Question

A coil of self inductance \(L\) is connected in series with a bulb B and an a.c. source. Brightness of the bulb decreases when

• A. frequency of a.c. source is decreased
• B. number of turns in the coil is reduced
• C. an iron rod is inserted in the coil
• D. a capacitance of reactance \(X_{c}\) is included in the same circuit

Hint:

Brightness depends on the current; the current falls when the inductive reactance of the coil rises.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The current in the series circuit is \(I=\dfrac{V}{\sqrt{R^{2}+X_{L}^{2}}}\), with \(X_{L}=\omega L\).

Step 2: Effect of an Iron Rod
Inserting an iron rod raises the permeability and hence the self inductance \(L\), so \(X_{L}\) increases.

Step 3: Result on the Bulb
A larger \(X_{L}\) reduces the current, so the bulb becomes dimmer.

Step 4: Compare the options
Decreasing frequency or turns reduces \(X_{L}\) and increases the brightness. A capacitor reduces the net reactance and raises the current. Only option (C) dims the bulb.

Final Answer:
The bulb dims when an iron rod is inserted in the coil. \[ \boxed{\text{(C) }\text{an iron rod is inserted in the coil}} \]