Question

A coil of self-inductance $L$ is connected in series with a bulb and an a. c. source. Brightness of the bulb decreases when

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

Hint:

Iron Rod = Higher Inductance = Higher Opposition to AC = Dimmer Bulb.

Answer 1

The Correct Option is A

Step 1: Concept
Brightness depends on the current ($I$) in the circuit. $I = V/Z$, where $Z = \sqrt{R^2 + X_L^2}$ and $X_L = 2\pi f L$.

Step 2: Meaning
Brightness decreases if impedance $Z$ increases, which happens if $L$ or $f$ increases.

Step 3: Analysis
- (A) Iron rod increases permeability ($\mu$), which increases $L$ ($L \propto \mu$). This increases $X_L$ and $Z$, decreasing current and brightness.
- (B) Decreasing frequency $f$ decreases $X_L$ and $Z$, increasing brightness.
- (C) Reducing turns decreases $L$, increasing brightness.
- (D) Resonance ($X_C = X_L$) minimizes $Z$ to $R$, maximizing brightness.

Step 4: Conclusion
Brightness decreases only in case (A). Final Answer: (A)