On direct current, volts times amps is the whole story. On alternating current it is not, because current and voltage do not always peak together — and only the part of the current that is in step with the voltage does any work.

$P = V I \, \text{PF}$, read aloud P equals V I times P F. $P$ is the real power in watts — what the wattmeter reads and what does the job. $V$ is the RMS supply voltage in volts and $I$ the RMS current in amperes, both as an ordinary meter reports them. $\text{PF}$ is the power factor: a bare ratio between 0 and 1, no units, equal to $\cos\varphi$ where $\varphi$ — phi — is the angle by which the current lags the voltage. Solve it for $P$ going down, or for $I$ coming back.

The product $V I$ has its own name: apparent power, $S$, measured in volt-amperes. It is what the supply has to deliver. And that gives the second relation of this lesson, $\text{PF} = \dfrac{P}{S}$ — P F equals P over S — the share of the delivered capacity that actually worked, with $P$ in watts and $S$ in volt-amperes.

Volt-amperes and watts are dimensionally identical and never interchangeable, which is the trap this whole chapter is built around. A 5 kVA supply feeding a 0.8 power factor load delivers 4 kW — and quoting that load as “5 kW” is how a generator gets ordered one size too small. Units guide, they do not confess: W, VA and var all reduce to the same algebra, so the unit will not catch this one for you. The reasoning has to.