Correcting power factor - worked example


An ac motor is connected to the supply of 230V(rms) and found to be causing an issue with the power factor of the supply. This will incur charges and require heavier switch gear to cope with the current. Your job as an electrical engineer is to determine the value of power factor correction capacitor required to improve the power factor.

When the coil is connected to an AC supply the current is measured as 6 Amps.

From this we can work out the impedance:

The coil is then connected to a DC supply of 230V.

From this we can work out the resistance of the coil:

Calculating Inductive Reactance

The equivalent circuit of the coil can therefore be drawn as shown below. We know the resistance and impedance of the circuit but we don’t yet know the inductive reactance of the coil:

Calculating the power factor of the coil

This can be done in two ways: by drawing an impedance triangle to scale or by calculation

Selecting a capacitor value

In an inductive load the current lags the voltage by 90 degrees, but in a capacitive load, the current leads the voltage. Because of this it is possible to correct the power factor by bringing the current back into phase with the voltage. The load will then effectively be purely resistive.

A capacitor is connected in parallel with the coil and must have the same reactance as the inductor, ie XC should be the same as XL.

The formula below proves how the capacitor makes the load purely resistive.

To calculate the capacitance required we must rearrange the formula:

The circuit below shows the capacitor added to the circuit to correct the power factor. For a challenge you could prove that the power factor is now unity.