Question

When $100V$ dc is applied across a solenoid, a current of $1.0A$ flows in it. When $100V$ ac is applied across the same coil, the current drops to $0.5A$. If the frequency of the ac source is $50Hz$, the impedance and inductance of the solenoid are

• A. $200\Omega and0.55H$
• B. $100\Omega and0.86H$
• C. $200\Omega and1.0H$
• D. $200\Omega and0.93H$

Answer 1

The correct option is A $200\Omega and0.55H$

impedance= $\frac{V_{ac}}{I_{ac}}=\frac{100V}{0.5A}=200\Omega$
R=$\frac{V_{dc}}{I_{dc}}=\frac{100}{1}=100\Omega$
$L\omega$ = $\sqrt{{impedance}^2-R^2}=\sqrt{{200}^2-{100}^2}=173.2$
$L=\frac{173.2}{\omega }=\frac{173.2}{2\pi 50}=0.55H$