Calculate three-phase power, current, and voltage for balanced and unbalanced loads. Handles star and delta configurations with full power triangle analysis.
Calculate power values for three-phase electrical systems including apparent, active, and reactive power
0 to 1
For PF correction calculation
Three-phase power is a method of alternating current (AC) electrical power generation, transmission, and distribution that uses three conductors, each carrying an alternating current of the same frequency and amplitude but displaced from one another by 120 degrees (one-third of a cycle). This phase separation is what gives three-phase power its key advantages over single-phase: higher power density, constant power delivery (no zero-crossing moments), and more efficient use of conductors.
In the UK, the standard three-phase supply provides 400 V between any two line conductors and 230 V between any line conductor and neutral. This is because the line voltage equals the phase voltage multiplied by the square root of 3 (approximately 1.732): 230 V x 1.732 = 398.4 V, rounded to 400 V. The nominal voltage tolerance is +10% / -6%, giving an actual range of 216.2 V to 253 V phase-to-neutral and 376 V to 440 V line-to-line.
Every electrician working on commercial and industrial installations needs to understand three-phase power calculations. Whether sizing cables for a three-phase sub-main, calculating the load on a three-phase distribution board, determining the current draw of a motor, or assessing voltage drop on a long cable run, the three-phase formulas are essential. The key difference from single-phase is the presence of the square root of 3 factor in the power and voltage/current relationships. Use the cable sizing calculator to size three-phase cables once the line current is known, and the voltage drop calculator to verify compliance with BS 7671 limits on three-phase distribution circuits.
Three-phase loads and generators can be connected in two fundamental configurations: star (also called wye, symbol Y) and delta (symbol triangle). The choice of configuration determines the relationship between line and phase voltages and currents.
The fundamental power formula for a balanced three-phase load is:
P = root 3 x VL x IL x cos phi
P = total three-phase real power in watts
VL = line-to-line voltage (400 V in UK)
IL = line current in amperes
cos phi = power factor (0 to 1)
To find the line current when you know the power:
IL = P / (root 3 x VL x cos phi)
For apparent power (kVA) and reactive power (kVAr), the power triangle relationships apply. Apparent power S = root 3 x VL x IL (without the power factor). Reactive power Q = S x sin phi. Real power P = S x cos phi. These three quantities form a right-angled triangle where S is the hypotenuse, P is the adjacent side, and Q is the opposite side.
A three-phase induction motor has a rated output of 15 kW with an efficiency of 90% and a power factor of 0.85. Calculate the line current drawn from a 400 V supply.
Input power = 15,000 / 0.9 = 16,667 W
IL = 16,667 / (1.732 x 400 x 0.85) = 16,667 / 588.9 = 28.3 A
The motor draws 28.3 A per phase from the supply. This determines the cable size, protective device rating, and contactor size needed.
A three-phase distribution board in a small office has the following loads: L1 = 35 A, L2 = 28 A, L3 = 42 A. All loads are resistive (power factor 1.0). Calculate the total power and assess the phase balance.
PL1 = 230 x 35 x 1.0 = 8,050 W
PL2 = 230 x 28 x 1.0 = 6,440 W
PL3 = 230 x 42 x 1.0 = 9,660 W
Total = 8,050 + 6,440 + 9,660 = 24,150 W (24.15 kW)
The imbalance between L2 (28 A) and L3 (42 A) is 14 A. The sub-main cable must be sized for the highest phase current (42 A) plus the neutral current arising from the imbalance. Good practice is to redistribute circuits to achieve better balance.
A 22 kW three-phase EV charger operates at 400 V with a power factor of 0.99. Calculate the line current and determine the cable and protective device size.
IL = 22,000 / (1.732 x 400 x 0.99) = 22,000 / 685.9 = 32.1 A
A 32 A Type C MCB and 6 mm² 5-core SWA cable would be appropriate (subject to voltage drop and derating calculations). The high power factor of 0.99 means the current draw is close to the minimum possible for this power level.
Purpose-built for UK electricians working on commercial and industrial three-phase installations.
Switch between balanced three-phase calculations (single formula) and per-phase unbalanced calculations.
Automatically handles the voltage and current relationships for both star (wye) and delta connected loads.
Shows real power (kW), reactive power (kVAr), and apparent power (kVA) with power factor. Understand the full power picture for any three-phase…
For unbalanced loads, shows the current on each phase and the neutral current. Highlights imbalances that could cause problems with voltage regulation or…
Calculate motor starting current (typically 6-8 times full load current) to check voltage drop and protective device suitability during direct-on-line…
All calculations follow the current 18th Edition wiring regulations including Amendment 4. Three-phase voltage drop uses the correct three-phase mV/A/m…
Balanced and unbalanced loads, star and delta configurations, motor starting currents — all calculated instantly on your phone.
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In a star (wye) connected three-phase system, the line voltage (measured between any two phases) is equal to the phase voltage (measured between a phase and the star point/neutral) multiplied by the square root of 3 (approximately 1.732). In the UK, the standard phase voltage is 230 V, giving a line voltage of 230 x 1.732 = 400 V. In a delta connected system, the line voltage equals the phase voltage, but the line current is the phase current multiplied by the square root of 3.
For a balanced three-phase load where all three phases carry equal current at the same power factor, use the formula: P = square root of 3 x VL x IL x cos phi, where VL is the line voltage (400 V in the UK), IL is the line current in amps, and cos phi is the power factor. For example, a balanced load drawing 50 A per phase at a power factor of 0.85 gives: P = 1.732 x 400 x 50 x 0.85 = 29,444 W or approximately 29.4 kW.
In a star (wye) connection, one end of each winding is connected to a common star point (which becomes the neutral). The line voltage is root 3 times the phase voltage, and the line current equals the phase current. In a delta connection, the windings form a closed triangle with no neutral point. The line voltage equals the phase voltage, but the line current is root 3 times the phase current. Star connection gives access to two voltages (230 V and 400 V in the UK) and is used for distribution. Delta connection is commonly used for motor windings and transformer secondaries.
Three-phase supplies are standard in commercial and industrial premises where the total load exceeds the capacity of a single-phase supply (typically above 15-20 kW). Common three-phase applications include: commercial distribution boards and sub-mains, three-phase motor supplies (lifts, air conditioning, industrial machinery), large EV charging installations (22 kW or 50 kW+ chargers), commercial solar PV inverters above 3.6 kW, electric heating systems in larger buildings, and any installation where load balancing across phases is required. Many newer domestic properties with heat pumps and EV chargers are also being connected to three-phase supplies.
When a three-phase load is unbalanced (different current or power on each phase), you cannot use the single balanced formula. Instead, calculate the power on each phase separately using P = VP x IP x cos phi, where VP is the phase voltage (230 V) and IP is the current on that specific phase. The total power is the sum of all three phases. The neutral current in an unbalanced star-connected system is the vector sum of the three phase currents, which can be significant in heavily unbalanced installations. This is why BS 7671 requires the neutral conductor to be sized appropriately in three-phase systems.
The power factor depends on the type of load. Resistive loads (heaters, kettles) have a power factor of 1.0. Induction motors typically have a power factor of 0.8 to 0.9 at full load, dropping to 0.3 to 0.5 at light load. LED lighting with power factor correction is typically 0.95 or above. Fluorescent lighting with magnetic ballasts is around 0.5 without correction. If the power factor is unknown, a value of 0.8 is commonly used as a conservative estimate for mixed commercial loads. Power factor correction equipment can improve the overall power factor to 0.95 or above, reducing the current drawn from the supply.
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