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

0.55 H

For DC circuit:

Here, Inductor will behave as if its short circuited, hence \( R = \frac{V}{I} = \frac{100}{1} = 100 Ω \)

For A.C. circuit:

Here, Total impedance is both due to resistor and inductor , \( Z=\frac{V}{I} = \frac{100}{0.5}=\sqrt{(\omega L)^2+(100)^2} \)

\( Z=200=\sqrt{(100 \pi L)^2+(100)^2} \)

\( 200=100 \sqrt{\pi^2 L^2+1} \)

\( 4=\pi^2 L^2+1 \)

\( \frac{\sqrt{3}}{\pi}=L \)

⇒ L = 0.55 H