Let us consider an inductive load taking a lagging current I at a power factor cosΦ1. For us to improve the power factor of such a circuit, the solution is to connect equipment in parallel with the load which takes a leading reactive component and partly cancels the lagging reactive component of the load. Figure 1.0 below shows a capacitor connected across the load.
The capacitor takes a current IC which leads the supply voltage V by 90°. The current IC partly cancels the lagging reactive component of the load current as demonstrated in the phasor diagram in Figure 1.1. The resultant circuit current becomes I’ and its angle of lag is Φ2. It is clear that Φ2 is less than Φ1 so that the new power factor cosΦ2 is more than the preceding power factor cosΦ1.
From the phasor diagram, we can clearly see that, after the power factor correction, the lagging reactive component of the load is reduced by I’ sin Φ2.
In other words,
I’ sin Φ2 = I sin Φ1 – IC
IC = I sinΦ1 – I’ sin Φ2
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We know that:
Therefore the capacitance of the capacitor to improve the power factor from cosΦ1 to cosΦ2 is:
Also Read: Power Measurement in AC Circuits (Single-phase & Polyphase Systems)
Power Factor Correction Demonstration Using a Power Triangle
Let’s consider Figure 1.3 below:
The power triangle OAB is for the power factor cos Φ1, whereas power triangle OAC is for the improved power factor cos Φ2. Though the active power (OA) can be seen not to change with the power factor improvement, the lagging kVAR of the load is reduced by the power factor correction equipment, hence improving the power factor to cosΦ2.
The leading kVAR supplied by power factor correction equipment
BC = AB – AC
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= kVAR1 – kVAR2
= OA (tan Φ1 – tan Φ2)
= kW (tan Φ1 – tan Φ2)
Therefore, if we know the leading kVAR supplied by the power factor correction equipment, the appropriate results can be achieved.
Also Read: AC Circuits with Resistors, Inductors and Capacitors
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