Power Factor Applied?

No, an electrical load with a power factor of .8 requires an electrical power source producing 25% more Volt-Amps than would be required for an electrical load with a power factor of 1. Here`s why.

Watts = Volt-Amps X pf

W = VA X .8

VA = W/.8 = 1.25W

Thus, the required Volt-Amps is 1.25 times the actual power (Watts) consumed by the load.

Roughly speaking an electrical power source produces only Volts when it has no load attached. It will produce Volt-Amps within it`s design capabilities when any given load impedance is connected across it`s output terminals.. It is the ratio of load reactance (X) to load resistance (R) that determines the phase angle whose cosine is the power factor. I think it can be said, without too much argument otherwise, that an electrical power source will produce Volt-Amps when connected to a load. Also that the Voltage amplitude is determined by the Voltage source design, and that the load`s Amperes amplitude for that given Voltage is determined by the load impedance (Z). (Amps = V / Z) The loads reactance (X) to resistance (R) ratio determines what portion of the total Volt-Amps that are coverted to power {Watts = [VA X cos (arc tan X/R)]} in the form of Watts and what portion of the total Volt-Amps that are reflected back toward the electrical power source in the form of Volt-Amps reflective (VARs) . VARs = (VA) X (sin arc tan X/R). Reactive power is not consumed by the load`s inductance as one answerer stated or implied.

For the examples given:

If the load is 100kW purely resistive then power factor = 1 and:

Volt-Amps = (Watts)/(pf) = 100kW/1 = 100kVA

But, if the load not only contains 100kW of resistive load but also contains enough inductive reactance such that the power factor is .8 then:

Volt-Amps = (Watts)/(pf) = (100kW) / (.8) = 125kVA

A power factor of 0.8 means that the generator must generate 125 KVA, VA represent the product of Volt Amperes, this is apparent power and it includes both real and imaginary power. 125 KVA * (0.8) equals the actual power that the load consumes which is 100 KW. The inverse cosine of 0.8 = 36.87°, the reactive power consumed by the load's inductance is 125KV* sin 36.87° = 125KVA *(0.6) = 75 KVAR. VAR stands for Volt Ampere Reactive, this is reactive power. The best way to see what is going on is to draw the power triangle which is a right triangle with the base leg being 100 (scale this to 2 inches,the height being 75 (scale this to 1.5 inches) and the hypotenuse being 125 (this 2.5 inches). Real power is drawn on the horizontal axis, the imaginary power is drawn on the vertical axis with inductive reactance drawn above the horizontal axis and capacitive reactance drawn below the horizontal power axis. Real power is always the product of the apparent power in VA times the power factor. The power factor is also the cosine of the phase angle between voltage and current sinusoidal waves. The sine of the phase angle times the apparent power is the reactive power of the load.