Three-phase alternating current (AC) is the global standard for distributing electrical power in factories, commercial buildings, data centres and utility grids. Unlike single-phase circuits, a three-phase system delivers power through three conductors whose voltages are displaced by 120°, which produces a constant, non-pulsating power flow and allows motors to self-start. Understanding the relationship between voltage, current and power — real, apparent and reactive — is essential for engineers sizing cables, specifying transformers, selecting circuit breakers and assessing energy efficiency.
This calculator handles all three variants of the core three-phase problem: computing power (S, P, Q) from voltage and current; computing line current from voltage and real power; and computing the required line voltage from current and real power. It covers balanced loads in both wye (star) and delta configurations and derives every phase quantity alongside the line quantities.
How it works
All three-phase power relationships for a balanced load follow from a single core identity. Given line-to-line voltage V_LL and line current I_L:
- Apparent power: S (VA) = √3 × V_LL × I_L
- Real power: P (W) = S × cos(θ) = √3 × V_LL × I_L × pf
- Reactive power: Q (VAR) = S × sin(θ) = √3 × V_LL × I_L × sin(arccos(pf))
The power factor pf = cos(θ) sits between 0 (purely reactive) and 1 (purely resistive). The three quantities form a right-angled power triangle: P on the horizontal leg, Q on the vertical leg and S as the hypotenuse, so S² = P² + Q².
Phase quantities depend on the winding topology:
| Quantity | Wye (star) | Delta |
|---|---|---|
| Phase voltage V_ph | V_LL ÷ √3 | V_LL |
| Phase current I_ph | I_L | I_L ÷ √3 |
For standard EU/UK supplies: 400 V line-to-line gives a 230.9 V phase voltage in a wye system — exactly the single-phase 230 V that domestic sockets use.
Worked example
A 15 kW three-phase induction motor on a 400 V supply runs at pf 0.85.
- Line current: I_L = P ÷ (√3 × V_LL × pf) = 15,000 ÷ (1.7321 × 400 × 0.85) = 25.5 A
- Apparent power: S = √3 × 400 × 25.5 = 17,647 VA ≈ 17.65 kVA
- Reactive power: Q = √(S² - P²) = √(17,647² - 15,000²) ≈ 9,260 VAR ≈ 9.26 kVAR
- Phase angle: θ = arccos(0.85) ≈ 31.8°
- Phase voltage (wye): V_ph = 400 ÷ √3 ≈ 231 V
To reduce reactive power demand you would add power factor correction capacitors until the kVAR drops and kVA approaches kW.
Formula note
The √3 multiplier arises because the line-to-line voltage spans two phases displaced by 120°. By the law of cosines the resultant is exactly √3 times the phase magnitude. This is a geometric fact, not an approximation. The calculator uses the full double-precision value √3 = 1.7320508… for all arithmetic.
The reactive power formula Q = S × sin(θ) is derived purely from the power triangle identity S² = P² + Q², which itself follows from the phasor representation of voltage and current. No additional measurement is required — the angle θ = arccos(pf) is computed analytically from the power factor you supply.