How this instrument works
Real power, P, is the rate at which a three-phase circuit does useful work — turning a shaft, making light, making heat — measured in watts. The formula multiplies the three quantities a clamp-on power logger reads directly at a switchboard: line-to-line voltage, line current, and power factor, then scales the product by √3, about 1.732. That constant is not a rounding convenience; it is what the arithmetic produces once three separate phase windings are folded into the line quantities an electrician actually measures.
A single-phase load pulls power in a pulse: instantaneous power swings from zero up to twice its average value and back, twice every cycle, because voltage and current cross zero together. Stack three identical loads a third of a cycle out of step and that pulsation cancels almost exactly — the sum of the three instantaneous powers holds constant rather than oscillating. That steadiness, not just the raw watts, is the real engineering reason large motors and generators are wound three-phase: constant torque needs no flywheel to smooth it out.
The formula assumes a balanced load — equal voltage, equal current, and the same power factor on all three lines, the normal condition on a healthy circuit and the only case a single P, V, I, and PF figure can describe. A genuinely unbalanced load, with one line drawing noticeably more current than the other two, has no single correct answer here; the honest fix is to measure real power on each of the three lines separately and add them, rather than clamp one line and multiply by three.
- Enter Line-to-line voltage — the reading between any two of the three supply lines, in volts, as shown on a multimeter or the switchboard display.
- Enter Line current — the amperage flowing in one line, read from a clamp-on ammeter; on a balanced circuit all three lines carry the same figure.
- Enter Power factor as a decimal between 0 and 1 — read it from a power-quality meter, a motor nameplate, or the utility bill's power-factor line.
- Read Real power — the watts, or kW, the circuit actually delivers, ready to compare against a generator's rating or a breaker's thermal limit.
Worked example — 400 V, 20 A at 0.90 power factor
A facilities engineer clamps a three-phase power logger onto the incoming feeder for a rooftop chiller and reads 400 V line-to-line, 20 A on each line, and a power factor of 0.90 off the meter's display — ordinary figures for a mid-size industrial motor load. Entering those three numbers gives P = √3 × 400 × 20 × 0.90 = 12,470.77 W, or about 12.47 kW of real power actually being delivered to the compressor.
That figure is what matters when sizing the standby generator against the load's real kilowatts, not the 13,856 VA of apparent power the same three readings also imply (S = √3 × 400 × 20 = 13,856 VA). Dividing power by apparent power returns exactly 0.90, the power factor already entered — confirmation that a generator or UPS rated in kVA needs about 11 percent more headroom than the kW figure alone would suggest for this same load.
Questions
Why does the formula multiply by √3 instead of by 3?
Because line and phase quantities differ in a balanced star (wye) connection. Per-phase real power is phase voltage times phase current times cosφ, and the total across three phases is three times that. Since line voltage equals √3 times phase voltage in a star connection while line current equals phase current, substituting the phase voltage back out leaves a single √3 in front of the line quantities — not a plain 3.
Why doesn't three-phase power pulsate the way single-phase power does?
In a single-phase circuit, instantaneous power swings from zero up to twice the average value and back, twice every cycle, because voltage and current cross zero together. Add three identical loads spaced a third of a cycle apart and those pulsations cancel almost perfectly, leaving a real power that stays constant instant to instant — a large part of why motors and generators built for serious capacity are wound three-phase rather than single-phase.
Does this formula still work if the three lines are not perfectly balanced?
No. P = √3 × V × I × cosφ assumes matching voltage, current, and power factor on all three lines, the normal healthy-circuit case. A genuinely unbalanced load — one line pulling noticeably more current than the other two — has no single correct answer from this equation; the reliable fix is to total the three lines individually rather than read one line's current and multiply by three.
What's the difference between the real power this calculator returns and apparent power?
Real power, P, is the watts that actually do work — heat, light, torque. Apparent power, S = √3 × V × I, is what the supply, transformer, and wiring must be sized to carry regardless of phase angle; the two are equal only at a power factor of 1. Dividing P by S recovers the power factor already entered here, which is exactly what a dedicated power-factor calculator computes starting from P and S instead of from V, I, and PF.
Can I use this formula for a single-phase circuit?
No. A single-phase circuit has only one voltage and current pair, so its real power is P = V × I × cosφ, with no √3 term at all. The √3 belongs specifically to three-phase systems, accounting for three voltages each offset from the next by a third of a cycle; applying it to a single-phase reading overstates the power by about 73 percent.