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 $\pi^2 = 10$)
Answer & explanation
Correct answer: option 2
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: $R = 50 \, \Omega, \; L \approx 0.312 \, \text{H}$