How the calculation works
Current depends on the power drawn, the voltage, whether the circuit is single-phase or three-phase, and the power factor (PF):
- Single-phase: amps = watts ÷ (volts × PF)
- Three-phase: amps = watts ÷ (√3 × volts × PF), where volts is the line-to-line voltage and √3 ≈ 1.732
Watts here means real power, the figure on most power meters and data sheets. The power factor is the ratio of real power (watts) to apparent power (volt-amperes, VA). A PF of 1.0 means the two are equal, which is true of resistive loads such as heaters. Equipment with a PF below 1.0 draws more current for the same watts, so the calculator also shows the apparent power in kVA (watts ÷ PF ÷ 1,000), the number UPS and generator ratings are often quoted in.
Line-to-line or line-to-neutral? A three-phase system has two voltages. Line-to-line is measured between two phase conductors; line-to-neutral is measured between one phase and neutral, and it is the line-to-line voltage divided by √3. A 208Y/120 V system is 208 V line-to-line and 120 V line-to-neutral. The three-phase formula above uses the line-to-line voltage. A single-phase circuit taken from a three-phase panel uses the voltage of that circuit: 120 V between a phase and neutral, or 208 V between two phases.
The last line of the result is context, not electrical advice. A circuit serving a continuous load is commonly limited to 80% of its breaker rating in North America, so the calculator divides the current by 0.8 to show the smallest circuit rating that keeps this load at or below 80%.
Worked example
A colocation cabinet will hold equipment drawing 5,000 W, with a power factor of 0.95, fed at 208 V.
On a single-phase 208 V circuit:
- amps = 5,000 ÷ (208 × 0.95) = 5,000 ÷ 197.6 = 25.30 A
- at 80% loading, the circuit needs a rating of at least 25.30 ÷ 0.8 = 31.63 A, so a 30 A circuit is too small and a 40 A circuit fits
On a three-phase 208 V circuit:
- amps = 5,000 ÷ (1.732 × 208 × 0.95) = 5,000 ÷ 342.3 = 14.61 A per phase
- at 80% loading that needs at least 14.61 ÷ 0.8 = 18.26 A, so a 20 A three-phase circuit fits
In both cases the apparent power is 5,000 ÷ 0.95 ÷ 1,000 = 5.26 kVA. With redundant A and B feeds, each feed has to carry the full load on its own if the other fails, so size each one for the whole 5,000 W.



