Question

A bulb is rated at $100\text{V}\&100\text{W}$ it can be treated as a resistor. Find out the inductance of an inductor (called choke coil) that should be connected in series with the bulb to operate the bulb at its rated power with the help of an ac source of $200\text{V}$ and $50\text{Hz}$

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

Answer 1

The correct option is D $\frac{\sqrt{3}}{\pi }\text{H}$



Given the rating of the bulb as,

$\text{Voltage of bulb =}100\text{V}$
$\text{Power of bulb =}100\text{W}$
$\text{Voltage of source =}200\text{V}$
$\text{Frequency of source=}50\text{Hz}$

$\because R=\frac{V_{rms}^2}{P}=\frac{{100}^2}{100}=100\Omega$

For the bulb to be operated at its rated value, the rms current through it should be $1\text{A}$

Also we know,

$I_{rms}=\frac{V_{rms}}{Z}$
$\therefore 1=\frac{200}{\sqrt{{100}^2+(2\pi 50L{)}^2}}$

$\Rightarrow L=\frac{\sqrt{3}}{\pi }\text{H}$

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