A lamp rating 8 V, 800 W is connected to a 100 V, 50 Hz a. c. source. The inductance of the choke required in the circuit is: (Take $π ≃ 3$)

2 mH

The correct answer is Option (1) → 2 mH

Given:

Lamp rating: $V_L = 8 \, \text{V}, P = 800 \, \text{W}$

Supply voltage: $V_s = 100 \, \text{V}, f = 50 \, \text{Hz}$

1. Resistance of the lamp:

$R = \frac{V_L^2}{P} = \frac{8^2}{800} = \frac{64}{800} = 0.08 \, \Omega$

2. Current through lamp at rated voltage:

$I = \frac{P}{V_L} = \frac{800}{8} = 100 \, \text{A}$

3. Total impedance needed to drop 100 V at 100 A:

$Z = \frac{V_s}{I} = \frac{100}{100} = 1 \, \Omega$

4. Series reactance of the choke:

$X_L = \sqrt{Z^2 - R^2} = \sqrt{1^2 - 0.08^2} = \sqrt{1 - 0.0064} = \sqrt{0.9936} \approx 0.9968 \, \Omega \approx 1 \, \Omega$

5. Inductance of the choke:

$X_L = 2 \pi f L \;\;\Rightarrow\;\; L = \frac{X_L}{2 \pi f} = \frac{1}{2 \cdot 3 \cdot 50} = \frac{1}{300} \, \text{H} \approx 0.002 \, \text{H} = 2 \, \text{mH}$

Answer: $L \approx 2 \, \text{mH}$