Industrial power arrives on three conductors carrying three voltages 120° apart. $P = \sqrt{3} \, V_{L} I_{L} \, \text{PF}$ — read aloud P equals root three V-L I-L times P F. $P$ is the real power in watts; $V_{L}$ is the line-to-line voltage in volts, measured between any two of the three conductors, and the number printed on every panel; $I_{L}$ is the line current in amperes, in one conductor; $\text{PF}$ is the same bare power factor as before. The subscript L means LINE throughout — outside the machine — as opposed to phase, which is inside it.

Drop the power factor and you have the apparent power, $S = \sqrt{3} \, V_{L} I_{L}$, in volt-amperes. Read backwards, that same line is how a kVA nameplate becomes full-load amps, which is the calculation that sizes a transformer's secondary breaker.

The $\sqrt{3}$ is not there because there are three phases. It is there because the voltage you measure spans two windings set 120° apart, while the current you measure belongs to one — and the geometry of that mismatch is $\sqrt{3} \approx 1.732$. Writing 3 instead is the single commonest arithmetic error in the trade: it overstates a load by 73%, and it looks completely reasonable on the page.