Power factor correction of predominantly inductive load component with capacitor

Question

I started preparing for the college next year, and I came across this question: How can I express the capacitance of a capacitor which needs to be connected parallel to a predominantly inductive load component to fully correct the power factor? I remember mentioning power factor correction in high school, I think in Physics class. But we just mentioned it. What does power factor mean and how can you correct it?

Answer 1

The power factor of a circuit is the ratio of real or useful power P (something we get out of circuit - heat, light, mechanical) to apparent power S. It is a kind of efficiency. The closer the pf is to 1 the better.

$$pf = \frac PS = cos\ \theta $$

Most loads are inductive in nature. They have a lagging power factor. This means that the source has to provide more current than is needed to drive the load (real power P).

If you look at the blue triangle. Source must provide S1. Q1 (\$ Q_L \$) is lagging inductive reactive power in VARs. I lags \$ V_S \$ by \$ \phi_1 \$.

Power factor correction adds a capacitor in parallel with the load. This adds leading reative power to circuit. The Qc. P and Q1 still exist.

The capacitor acts like a source decreasing the reactive power the source must provide (Q2 = Q1 - Qc). This means S2 will be less than S1, so the current required will be less. I lags \$ V_S \$ by \$ \phi_2 \$.

The power factor has improved. Best case scenario is when \$ Q_L = Q_C \$. The triangle disappears and pf = 1.

Answer 2

Daer friend, for the same power, the current will be more if the pf becomes low. so to improvw pf in an inductive circuit, we use capacitor bank. the capacity of the bank depends on the normal pf, value you want to acieve, the voltage rating, motor or inductive load capacity.