Physics - EMI/AC Question with Solution | TestHub

Here, $P=50\mathrm{W},V=100\mathrm{V}$ $I=\frac{P}{V}=\frac{50}{100}=0.5\mathrm{A},R=\frac{V}{I}=\frac{100}{0.5}=200\Omega$ Let $L$ be the inductance of the choke coil $$\begin{array}{rl}& \therefore I_v=\frac{E_v}{Z}\text{ or }Z=\frac{E_v}{I_v}=\frac{200}{0.5}=400\Omega \\ & =100\sqrt{12}\Omega \\ & \end{array}$$ Now, $X_L=\sqrt{Z^2-R^2}=\sqrt{400^2-200^2}$ $=100\sqrt{12}\Omega$ or, $\omega L=100\times 2\sqrt{3}\Omega$ $\begin{array}{rl}\Rightarrow L=\frac{200\sqrt{3}}{\omega }=\frac{200\sqrt{3}}{2\pi v}=\frac{200\sqrt{3}}{100\pi }& =\frac{2\times 1.732}{3.14}\\ & =1.1\mathrm{H}\end{array}$