Question

A bulb is rated at $100V,100W$. It can be treated as a resistor. If 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 $200V$ and $50Hz$ is $\frac{\sqrt{x}}{\pi }H$. Find $x$

Answer 1

$\frac{V^2}{R}=100\Rightarrow R=\frac{{100}^2}{100}=100\Omega$
now,
$V=200\sqrt{2}sin\left(2\pi 50t\right)V\omega =2\pi 50rad/secinphasor,I_{rms}=\frac{200}{\sqrt{R^2+L^2{\omega }^2}}Atherefore,I_{rms}^{2}R=100\Rightarrow \frac{{200}^2}{R^2+L^2{\omega }^2}R=100\Rightarrow \frac{{200}^2}{{100}^2+L^2{\left(2\pi 50\right)}^2}100=100\Rightarrow \frac{1}{{100}^2}\frac{{200}^2}{1+L^2{\pi }^2}100=100\Rightarrow \frac{4}{1+L^2{\pi }^2}=1\Rightarrow 4=1+L^2{\pi }^2\Rightarrow L=\frac{\sqrt{3}}{\pi }H$