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:

200 Ω and 0.55 H

The correct answer is Option (1) → 200 Ω and 0.55 H

In the DC case the current is purely due to the resistance of solenoid,

$R=\frac{V_{DC}}{I_{DC}}=\frac{100}{1}=100Ω$

In the AC case,

Impedance, $Z=\frac{V_{AC}}{I_{AC}}=\frac{100}{0.5}=200Ω$

Impedance, $Z=\sqrt{R^2+{X_L}^2}$

$200=\sqrt{100^2+{X_L}^2}$

$⇒{X_L}^2=30000$

$≃173.2Ω$

and,

Inductive resistance, $X_L=2πfL$

$⇒L=\frac{X_L}{2πf}=\frac{173.2}{2×3.14×50}≃0.55H$