Return to power (idea)

Power Factor

We were all¹ taught in the [electicity] module of our [Physics] [class] that:

P = VI (Power = [Volts] × [Amps])

This is [true], but with a clarification which is glossed over in all but the most advanced courses.

For AC power, I and V are [vector] quantities, not scalar. The [peak] [current] may not be drawn at the same time as the [potential difference] peaks, especially if the [load] is [reactive], rather than just [resistive].

Imagine two [sine] waves which may or may not be out of [phase] with each other. If the waves are in phase, then our P=IV equation holds. If the waves are [exactly] out of phase, then P=0. However much power we supply, no [work] will get done. All electrical equipment has a power factor between these extremes, with [motor|motors] generally being worse than solid-state equipment.

This [power factor] is very important in machines which use a large amount of [electricity]; but can generally be [ignore|ignored] on small equipment. A low power factor means that the [power station] has to supply current at times when potential is low, and can [damage] their equipment. For this reason, large [factory|factories] have to [measure] and [declare] their power factor.

Equipment Rating

Electrical equipment may be [rate|rated] in one or more ways.

  • VA (Volt-amps) - The peak voltage × peak current of the supply.
  • W (watts) - The power [dissipate|dissipated] within the unit. This can be useful for measuring the expected heat generated if you're designing an air conditioning system.
If the power factor is supplied, then one value can be deduced from the other.

W = PowerFactor × VA

If VA, but not the power factor is supplied, then we can still work out a [worst-case] heat dissipation; it will always be VA or better.


¹ Those who took Physics, at any rate.
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