A circuit draws a power 550 watt from a source of 220 volt, 50 Hz. The power factor of the circuit is 0.8 and the current lags in phase being the potential difference. To make the power factor of the circuit as 1.0 capacitance of $1.5\mathrm{n\mu F}$ will have be connected with it. Find the value of ‘n’.

I lags V circuit is R L , $\mathrm{P}={\mathrm{v}}_{\mathrm{rms}}{\mathrm{i}}_{\mathrm{rms}}\cos \varphi =550$

$$\begin{gathered} \begin{array}{l}{\mathrm{I}}_{\mathrm{rms}}=\frac{{\mathrm{V}}_{\mathrm{rms}}}{\mathrm{Z}}, \mathrm{Z}=\sqrt{\left[\left({\mathrm{R}}^2+(\mathrm{\omega L}{)}^2\right)\right]}\\ \therefore \mathrm{P}=\frac{{\mathrm{V}}_{\mathrm{rms}}^2\times \cos \mathrm{\varphi }}{\mathrm{Z}}\left(\text{ or }\right)\mathrm{Z}=\frac{{\mathrm{V}}_{\mathrm{rms}}^2\times \cos \varphi }{\mathrm{P}}\\ \mathrm{Z}=\frac{{220}^2\times 0.8}{550} = 70.4 \mathrm{\Omega }\\ \cos \varphi =\frac{\mathrm{R}}{\mathrm{Z}}\Rightarrow 0.8 =\frac{\mathrm{R}}{70.4}; \\ \Rightarrow \mathrm{R}= 56.32 \mathrm{\Omega }\\ {\mathrm{Z}}^2={\mathrm{R}}^2+(\mathrm{\omega L}{)}^2; \mathrm{\omega L} = 42.24 \mathrm{\Omega }\\ \end{array} \end{gathered}$$

When C in circuit,

$$\begin{gathered} \begin{array}{l}\mathrm{Z}=\sqrt{\left[{\mathrm{R}}^2+{\left(\mathrm{\omega L}-\frac{1}{\mathrm{\omega C}}\right)}^2\right]}\text{ and }\\ \cos \mathrm{\varphi }=\frac{\mathrm{R}}{\sqrt{\left[{\mathrm{R}}^2+{\left(\mathrm{\omega L}-\frac{1}{\mathrm{\omega C}}\right)}^2\right]}}\\ \cos \mathrm{\varphi }=\mathrm{I}, \mathrm{\omega L}=\frac{1}{\mathrm{\omega C}}\therefore \mathrm{C}=\frac{1}{\mathrm{\omega }\left(\mathrm{\omega L}\right)}=\frac{1}{2\mathrm{\pi f}\left(\mathrm{\omega L}\right)}\end{array} \\ \Rightarrow \mathrm{C} = \frac{1}{2\times 3.14\times 50 \times 42.24} \\ \Rightarrow \mathrm{C} =75.4 \times {10}^{-6} \mathrm{F} \end{gathered}$$