From the given data, we have \(cos\ \theta=0.8\),
Or \(\theta=\cos^{-1}{0.8}={36.86}^0\)
The component of current in phase with voltage
\(I_x=I\ cos\ \theta=50\times0.8=40Amp\).
The component of current lagging the voltage by \({90}^0\),
Or
\(I_x=I\ sin\ \theta=50\times0.6=30Amp\).
Or \(C\omega=\frac{I}{V}=\frac{30}{250}Siemens\)
Therefore \(C=\frac{30}{250\times2\times\pi\times50}=382\ \mu F\)
Therefore required capacitance to make the power factor unity is 382\(\mu F\)
Fig.23.png)
Fig.22(a)