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How to calculate the Power Factor Correction

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.

A capacitor connected across the load
Figure 1.0 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.

Phasor diagram - power factor calculation
Figure 1.1 Phasor diagram

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

We know that:

Power factor calculations

            

Therefore the capacitance of the capacitor to improve the power factor from cosΦ1 to cosΦ2 is:

Power factor calculations

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:

Power triangle
Figure 1.3 Power triangle

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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