Here is an example. A steel stamping operation runs at 100 kW (Working Power) and the Apparent Power meter records 125 kVA. To find the PF, divide 100 kW by 125 kVA to yield a PF of 80%. This means that only 80% of the incoming current does useful work and 20% is wasted through heating up the conductors. Because Laurens Electric must supply both the kW and kVA needs of all customers, the higher the PF is, the more efficient our distribution system becomes. Improving the PF can maximize current-carrying capacity, improve voltage to equipment, reduce power losses, and lower electric bills. The simplest way to improve power factor is to add PF correction capacitors to the electrical system. PF correction capacitors act as reactive current generators. They help offset the non-working power used by inductive loads, thereby improving the power factor. The interaction between PF capacitors and specialized equipment, such as variable speed drives, requires a well designed system. PF correction capacitors can switch on every day when the inductive equipment starts. Switching a capacitor on can produce a very brief “over-voltage” condition. If a customer has problems with variable speed drives turning themselves off due to “over-voltage” at roughly the same time every day, investigate the switching control sequence. If a customer complains about fuses blowing on some but not all, of their capacitors, check for harmonic currents.