To light a 50 W, 100 V lamp it is connected, in series with a capacitor of capacitance $\frac{50}{\mathrm{\pi }\sqrt{\mathrm{x}}}\mathrm{\mu F}$, with 200 V, 50Hz AC source. The value of x will be ____ .

$\mathrm{P}=\frac{{\mathrm{V}}^2}{\mathrm{R}}\Rightarrow \mathrm{R}=\frac{{\mathrm{V}}^2}{\mathrm{P}}$

$\begin{array}{l}\mathrm{R}=\frac{100\times {10}^2}{50}=\mathrm{R}=200\mathrm{\Omega }\\ {\mathrm{V}}_{\mathrm{R}}^2+{\mathrm{V}}_{\mathrm{C}}^2={\mathrm{V}}^2\\ (100{)}^2+{\mathrm{V}}_{\mathrm{C}}^2=(200{)}^2\\ {\mathrm{V}}^2=40000\\ {\mathrm{V}}_{\mathrm{C}}^2=30000\\ \mathrm{i}=\frac{100}{200}=\frac{1}{2} ; \\ {\mathrm{V}}_{\mathrm{C}}=\mathrm{I}\times {\mathrm{X}}_{\mathrm{C}} ; \\ {\mathrm{V}}_{\mathrm{C}}=100\sqrt{3}\\ {\mathrm{X}}_{\mathrm{C}}=200\sqrt{3}\\ 200\sqrt{3}=\frac{1}{\mathrm{\omega C}}\\ \mathrm{C}=\frac{1}{2\times \pi \times 50\times 200\sqrt{3}}=\frac{50\times {10}^{-6}}{\pi \sqrt{\mathrm{x}}}\\ \\ \Rightarrow \mathrm{X}=3\end{array}$