The current and power dissipation in an inductor coil are 2 A and 200 W, respectively, when connected to an alternating source of 220 V, 50 Hz. The resistance and inductance of the coil are: (take $\pi^2 = 10$)

50 Ω and 0.312 H

The correct answer is Option (2) → 50 Ω and 0.312 H

Given:

AC voltage: $V = 220 \, \text{V}$

Current: $I = 2 \, \text{A}$

Power: $P = 200 \, \text{W}$

Frequency: $f = 50 \, \text{Hz}$

1. Resistance of coil:

Power dissipated: $P = I^2 R \;\;\Rightarrow\;\; R = \frac{P}{I^2} = \frac{200}{2^2} = \frac{200}{4} = 50 \, \Omega$

2. Impedance of coil:

$Z = \frac{V}{I} = \frac{220}{2} = 110 \, \Omega$

3. Inductive reactance:

$X_L = \sqrt{Z^2 - R^2} = \sqrt{110^2 - 50^2} = \sqrt{12100 - 2500} = \sqrt{9600} = 97.98 \approx 98 \, \Omega$

4. Inductance of coil:

$X_L = 2 \pi f L \;\;\Rightarrow\;\; L = \frac{X_L}{2 \pi f} = \frac{98}{2 \cdot 3.1416 \cdot 50} \approx \frac{98}{314.16} \approx 0.312 \, \text{H}$

Answer: $R = 50 \, \Omega, \; L \approx 0.312 \, \text{H}$