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Physics

Three-Phase Power Calculator

Enter the line voltage, line current, and power factor for a balanced three-phase AC system to calculate apparent power (S in kVA), real power (P in kW), and reactive power (Q in kVAR). Choose Wye or Delta connection to see the related phase quantities. Results update as you type.

Your details

Wye (star) connects each load to a common neutral point; Delta connects loads in a closed triangle. The relationship between line and phase quantities differs between them.
Choose which quantity to calculate. The remaining inputs change accordingly.
The voltage measured between any two line conductors (line-to-line). Common values: 208 V (US low-voltage), 400 V (European standard), 480 V (US industrial), 11 kV / 33 kV (medium-voltage distribution).
V
Current flowing in each supply conductor. In a Wye system this equals the phase current; in Delta it is the phase current multiplied by the square root of 3.
A
The cosine of the phase angle between voltage and current. Unity (1.0) means the load is purely resistive; motors and transformers typically have a lagging power factor of 0.7 to 0.95.
Apparent power (S)Good power factor
69.282kVA

Total power drawn from the supply, including both useful and reactive components

Real power (P)58.89kW
Reactive power (Q)36.497kVAR
Phase angle (φ)31.79°
Phase voltage (Vₚʰ)230.94V
Phase current (Iₚʰ)100A
Line current (Iₗ)-
Line voltage (Vₗₗ)-
S (kVA)69.282
P (kW)58.89
Q (kVAR)36.497
034.6469.28111
Power Factor
  • Real power P (kW)
  • Reactive power Q (kVAR)

69.28 kVA apparent power in a balanced Wye (Star) system.

  • Your Wye (Star) circuit delivers 58.89 kW of real power out of 69.28 kVA of apparent power.
  • 36.50 kVAR of reactive power is circulating in the system, which stresses cables and transformers without doing useful work.
  • A power factor of 0.85 is good but has room for improvement. Capacitor banks can push it toward unity and cut losses.
  • The system is transferring 85.0% of apparent power as useful work. The remaining 15.0% is reactive overhead.

Next stepIf your power factor is below 0.9, consult an electrical engineer about capacitor bank sizing to reduce reactive demand charges.

Formula

S=3VLLILP=ScosφQ=SsinφWye: Vph=VLL3,  Iph=ILDelta: Vph=VLL,  Iph=IL3S = \sqrt{3}\,V_{LL}\,I_L \quad P = S\cos\varphi \quad Q = S\sin\varphi \quad \text{Wye: }V_{ph}=\frac{V_{LL}}{\sqrt{3}},\;I_{ph}=I_L \quad \text{Delta: }V_{ph}=V_{LL},\;I_{ph}=\frac{I_L}{\sqrt{3}}

Worked example

A 400 V, 100 A Wye-connected motor load with power factor 0.85: S = sqrt(3) x 400 x 100 / 1000 = 69.28 kVA. Real power P = 69.28 x 0.85 = 58.89 kW. Reactive power Q = 69.28 x sin(arccos(0.85)) = 69.28 x 0.527 = 36.52 kVAR. Phase voltage = 400 / sqrt(3) = 231 V. Phase current equals line current = 100 A.

What is three-phase power?

Three-phase power is the standard way electricity is generated, transmitted, and distributed for industrial and commercial use. A three-phase generator produces three sinusoidal voltages, each separated by 120 degrees in time. Together they deliver power more smoothly than single-phase, with no instantaneous zero-crossings of total power, which makes three-phase motors run more evenly and efficiently. The three conductors carrying these voltages are called line conductors, and in many systems a fourth neutral conductor is added to carry any unbalanced current back to the source.

Wye versus Delta connections

The two ways to connect a three-phase load are Wye (star) and Delta (triangle). In a Wye connection, one end of each impedance connects to a common neutral point and the other end connects to a line conductor. This means the voltage across each impedance (phase voltage) is the line-to-line voltage divided by the square root of 3, and the line current equals the phase current. In a Delta connection, the impedances form a closed triangle with no neutral point, so phase voltage equals line voltage, but line current is the phase current multiplied by the square root of 3. Both connections carry the same total power for the same line quantities - only the internal voltage and current distributions differ.

