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
A real coil has resistance as well as inductance. Power is dissipated only in the resistance, so \( P = I^{2}R \) with rms values, and the impedance of the coil is \( Z = V/I = \sqrt{R^{2} + X_L^{2}} \).
\( \displaystyle R = \frac{P}{I^{2}} = \frac{200}{2^{2}} = 50\,\Omega \)
\( \displaystyle Z = \frac{V}{I} = \frac{220}{2} = 110\,\Omega \)
\( \displaystyle X_L = \sqrt{Z^{2} - R^{2}} = \sqrt{110^{2} - 50^{2}} = \sqrt{9600} = 97.98\,\Omega \)
With \( \pi^{2} = 10 \), \( \pi = \sqrt{10} = 3.162 \):
\( \displaystyle L = \frac{X_L}{2\pi f} = \frac{97.98}{2 \times 3.162 \times 50} = \frac{97.98}{316.2} \approx 0.31\,\mathrm{H} \)
This is the value 0.312 H given in Option (2).