A circuit receives 50 A current at a power factor of 0.8 lag from a 250, 50 Hz, 1-ph A.C. supply. Calculate the capacitance of the capacitor which is required to be connected across the circuit to make the power factor unity.
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\(\mu F\)
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Fig. 20(a)
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Fig.22(a)