Three-phase power underpins virtually every commercial and industrial electrical installation. Understanding the relationship between line voltage, phase voltage, and current — and how power factor affects everything — lets you size cables, protection, and transformers correctly. This guide covers the essential formulas and applies them to real scenarios.
Line voltage vs phase voltage
In a three-phase system, there are two voltages to consider:
Phase voltage (Vφ)— the voltage between one line conductor and neutral. In the UK (and Australia/NZ), this is 230 V nominal. In North America it is 120 V (for the 120/208 V system) or 277 V (for the 277/480 V system).
Line voltage (V⊂L)— the voltage between any two line conductors. Line voltage is always phase voltage multiplied by √3 (approximately 1.732):
V⊂L = Vφ × √3
For a UK 230 V phase voltage: V⊂L = 230 × 1.732 = 400 V (rounded to the nearest volt — the nominal UK three-phase voltage).
For a 120 V North American system: V⊂L = 120 × 1.732 = 208 V (the 208 V service common in commercial buildings).
For a 277 V North American system: V⊂L = 277 × 1.732 = 480 V (the 480 V service used for larger motors and HVAC equipment).
Three-phase power formulas
For a balanced three-phase load (where all three phases carry the same current):
- Apparent power (S):
S = √3 × V⊂L × I⊂L— in volt-amperes (VA) - Real power (P):
P = √3 × V⊂L × I⊂L × pf— in watts (W) - Reactive power (Q):
Q = √3 × V⊂L × I⊂L × sin(θ)— in VAR
Where I⊂L is the line current (the current in each phase conductor), pf is the power factor, and θ is the phase angle (cosθ = pf).
Rearranging to find line current:
I⊂L = S ÷ (√3 × V⊂L)
or with power factor:
I⊂L = P ÷ (√3 × V⊂L × pf)
Worked example 1: sizing a motor circuit
A 15 kW three-phase motor runs at 400 V with a power factor of 0.86 and an efficiency of 92%. What is the full-load line current?
- Input power = output power ÷ efficiency = 15,000 ÷ 0.92 = 16,304 W
- I⊂L = P ÷ (√3 × V⊂L × pf) = 16,304 ÷ (1.732 × 400 × 0.86) = 16,304 ÷ 597 = 27.3 A
You would size the cable and protection for at least 27.3 A, then apply the relevant motor circuit protection rules (125% of full-load current for the cable under BS 7671 overload protection; overcurrent protection per motor starting characteristics). A 32 A cable and 32 A type C MCB (or a motor circuit breaker rated for the motor) would be typical.
Worked example 2: three-phase transformer sizing
A commercial unit has a total three-phase load of 45 kW at a power factor of 0.80. What kVA transformer is required?
- S = P ÷ pf = 45,000 ÷ 0.80 = 56,250 VA = 56.25 kVA
Select the next standard transformer size above 56.25 kVA. Common sizes are 50, 63, 75, 100, 125, 160, 200 kVA. The correct choice here is a 63 kVA unit, providing adequate headroom for load growth without gross over-specification.
Unbalanced loads
When a three-phase system carries unbalanced loads — different currents on each phase — the neutral conductor carries the vector sum of the three phase currents, which in a balanced system cancels to zero. In an unbalanced system, the neutral current can be significant and must be accounted for in conductor sizing.
A common cause of severe neutral loading is heavy single-phase electronic loads (computers, variable-speed drives) which draw non-sinusoidal current. The third harmonic (and multiples thereof) from these loads does not cancel at the neutral — it adds. In a building with many computers or switch-mode power supplies, the neutral can carry more current than any individual phase. BS 7671 Regulation 523.6.3 requires oversizing the neutral in such cases.
Delta vs star (wye) connections
Three-phase loads connect in one of two ways:
In a star (wye) connection, one end of each phase winding connects to a common neutral point. The phase voltage across each winding is Vφ; the line voltage between any two line terminals is Vφ × √3. Most distribution systems are star-connected.
In a delta connection, the windings form a loop — there is no neutral point. Each winding sees the full line voltage. Delta connections are common for motors and some transformer configurations. The line current in a delta connection is:
I⊂L(delta) = √3 × I⊂phase
Where I⊂phase is the current flowing through each winding. The line current is larger than the phase current — the opposite of the voltage relationship in a star system.
Power factor and its consequences
Power factor below 1.0 means a load draws more current than its real power demand would suggest. A 10 kW load at 0.7 power factor draws the same current as a 14.3 kW resistive load. That extra current flows through every conductor, transformer, and switchgear — contributing to losses and reducing the capacity available for productive loads.
Power factor correction capacitors are installed in parallel with inductive loads (motors, fluorescent fittings with magnetic ballasts) to supply the reactive power locally rather than drawing it from the supply. A distribution board serving a large motor load with poor power factor may see its current fall significantly after capacitor installation — sometimes enough to defer a service upgrade.
Related calculators
For three-phase cable sizing, use the voltage drop calculator(which applies the √3 factor correctly for three-phase circuits) and the cable sizing calculator. For breaker and OCPD selection on a motor circuit, the breaker size calculator covers NEC motor circuit rules under Article 430.