How this instrument works
A three-phase motor's nameplate power rating describes what comes out of the shaft, not what goes in at the terminals. Between those two figures sit two separate physical losses: efficiency, η, which accounts for the heat, friction, and magnetic losses inside the machine that mechanical output never captures; and power factor, PF, which accounts for the phase lag between voltage and current that reactive components — the motor's own magnetising winding among them — introduce. This instrument chains both: it first divides rated power by efficiency to recover the electrical power actually drawn from the supply, then divides that by √3 × line voltage × power factor to solve for the current in each supply line.
The √3 term is not a fudge factor; it falls directly out of the algebra of a balanced three-phase system. Total real power delivered across three lines is P = √3 × V (line-to-line) × I (line) × PF, whether the motor's windings are wired in star or delta internally — the line quantities absorb that detail. Solve that identity for current and efficiency's division is the only piece borrowed from outside pure electromagnetism: it converts the mechanical watts on the nameplate into the electrical watts the identity actually needs. An electrician sizing a feeder ahead of equipment delivery, or a panel builder cross-checking a manufacturer's stated full-load amps against a datasheet's efficiency and power-factor figures, runs exactly this chain by hand or reaches for a sheet like this one instead.
The result is full-load current: the steady amperage a properly loaded motor draws once it is running normally, not the surge at start-up. Locked-rotor current when a motor first energises commonly reaches five to eight times this figure for a second or two, which is why motor branch circuits use time-delay fuses or motor-rated circuit breakers rather than devices sized for the running current alone. The other common mistake runs the other way: treating the nameplate kilowatt figure as electrical input and skipping the efficiency division entirely returns a current that is too low, because mechanical output is always smaller than the electrical power that produced it — an error that looks conservative and is actually the opposite.
- Enter Motor power — the nameplate's rated shaft output, in kW or hp.
- Enter Line-to-line voltage as measured or specified at the supply, in volts.
- Enter Power factor from the nameplate or a test report, typically 0.80 to 0.95 at full load.
- Enter Motor efficiency as a percentage, for example 90 for 90%, straight from the nameplate.
- Read Full-load line current, then compare it against the manufacturer's stated rated current before sizing cable or protection.
Worked example — 15 kW motor at 400 V, PF 0.85
Take a 15 kW three-phase motor rated for 400 V line-to-line supply, with a power factor of 0.85 and an efficiency of 90% at full load — ordinary nameplate figures for an industrial induction motor of that size. Electrical input power comes first: P_elec = 15,000 W ⁄ 0.90 = 16,666.7 W, the wattage the motor actually pulls from the supply once its own internal losses are accounted for. Current follows directly: I = 16,666.7 ⁄ (√3 × 400 × 0.85) = 16,666.7 ⁄ 588.90 = 28.30 A, the full-load line current a designer would carry forward to conductor and breaker sizing.
That 28.3 A is the number a panel builder actually works from: a thermal overload relay gets set close to it, and IEC 60034-1 requires a motor's nameplate to state rated current alongside efficiency and power factor precisely so a figure like this one can be checked against the manufacturer's own test data rather than taken on faith. Let the same motor's supply sag to 380 V while everything else holds steady and current climbs to about 29.8 A — current rises as line voltage falls, because the motor still needs the same electrical power to turn the same shaft load.
Questions
Why does the formula divide by efficiency as well as by power factor?
Because they account for two different losses. Power factor describes the phase lag between voltage and current in an AC circuit, changing how much current is needed to deliver a given electrical wattage. Efficiency describes a separate conversion: how much of the electrical wattage drawn from the supply actually leaves the shaft as mechanical work, versus being lost to heat, friction, and stray magnetic effects inside the motor. Dividing rated shaft power by efficiency first recovers the electrical power the supply must provide; dividing that by √3 × V × PF then converts electrical power into line current.
What is the difference between full-load current and starting current?
Full-load current, the figure this instrument returns, is the steady amperage a motor draws once it is running at its rated mechanical load. Starting current is far higher — a standard induction motor typically inrushes at five to eight times its full-load current for the first second or two while it comes up to speed, which is why motor circuits use time-delay or motor-rated protective devices rather than fuses sized for the running figure alone.
Should Line-to-line voltage be the supply voltage or the voltage at the motor terminals?
Use the voltage actually present at the motor terminals under load, which on a well-designed feeder sits close to the supply voltage but is not always identical to it. Voltage drop along a long or undersized cable run lowers the terminal voltage below the source, and because current must rise to deliver the same electrical power at a lower voltage, an overstated Line-to-line voltage figure understates the true full-load current.
My motor's nameplate current doesn't match what this calculator gives me. Why?
Small differences are normal. Manufacturers measure efficiency and power factor on a dynamometer under controlled conditions and round the results before printing a nameplate, so a calculation built from those rounded figures will not reproduce the exact test-bench current to the last decimal. Larger gaps usually mean the power figure entered is a different rating than the nameplate intends — rated output is not always the same quantity as an 'input power' figure printed alongside it — so confirm which value each number represents before comparing.
Why does three-phase current use a √3 factor instead of just multiplying by 3?
Because the three phases are not aligned; each line voltage and its corresponding current sit 120° apart from one another, and summing three sinusoids at that spacing yields total power of √3 × V (line) × I (line) × PF for a balanced system, not 3 × V × I × PF. A plain factor of 3 would only apply to three independent single-phase circuits acting in step, which is not how a three-phase supply behaves. √3, about 1.732, is what the trigonometry of that 120° spacing actually produces.
Does this current change with the motor's internal connection, star or delta?
Not directly — the formula works entirely in line quantities, so line-to-line voltage and full-load line current come out the same regardless of whether the windings are wired in star or delta internally. What changes is the current inside each phase winding: in star, phase current equals line current, while in delta it is line current divided by √3. That distinction matters for checking a winding's own rating, not for sizing the supply conductors, which carry the line current this instrument returns.