Apparent, real, and reactive power explained

Apparent power (S, in volt-amperes or kVA) is the product of the rms voltage and rms current and represents the total electrical loading on the supply. Real power (P, in watts or kW) is the portion that does useful work: turning motors, creating heat, or running lights. Reactive power (Q, in var or kVAR) is power that cycles back and forth between the supply and inductive or capacitive loads without doing useful work, but it must still be supplied by generators and transmitted through cables. The three form a right triangle: S squared equals P squared plus Q squared. The ratio P/S is the power factor, and maximising it reduces the reactive component, improving efficiency and reducing cable and transformer loading.

Power factor and why it matters

Power factor (cos phi) is the cosine of the angle between the voltage and current waveforms. A purely resistive load like a heater has a power factor of 1.0: all apparent power becomes real power. An inductive load such as an induction motor has a lagging power factor less than 1 because the magnetic field requires reactive energy. Most utilities penalise industrial customers whose power factor falls below about 0.9 to 0.95, because poor power factor forces them to build extra generation and distribution capacity to supply reactive current that does no useful work. Capacitor banks are connected in parallel with inductive loads to supply reactive power locally, raising the overall power factor seen by the utility.

Common three-phase system voltages worldwide

RegionV_LL (line-to-line)V_LN (line-to-neutral)Typical use
North America (low)208 V120 VCommercial buildings, light industrial
North America (industrial)480 V277 VHeavy motors, HVAC, industrial plant
Europe / International400 V230 VAll general-purpose industrial use
UK (HV distribution)11 kV6.35 kVPrimary distribution feeders
Medium voltage33 kV19.05 kVRegional sub-transmission
High voltage132 kV76.2 kVBulk transmission

Standard line-to-line (V_LL) voltages used in industrial and commercial power distribution.

Frequently asked questions

What is the difference between line voltage and phase voltage in a Wye system?

In a Wye (star) system, the phase voltage is the voltage measured from one line conductor to the neutral point. The line voltage is measured between any two line conductors. The relationship is V_LL = sqrt(3) x V_LN, so for a standard 400 V supply the phase voltage is 400 / 1.732 = 231 V. This is why European homes get 230 V single-phase from a 400 V three-phase supply.

How does line current relate to phase current in a Delta connection?

In a Delta connection there is no neutral wire. Each branch of the delta carries phase current, and at each corner of the triangle the currents from two branches combine into the line current. Because of the 120-degree phase shift between them, the line current is sqrt(3) times the phase current, not twice it.

Why is the formula sqrt(3) x V_LL x I_L used for three-phase power?

A three-phase system has three conductors each contributing power. You might expect total power to be 3 x V_phase x I_phase. Substituting the Wye relationships (V_phase = V_LL / sqrt(3) and I_phase = I_L) gives 3 x (V_LL / sqrt(3)) x I_L = sqrt(3) x V_LL x I_L. The same result comes from the Delta substitution, so the formula holds for both connection types as long as you use line quantities.

What is a balanced three-phase circuit?

A balanced circuit has identical impedances on all three phases. This means the three phase currents are equal in magnitude and separated by exactly 120 degrees, which causes the neutral current to be zero in a Wye system. This calculator assumes balanced loading. Real systems often have some imbalance from single-phase loads tapped off different phases, which requires a more complex analysis.

How do I improve a poor power factor?

The most common approach is to install capacitor banks in parallel with the inductive load. Capacitors supply the reactive current that inductive loads demand, so the reactive power measured at the supply point falls. The required capacitor size in kVAR equals Q_current minus Q_target, where Q_target = S x sin(arccos(target_PF)). Variable-speed drives and synchronous condensers are other options for larger installations.

What kVA rating should a transformer be for a three-phase motor?

Divide the motor's rated output in kW by its efficiency and power factor to find the apparent power demand in kVA. For example, a 50 kW motor with 92% efficiency and 0.88 power factor draws 50 / (0.92 x 0.88) = 61.7 kVA. Add headroom for starting currents and any other loads sharing the transformer, and round up to the next standard kVA rating.

Sources

Written by Dr. Tomás Okafor, PhD Physicist · Lagos, Nigeria

Physicist specializing in classical mechanics, bringing 17 years of research and applied dynamics expertise to every calculator he reviews.

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