Question

An electric bulb, an open coil inductor, an ac source and a key are all connected in series to form a closed circuit. The key is closed and after some time an iron rod is inserted into the interior of the inductor, then

• A. The glow of the bulb increases
• B. The glow of the bulb remains unchanged
• C. The glow of the bulb decreases
• D. The bulb does not glow

Hint:

Remember the role of core materials in inductors: inserting a soft iron core into an inductor greatly increases its inductance. In a DC circuit, an inductor acts as a short circuit after the initial transient, so inserting an iron core would have no effect on the steady-state brightness. However, in an AC circuit, its reactance is crucial.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The brightness of the bulb depends on the current flowing through it. The circuit is an AC series circuit containing a resistor (the bulb) and an inductor. The inductor offers opposition to the flow of alternating current, known as inductive reactance. This reactance depends on the inductor's self-inductance.
Step 2: Analyzing the Circuit and the Effect of the Iron Rod:
1. Initial State: The circuit consists of an AC source, a bulb (resistance \(R\)), and an air-cored inductor (inductance \(L_{air}\)). The total opposition to the current, called impedance (\(Z\)), is given by \(Z = \sqrt{R^2 + X_L^2}\), where \(X_L = \omega L_{air}\) is the inductive reactance (\(\omega\) is the angular frequency of the AC source). The current in the circuit is \(I = V/Z\).
2. Inserting the Iron Rod: An iron rod is a ferromagnetic material with a high relative magnetic permeability (\(\mu_r \gg 1\)). The self-inductance of a coil is directly proportional to the permeability of its core material (\(L \propto \mu\)). When the iron rod is inserted into the inductor, its self-inductance increases significantly: \(L_{iron} = \mu_r L_{air} \gg L_{air}\).
3. Consequences:
- The inductance \(L\) increases.
- The inductive reactance \(X_L = \omega L\) increases.
- The total impedance of the circuit \(Z = \sqrt{R^2 + X_L^2}\) increases.
- According to Ohm's law for AC circuits (\(I = V/Z\)), if the impedance \(Z\) increases while the source voltage \(V\) remains constant, the current \(I\) in the circuit must decrease.
- The power dissipated by the bulb, which determines its glow, is \(P = I^2 R\). Since the current \(I\) decreases, the power dissipated also decreases.
Step 3: Final Answer:
As the current through the bulb decreases, its glow diminishes. Therefore, the glow of the bulb decreases. Option (C) is correct.