A circuit draws a power of 550 watt from a source of 220volt, 50 Hz. The power factor of the circuit is 0.8 and the current lags in phase behind the potential difference. To make the power factor of the circuit as 1.0, capacitance of $15n\mu F$  will have to be connected with it. Find the value of n to the nearest integer

As the current lags behind the potential difference, the circuit contains resistance and inductance. 
Power, P = vrms irms  $\cos \varphi$
Here, $i_{rms}=\frac{V_{rms}}{Z},$ where Z
 $=\sqrt{\left[R^2+{\left(\omega L\right)}^2)\right]}$
$P=\frac{V_{rms}^2\times \cos \varphi }{Z}$ or  $Z=\frac{V_{rms}^2\times \cos \varphi }{P}$
So,  $Z=\frac{{\left(220\right)}^2\times 0.8}{550}=70.4ohm$
Now, power factor cos  $\varphi =\frac{R}{Z}$
or R = Z  $\cos \varphi$
 R = 70.4 0.8 = 56.32 ohm 
Further,  $Z^2=R^2+{\left(\omega L\right)}^2$
or   $\left(\omega L\right)=\sqrt{(Z^2-R^2)}$
or  $\omega L=\sqrt{{\left(70.4\right)}^2-{\left(56.32\right)}^2}=4.22\Omega$
When the capacitor is connected in the circuit, 
$Z=\sqrt{[R^2+{(\omega L-\frac{1}{\omega C})}^2]}$

and  $\cos \varphi =\frac{R}{\sqrt{[R^2+{(\omega L-\frac{1}{\omega C})}^2]}}$

when  $\cos \varphi =1, \omega L=\frac{1}{\omega C}$

$$\begin{gathered} \therefore C=\frac{1}{\omega \left(\omega L\right)}=\frac{1}{2\pi f\left(\omega L\right)}=\frac{1}{(2\times 3.14\times 50)\times \left(42.2\right)} \\ =75\times {10}^{-6}F=75\mu F \end{gathered}$$