Calculation
P = √3 · U · I · cosφ
√3 applies to line voltage and balanced load; cosφ accounts for phase shift.
P = √3·U·I·cosφ
Three-Phase Power Calculator
√3 applies to line voltage and balanced load; cosφ accounts for phase shift.
MINTSI
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Result
Select a target and calculate.
Understand the inputs
- Real power P — The real power actually consumed or delivered by the three-phase load – the power that is actually converted into work or heat, as distinct from apparent power.
- Line voltage U — The line voltage – the voltage between two of the three phase conductors, usually 400 V in Europe. Not the same as the phase voltage (conductor to neutral, usually 230 V).
- Line current I — The RMS current in each of the three phase conductors (equal in all three for a balanced load). Source: a current measurement (clamp meter) or the nameplate.
- Power factor cos φ — The power factor – the ratio of real power used to the total apparent power flowing. 1 means a purely resistive load (heater); motors are often around 0.8–0.9. Source: the load's nameplate, or a power-meter measurement.
Example
√3 · 400 V · 10 A · 1 = 6.93 kW.
Assumptions and limits
Balanced 50 Hz system; harmonics and imbalance are excluded.
Understand Three-Phase Power Calculator
This calculator finds the missing quantity among real power P, line voltage U, line current I and power factor cos φ from the other three, for a balanced three-phase system. It explains where the characteristic √3 factor comes from and why European three-phase grids carry both 230 V and 400 V as rated voltages.
What does this quantity describe?
The real power of a balanced three-phase system follows from line voltage U (between two line conductors), line current I and power factor cos φ: P = √3·U·I·cos φ. The √3 factor accounts for the 120° phase separation between the three line conductors.
A four-wire three-phase supply carries two voltage levels at once: between one line conductor and neutral sits 230 V (phase voltage), while between two line conductors sits 400 V (line voltage). The two are linked by the √3 factor (400 V ≈ 230 V · √3), since the three phase voltages are offset by 120° from each other and so don't simply double but add geometrically.
Formula and variables
P = √3 · U · I · cosφ
P = √3 · U · I · cos φI = P / (√3 · U · cos φ)U = P / (√3 · I · cos φ)
| Symbol / input | Meaning |
|---|---|
| Real power P | The real power actually consumed or delivered by the three-phase load – the power that is actually converted into work or heat, as distinct from apparent power. |
| Line voltage U | The line voltage – the voltage between two of the three phase conductors, usually 400 V in Europe. Not the same as the phase voltage (conductor to neutral, usually 230 V). |
| Line current I | The RMS current in each of the three phase conductors (equal in all three for a balanced load). Source: a current measurement (clamp meter) or the nameplate. |
| Power factor cos φ | The power factor – the ratio of real power used to the total apparent power flowing. 1 means a purely resistive load (heater); motors are often around 0.8–0.9. Source: the load's nameplate, or a power-meter measurement. |
Choose the inputs correctly
Enter three of the four quantities to find the fourth. U is always the line voltage between two line conductors (usually 400 V in Europe), not the phase voltage against neutral (230 V).
How to use the calculator
Select the target quantity and enter the other known values with units. For purely resistive loads set cos φ = 1; for motors and other inductive loads, use the power factor stated on the nameplate.
Worked example
Given U = 400 V, I = 10 A and cos φ = 1, substitution gives P = √3 · 400 V · 10 A · 1 ≈ 6.93 kW.
Understand the result and units
A real power of 6.93 kW shows the actual work-producing power of the balanced three-phase system. At cos φ < 1, apparent power (√3·U·I) would exceed real power, with the difference oscillating as reactive power between source and load without doing usable work.
Line voltage in volts (V), line current in amperes (A), power in watts (W) or kW. Power factor cos φ is dimensionless and ranges between 0 and 1.
Where does the √3 factor in three-phase systems come from?
In a balanced three-phase system, the three phase voltages are offset by 120° from each other. The voltage between two line conductors (line voltage) results geometrically from the vector difference of two phase voltages offset by 120°, not their simple sum. This vector difference produces the √3 factor via the sine of 120°, so line voltage is always √3 times phase voltage (the voltage against neutral). The same factor consequently appears throughout the power formula for balanced three-phase systems.
Typical applications
The formula is used to size three-phase motors, industrial three-phase connections, EV charging stations, and to estimate the power demand of three-phase-supplied installations.
Assumptions, limits and common mistakes
The formula applies to a balanced three-phase system with sinusoidal 50/60 Hz quantities and no harmonics. For unbalanced loads (differing currents per line conductor) or non-sinusoidal currents, the simple formula gives only an approximation; accurate analysis then needs a per-phase treatment or harmonic analysis.
Common mistake: Do not confuse line voltage (400 V, between two line conductors) with phase voltage (230 V, against neutral) — the two differ by a factor of √3 and must not be swapped uncritically. Also, don't forget power factor cos φ: without it, inductive loads yield apparent power instead of actual real power.
Frequently asked questions
What is the difference between line voltage and phase voltage?
Phase voltage (usually 230 V in Europe) is between one line conductor and neutral; line voltage (usually 400 V) is between two line conductors. Line voltage = √3 · phase voltage.
Do I need to account for power factor with purely resistive loads?
For purely resistive loads, cos φ = 1, so the formula gives real power directly. For motors or other inductive loads, cos φ is usually below 1.
Does the formula still apply under unbalanced loading?
Only approximately. With noticeably different currents per line conductor, each phase needs separate treatment.