A bulb is rated at $100\mathrm{ }\mathrm{V},\mathrm{ }100\mathrm{ }\mathrm{W}$, 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 $\mathrm{200 }\mathrm{V}$ and $50\mathrm{ }\mathrm{H}\mathrm{z}$.

• A

  $\frac{\mathrm{\pi }}{\sqrt{3}} \mathrm{H}$
• B

  $100\mathrm{ }\mathrm{H}$
• C

  $\frac{\sqrt{2}}{\mathrm{\pi }} \mathrm{H}$
• Correct answer
  D

  $\frac{\sqrt{3}}{\mathrm{\pi }} \mathrm{H}$

Solution

From the rating of the bulb, the resistance of the bulb can be calculated.

$R=\frac{{V_{\mathrm{r}\mathrm{m}\mathrm{s}}}^2}{P}=$$\frac{{\left(100\right)}^2}{100}=100\mathrm{\Omega }$



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

Also,

$I{ }_{\mathrm{r}\mathrm{m}\mathrm{s}}=\frac{V_{\mathrm{r}\mathrm{m}\mathrm{s}}}{Z}$

$\therefore 1=\frac{200}{\sqrt{{100}^2+{\left(2\pi 50L\right)}^2}}\Rightarrow L=\frac{\sqrt{3}}{\pi } \mathrm{H}$

Hence, the correct answer is option: 'D'.