The complete guide to three-phase electrical calculations. Power formula (P = root 3 x VL x IL x cos phi), line vs phase values, star vs delta configurations, phase balancing, three-phase voltage drop, and cable sizing — with worked examples for motors, distribution boards, and commercial loads.
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Written and reviewed by Andrew Moore, founder of Elec-Mate, against BS 7671:2018+A4:2026, IET Guidance Note 3 and the IET On-Site Guide.
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A three-phase electrical supply consists of three alternating current conductors (L1, L2, L3), each carrying a sinusoidal voltage that is offset by 120 electrical degrees from the other two. This arrangement has significant advantages over a single-phase supply: it can deliver more power for the same conductor size, it can power three-phase motors directly (which are simpler, more efficient, and more reliable than single-phase motors), and in a balanced system the power delivery is constant (whereas single-phase power pulsates at twice the supply frequency).
In the UK, the standard three-phase supply provides a line voltage of 400V (between any two phases) and a phase voltage of 230V (between any phase and neutral). Most domestic properties have a single-phase supply (one phase and neutral from the three-phase distribution network), but larger domestic properties, commercial premises, and industrial installations typically have a full three-phase supply.
Three-phase calculations are essential for cable sizing, protective device selection, and voltage drop verification on commercial and industrial installations. They are also needed for prospective fault current calculations on three-phase supplies.
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The fundamental three-phase power formula for a balanced load is:
P = √3 × VL × IL × cosφ
P = total three-phase power (W) | VL = line voltage (400V) | IL = line current (A) | cosφ = power factor
The root 3 factor (approximately 1.732) appears because of the 120-degree phase relationship between the three phases. It is a mathematical constant that relates the line values (measured between phases) to the individual phase contributions.
Real power: P = √3 × VL × IL × cosφ (watts)
Apparent power: S = √3 × VL × IL (volt-amperes)
Reactive power: Q = √3 × VL × IL × sinφ (VAr)
Line current: IL = P ÷ (√3 × VL × cosφ)
The line current formula — rearranged from the power formula — is the most commonly used form in practice. You know the load power (from the equipment nameplate, circuit design, or diversity calculation), the supply voltage (400V), and the power factor (from the equipment data or assumed), and you need to calculate the current for cable sizing and device selection.
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One of the most common sources of confusion in three-phase calculations is the difference between line values and phase values. Getting these mixed up results in calculations that are wrong by a factor of root 3 (1.732), leading to cables that are massively oversized or dangerously undersized.
Line voltage (VL) is measured between any two phase conductors: L1 to L2, L2 to L3, or L3 to L1. In the UK, VL = 400V. Line current (IL) is the current flowing in each phase conductor of the supply cable. This is the current you measure with a clamp meter on any one of the three phase conductors. Line current is the value used for cable sizing and protective device selection.
Phase voltage (Vp) is measured between any phase conductor and the neutral: L1 to N, L2 to N, or L3 to N. In the UK, Vp = 230V. Phase current (Ip) is the current flowing through each individual load element or winding. In a star configuration, the phase current equals the line current (Ip = IL). In a delta configuration, the phase current is the line current divided by root 3 (Ip = IL / 1.732).
Star and delta are the two fundamental ways of connecting three-phase loads or transformer windings. Understanding the difference is essential for motor installations, transformer connections, and three-phase load analysis.
In a star configuration, one end of each of the three load elements (or windings) is connected together at a central "star point" (also called the neutral point). The other end of each element is connected to one of the three phase conductors.
In a delta configuration, the three load elements are connected end-to-end in a closed triangle. Each element is connected between two phase conductors. There is no neutral point.
The star-delta starter is a common motor starting method. The motor windings are first connected in star (reducing the voltage across each winding to 230V and the starting current to approximately one-third of the delta starting current), and then switched to delta for normal running (full 400V across each winding). This reduces the mechanical stress on the drive train and the electrical stress on the supply during starting.
Elec-Mate's star-delta calculator computes line and phase currents for both configurations, calculates the current reduction from star starting…
Try it free for 7 daysThe most common three-phase calculation in practice is determining the line current from the total load power. This line current is the design current (Ib) used for cable sizing and protective device selection.
