Question

An inductor (called choke coil) of $L$ inductance used in series with a bulb rated as $100V-100W$ . The bulb can be treated as a resistor. Given that, AC source of $200V$ and $50Hz$ . Find the value of inductance $L$

• A. $\frac{\pi }{\sqrt{3}}H$
• B. $100H$
• C. $\frac{\sqrt{2}}{\pi }H$
• D. $\frac{\sqrt{3}}{\pi }H$

Answer 1

Answer: (D)

For bulb,
$I=\frac{P}{V}=\frac{100}{100}=1A$
When an inductor is connected to the bulb in series the circuit becomes an LR circuit.
$\therefore I_{rms}=\frac{V_{rms}}{\sqrt{R^2+X_L^2}}$
$\Rightarrow 1=\frac{200}{\sqrt{(100{)}^2+X_L^2}}$
$\Rightarrow (100{)}^2+X_L^2$
$=(200{)}^2\Rightarrow X_L^2=(200{)}^2-(100{)}^2$
$\Rightarrow X_L$
$=\sqrt{(200{)}^2-(100{)}^2}$
$\Rightarrow X_L=100\sqrt{3}$
Now,
$X_L=\omega L\Rightarrow \omega L=100\sqrt{3}$
$\Rightarrow L=\frac{100\sqrt{3}}{\omega }=\frac{100\sqrt{3}}{2\pi \nu }$
$\Rightarrow L=\frac{100\sqrt{3}}{2\pi \times 50}=\frac{\sqrt{3}}{\pi }H$