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.

The capacitor takes a current I_C which leads the supply voltage V by 90°. The current I_C 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 – I_C

I_C = I sinΦ₁ – I’ sin Φ₂

We know that:

Therefore the capacitance of the capacitor to improve the power factor from cosΦ₁ to cosΦ₂is:

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                  

= 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