IL = P ÷ (√3 × VL × cosφ)
Based on VL = 400V, balanced three-phase load
A useful rule of thumb for three-phase at 400V with unity power factor: approximately 1.44 amperes per kilowatt. At power factor 0.8 (typical for motors), it is approximately 1.80 amperes per kilowatt. These quick estimates are useful for on-site sanity checks but should not replace a proper calculation for cable sizing.
Power factor (cos phi) is the ratio of real power (watts) to apparent power (volt-amperes). A power factor of 1.0 (unity) means all the current drawn from the supply is doing useful work. A power factor less than 1.0 means additional current is flowing to supply the reactive power demand, without contributing to useful work.
Resistive loads (heaters, kettles): cos phi = 1.0 (unity)
Fluorescent lighting: cos phi = 0.5 to 0.9 (varies with ballast type)
LED lighting (with driver): cos phi = 0.9 to 0.95
Induction motors (full load): cos phi = 0.8 to 0.9
Induction motors (light load): cos phi = 0.3 to 0.5
Welding equipment: cos phi = 0.4 to 0.6
Computer loads: cos phi = 0.65 to 0.9 (depends on PSU type)
Power factor matters in three-phase calculations because a low power factor increases the line current for the same real power output. This means larger cables, larger protective devices, and higher electricity costs (commercial consumers are often penalised for poor power factor by the DNO). Power factor correction — typically using capacitor banks — reduces the reactive current and brings the power factor closer to unity.
For cable sizing purposes, the design current must be calculated using the actual power factor of the load. Using cos phi = 1.0 when the actual power factor is 0.8 will give a design current that is 20% too low, potentially resulting in an undersized cable.
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Start free trialWhen a three-phase supply feeds a mixture of single-phase and three-phase loads, the single-phase loads must be distributed across the three phases as evenly as possible. Perfect balance is rarely achievable because loads vary throughout the day, but the design should aim for the best possible balance at the expected peak demand.
In a perfectly balanced three-phase system, the neutral current is zero because the three phase currents cancel each other out. As the imbalance increases, the neutral current increases. The neutral current can be calculated from the three phase currents using vector addition, but for practical purposes on a distribution board with predominantly resistive loads, a reasonable approximation is that the neutral current does not exceed the current of the most heavily loaded phase.
The BS 7671 does not specify a maximum permissible imbalance, but good practice aims for the difference between the most and least loaded phases to be no more than 10-15% of the total load per phase.
In commercial three-phase installations with modern electronic loads — variable-frequency drives (VFDs), switch-mode power supplies, and LED drivers with poor power factor — triplen (third-order and multiple) harmonics add in the neutral conductor rather than cancelling. This means the neutral can carry a current that exceeds the line current even in an otherwise balanced system.
Reg 523.6.1 treats the neutral as loaded when the total harmonic distortion (THD) of the line current exceeds 15%, and requires the neutral to be included in the conductor count when selecting cable current-carrying capacity from the Appendix 4 tables (which reduces the rated capacity). Appendix 4 Section 5.5 rating factors account for the additional thermal effect of third-harmonic current in both the neutral and the line conductors.
Reg 524.2.1(b) requires the neutral conductor to be not less than the line conductor cross-sectional area in polyphase circuits where the line conductors are 16 mm² or smaller (copper). For larger conductors, the neutral may be reduced only where harmonic loading and unbalanced loading are assessed as not causing the neutral current to exceed the line current.
Elec-Mate's EIC circuit schedule lets you assign each circuit to a phase (L1, L2, L3) and automatically calculates the load per phase…
Try it free for 7 daysVoltage drop for three-phase circuits is calculated using the same formula as single-phase but with the three-phase mV/A/m values from the Appendix 4 (Section 6.4) tables:
VD = mV/A/m (three-phase) × Ib × L ÷ 1000
VD = voltage drop (volts) | mV/A/m = three-phase value | Ib = line current (A) | L = cable length (m)
The three-phase mV/A/m values in the tables are different from (and lower than) the single-phase values for the same cable. This is because the three-phase voltage drop formula inherently accounts for the phase relationships in a balanced three-phase system. You must always use the three-phase column when calculating voltage drop for a three-phase circuit — using the single-phase values would give an incorrect result.
Lighting circuits
12V
3% of 400V
Other circuits
20V
5% of 400V
On long three-phase cable runs — such as submain feeds to remote distribution boards in large commercial buildings or external supplies to outbuildings — voltage drop is often the governing factor in cable selection, requiring a larger cable than would be needed for current-carrying capacity alone.
