# How can a parallel capacitor improve the power factor of an inductive load?

In phasor or vector diagram, a capacitor that is parallel to the supply can improve power factor. I know this is practically true but I don't understand the mathematical equation:

The total impedance (Z) of the following circuit has imaginary part i=root(-1). That means it has a reactants and it will consume reactive power.

If XL = Xc , the reacance should be infinity or has a very large value so it will consume large reactive power.

I feel like algebra does not support phasor diagram... Would you tell me what I'm missing?

Thank you,

Edit:

Here is the opposite case:

If a very high reactance is good for power factor, Here is a circuit with a very low reactance. Does it also improve power factor? If yes, which one is better?

• Another way to look at this -- put just a plain resistor supplied by the voltage source, what would happen if the resistance goes to infinity (which would be a open circuit). That is, power goes down when impedance goes up. Commented Apr 9, 2016 at 19:09
• You need a load resistance in the circuit to appreciate the implications of power factor fully
– Chu
Commented Apr 9, 2016 at 22:00