Question

When $100\text{V}$ of d.c. is applied across a solenoid a current of $1.0\text{A}$ flows in it. When$100\text{V}$ of a.c. is applied across the same coil, the current drops to $0.5\text{A}.$ The frequency of the ac source is $50\text{Hz}.$ The impedance and inductance of the solenoid is

• A. $100\Omega \text{and}0.93\text{H}$
• B. $200\Omega \text{and}1.0\text{H}$
• C. $100\Omega \text{and}0.86\text{H}$
• D. $200\Omega \text{and}0.55\text{H}$

Answer 1

The correct option is D $200\Omega \text{and}0.55\text{H}$

When d.c. is applied,

$R=\frac{V}{I}=\frac{100}{1}=100\Omega$

When a.c. is applied, the impedance $Z$ is,

$Z=\frac{V_{rms}}{I_{rms}}=\frac{100}{0.5}=200\Omega$

$V_{rms}=I_{rms}Z=I_{rms}\sqrt{R^2+X{{}_L}^2}$

$100=0.5\sqrt{(100{)}^2+X{{}_L}^2}$

$(200{)}^2=(100{)}^2+X{{}_L}^2$

$X_L=\sqrt{(200{)}^2-(100{)}^2}$

$X_L=100\sqrt{3}[\because X_L=L\omega ]$

$L\omega =100\sqrt{3}$

$L=\frac{100\sqrt{3}}{\omega }=\frac{100\sqrt{3}}{2\pi f}=\frac{100\sqrt{3}}{2\times 3.14\times 50}=\frac{\sqrt{3}}{3.14}$

$\therefore L=0.55\text{H}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.