Three-Phase Apparent Power

S = \sqrt{3} \, V_{L} I_{L}

S =

V_L =

I_L =

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

Constant used — built into this formula, no need to enter

\sqrt{3} = 1.7320508075688772

Square Root of Three · exact

Learning zone

Apparent power is the product the copper actually feels: every amp heats the conductor whether or not it is in phase with the voltage. Volt-amperes and watts have identical dimensions — this solver's watt fields double as volt-amperes, and kW as kVA — but tradition keeps the names separate to remind you that S ≥ P always. A 208 V panel drawing 50 A per line is handling 1.732 × 208 × 50 ≈ 18 kVA regardless of what the loads are doing.

This is the number on a transformer's nameplate, and the reason it is: the transformer's limits are its winding heat (amps) and its core saturation (volts), neither of which knows anything about power factor. To size a service, work in kVA; to size a bill or a generator's engine, work in kW. Drop the √3 for single-phase and the formula is just S = VI.

Three-Phase Apparent Power

S = \sqrt{3} \, V_{L} I_{L}

Where

$S$ = Apparent power (VA)
$V_{L}$ = Line-to-line voltage
$I_{L}$ = Line current

Missing one of these? Work it out first, then come back

Apparent power (VA) — Power Factor from Real and Apparent Power, Reactive Power (Power Triangle)
Line-to-line voltage — Three-Phase Real Power, Single-Phase Real Power with Power Factor
Line current — Three-Phase Real Power, Single-Phase Real Power with Power Factor

Part of this formula set

Three-phase power and power factor

8 formulasReal, reactive and apparent power, power factor, phase angle and correction kvar — the power triangle as an electrician uses it.