Cable sizing for three-phase circuits follows the same six-step process as single-phase, but with three-phase-specific values at each step.
Calculate using the three-phase formula: Ib = P / (√3 × VL × cosφ). This gives the current per line conductor.
Select a three-phase MCB, MCCB, or fuse with In greater than or equal to Ib. Three-phase devices have linked poles that disconnect all three phases simultaneously.
Apply Ca, Cg, Ci, and Cf exactly as for single-phase. It = In / (Ca × Cg × Ci × Cf).
Use the three-phase (three loaded conductors) column of the appropriate Appendix 4 table. The three-conductor column gives lower capacity than the two-conductor column because three current-carrying conductors generate more heat than two.
A4:2026 note — buried cables: Tables 4A2, 4D4A, 4E4A, 4H4A, and 4J4A in Appendix 4 were revised in A4:2026 to reflect updated methods for buried cable installation. If the three-phase cable run is buried (direct in soil or in ducts), you must use the revised A4:2026 table values — pre-A4 values are no longer valid for these installation methods (Reg 653.2).
Use the three-phase mV/A/m values from Appendix 4 (Section 6.4). Check against 12V (lighting) or 20V (other) limits for a 400V supply (Reg 525.202).
Apply the adiabatic equation (k²S² ≥ I²t) using the three-phase prospective fault current at the point of installation.
PVC twin and earth (T+E) 70°C — single-phase: Table 4D1A. Standard flat grey cable for domestic and single-phase commercial circuits.
XLPE twin and earth (T+E) 90°C — single-phase: Table 4D5A. Higher-rated T+E cable; still single-phase (two loaded conductors), 230V. Do not use for three-phase motor or heating circuits.
3-core XLPE SWA 90°C — three-phase (400V): Table 4D4A. The correct table for three-phase submains, motor feeds, and heating circuits wired in multicore SWA. Revised in A4:2026 for buried installation methods.
Using the wrong table — for example T+E single-phase values for a three-phase SWA circuit — will give an incorrect Iz and mV/A/m, leading to an undersized or wrongly-assessed cable.
Elec-Mate's cable sizing calculator has a three-phase mode. Enter the total load, power factor, and cable route details.
Try it free for 7 daysAn 11kW three-phase induction motor with a power factor of 0.85 at full load. 400V supply, cable run of 25 metres, clipped direct (Method C), ambient 30 degrees Celsius, no grouping.
Line current: IL = 11,000 ÷ (1.732 × 400 × 0.85) = 11,000 ÷ 588.9 = 18.7A
Protective device: 20A Type C MCB (Type C for motor inrush)
Correction factors: Ca = 1.0 | Cg = 1.0 | Ci = 1.0 | Cf = 1.0
Required It: 20 ÷ 1.0 = 20A
From Table 4D4A (3-core XLPE SWA 90°C, Method C, 400V three-phase): 2.5mm² has Iz = 27A
Voltage drop: mV/A/m (3-phase) for 2.5mm² = 16 mV/A/m
VD = 16 × 18.7 × 25 ÷ 1000 = 7.5V (1.9% of 400V — within 5% limit)
A submain feed to a three-phase distribution board with a total demand of 60kW after diversity. Power factor 0.9, cable run of 40 metres in trunking (Method B), 4 circuits grouped in the trunking, ambient 30 degrees Celsius.
Line current: IL = 60,000 ÷ (1.732 × 400 × 0.9) = 60,000 ÷ 623.5 = 96.2A
Protective device: 100A TP MCCB
Correction factors: Ca = 1.0 | Cg = 0.65 (4 circuits) | Ci = 1.0 | Cf = 1.0
Required It: 100 ÷ 0.65 = 153.8A
Cable selection: 70mm² multicore SWA required for this current rating
Voltage drop check: Must verify using three-phase mV/A/m values for the selected cable
A 24kW three-phase heating element (resistive load, cos phi = 1.0). 400V supply, cable run 15 metres, clipped direct.
Line current: IL = 24,000 ÷ (1.732 × 400 × 1.0) = 24,000 ÷ 692.8 = 34.6A
Protective device: 40A Type B MCB (three-pole)
No derating factors apply. It = 40A
Cable: 6mm² four-core SWA (Iz = 47A for 3 loaded conductors, Method C — Table 4D4A, 3-core XLPE SWA 90°C, 400V three-phase)
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