When a 100 V dc is applied across a solenoid, a current of 1 A flows in it. When a 100 V, 50 Hz ac is applied across the same solenoid, the current drops to 0.5 A. The inductance of the solenoid is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.55 H
With dc the inductor offers no reactance, so the current is limited only by the resistance of the wire. With ac the solenoid behaves as a series \( LR \) circuit, whose impedance is \( Z = \sqrt{R^{2} + X_{L}^{2}} \).
\( \displaystyle R = \frac{V}{I} = \frac{100\,\mathrm{V}}{1\,\mathrm{A}} = 100\,\Omega \)
\( \displaystyle Z = \frac{V}{I} = \frac{100\,\mathrm{V}}{0.5\,\mathrm{A}} = 200\,\Omega \)
\( \displaystyle X_{L} = \sqrt{Z^{2} - R^{2}} = \sqrt{200^{2} - 100^{2}} = 100\sqrt{3}\,\Omega \approx 173.2\,\Omega \)
\( \displaystyle L = \frac{X_{L}}{2\pi\nu} = \frac{173.2\,\Omega}{2\pi \times 50\,\mathrm{Hz}} = \frac{173.2}{314.2}\,\mathrm{H} \approx 0.55\,\mathrm{H} \)