When 100 V dc is applied across a solenoid, a current of 1 A flows in it. When 100 V ac is applied across the same solenoid the current drops to 0.5 A. If the frequency of the ac source is 50 Hz, the impedance and inductance of the solenoid are:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 200 Ω and 0.55 H
A solenoid behaves as a resistance R in series with an inductance L. For d.c. there is no inductive opposition, so the d.c. current gives R. For a.c. the opposition is the impedance \( Z = \sqrt{R^{2} + X_L^{2}} \), with \( X_L = 2\pi fL \).
From the d.c. reading
\( \displaystyle R = \frac{100\,\mathrm{V}}{1\,\mathrm{A}} = 100\,\Omega \)
From the a.c. reading
\( \displaystyle Z = \frac{100\,\mathrm{V}}{0.5\,\mathrm{A}} = 200\,\Omega \)
\( \displaystyle X_L = \sqrt{Z^{2} - R^{2}} = \sqrt{200^{2} - 100^{2}}\,\Omega = 100\sqrt{3}\,\Omega \approx 173.2\,\Omega \)
\( \displaystyle L = \frac{X_L}{2\pi f} = \frac{173.2}{2\pi\times50}\,\mathrm{H} = \frac{173.2}{314.2}\,\mathrm{H} \approx 0.55\,\mathrm{H} \)