The current $I = V / Z$, where $Z = \sqrt{R^2 + X_L^2}$ (the bulb's resistance $R$ is fixed, and the coil's own resistance is negligible compared to its reactance).
2. When the iron rod is inserted, the core's magnetic permeability $\mu$ increases sharply, so $L$ increases. Since $X_L = 2\pi f L$, the reactance rises. …

Why this approach works — the concept first

The key idea here is that an inductor in an AC circuit opposes changes in current through a property called inductive reactance, $X_L$. Unlike a resistor, which opposes current steadily, an inductor's opposition depends on how quickly the current is changing — and that's set by the AC frequency. But the inductor's design also matters: its inductance $L$ depends on the core material.

When you slide an iron rod into the coil, you're changing the core from air (low permeability) to iron (high permeability). That dramatically increases $L$. And since $X_L = 2\pi f L$, the reactance jumps up. With a fixed AC voltage source, more reactance means less current — and the bulb, which glows based on the power it dissipates ($P = I^2 R$), gets dimmer.

Step-by-step reasoning

1. Identify the circuit elements and their roles.

The bulb is a resistive load (its filament has resistance $R$). The open-coil inductor is a pure inductor (ideally, zero resistance) with inductance $L$. They're in series with an AC source of fixed voltage $V_{\text{rms}}$ and fixed frequency $f$. The total impedance of the series combination is:

2. Understand what inserting the iron rod does.

The inductance of a coil depends on the magnetic permeability $\mu$ of the core material:

where $N$ is the number of turns, $A$ is the cross-sectional area, and $l$ is the length. Air has $\mu \approx \mu_0$ (permeability of free space). Iron has a relative permeability $\mu_r$ that can be hundreds or thousands — so $\mu = \mu_r \mu_0$ becomes huge. Therefore, $L$ increases dramatically.

3. Trace the effect on current.

The RMS current in the circuit is:

When $L$ increases, $X_L$ increases, so $Z$ increases. Since $V_{\text{rms}}$ and $f$ are fixed, $I_{\text{rms}}$ decreases.

4. Connect current to bulb brightness. …

Method: Predicting How a Circuit Responds When a Reactive Element Changes

This method applies whenever a question changes one physical property of an inductor or capacitor (inserting a core, changing frequency, adding a dielectric) and asks how the current or brightness of another element responds.

Steps

Step 1: Identify what stays fixed and what changes.

Write down the source voltage and frequency (usually fixed) separately from the component property being altered (here, the core material, which changes $L$). Never let the changing quantity be confused with the fixed ones.

Step 2: Express the total opposition to current (impedance) in terms of the changing quantity.

For a resistor in series with an inductor,

Identify which reactance formula governs the changing element ($X_L = \omega L$ for an inductor, $X_C = 1/(\omega C)$ for a capacitor) and note whether the change increases or decreases that reactance.

Step 3: Track the chain of cause and effect through to the quantity being asked about. …