Question

An inductor (called choke coil) of $L$ inductance used in series with a bulb rated as $100\, V-100\, W$ . The bulb can be treated as a resistor. Given that, AC source of $200\, V$ and $50\, Hz$ . 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,
$I=\frac{P}{V}=\frac{100}{100}=1 A$
When an inductor is connected to the bulb in series the circuit becomes an LR circuit.
$\therefore I_{r m s}=\frac{V_{r m s}}{\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} $
$\Rightarrow L =\frac{100 \sqrt{3}}{2 \pi \times 50}=\frac{\sqrt{3}}{\pi} H$