In a series LR circuit, power of 400   W is dissipated from a source of 250   V ,   50  …

$\mathrm{P}={\mathrm{V}}_{\mathrm{m}}$. in $\mathrm{cos\varphi }$

$400=250\times 1\mathrm{m}\times 0.8$

${\mathrm{i}}_{\mathrm{rms}}=2\mathrm{A}$

${\left(1\mathrm{m}\right)}^2.\mathrm{R}=\mathrm{P}$

$4\times \mathrm{R}=400$

$\Rightarrow$ $\mathrm{R}=100\mathrm{\Omega }$.

$\mathrm{cos\varphi }=\frac{\mathrm{R}}{\sqrt{{\mathrm{R}}^2+{\mathrm{X}}_{\mathrm{L}}^2}}$

${100}^2+{\mathrm{X}}_{\mathrm{L}}^2={\left(\frac{100}{0.8}\right)}^2$

${100}^2+{\mathrm{X}}_{\mathrm{L}}^2={\left(\frac{100}{0.8}\right)}^2$

${\mathrm{X}}_{\mathrm{L}}=75\mathrm{\Omega }$

Power factor is unity

${\mathrm{X}}_{\mathrm{C}}={\mathrm{X}}_{\mathrm{L}}=75$

$\frac{1}{\mathrm{\omega }}=75$

$\Rightarrow$ $\mathrm{C}=\frac{1}{75\times 2\mathrm{H}\times 50}=\frac{1}{7500\mathrm{\pi }}\mathrm{F}$

$3\mathrm{\pi }\times 2500$

$=\frac{1}{3\mathrm{\pi }}\times 4\times {10}^2\mathrm{mF}$

$=\frac{400}{3\mathrm{\pi }}\mathrm{\mu F}$

$\mathrm{N}=400$