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Three-Phase Power (Line Values)

In a balanced three-phase system, the total real power delivered to the load equals the square root of three times the line voltage times the line current times the power factor (cosine of the phase angle). This formula is used by engineers to size cables, transformers, and generators. For example, a 400 V, 50 A load at 0.9 power factor draws about 31.2 kW.

Power Systems Three-Phase

Three-Phase Power Calculator

Real power of a balanced three-phase load from line quantities: P = √3 · VL · IL · cos θ.

Inputs

Drag the handle (or focus it and press ↑/↓) to reorder. Scientific notation such as 1.2e3 is accepted. Pick any unit; conversion is automatic. Use line-to-line voltage and line current for a balanced load. Pick any unit; conversion is automatic.

Presets

Results

Enter values and press Calculate. Your answer will appear here.

Real and reactive power vs. phase angle

For your VL and IL the apparent power S is fixed. P = S·cos θ falls and Q = S·sin θ rises as the angle grows. The dot is your operating point; the amber line is S.

What is being calculated

Three-phase supply feeding a load A three-phase source feeds a load through three line conductors L1, L2 and L3. The line current IL flows in each conductor, and the line-to-line voltage VL appears between any two lines. Real power is the square root of 3 times VL times IL times the power factor. 3~ Loadcos θ L1L2L3 IL VL line-to-line P = √3 · VL · IL · cos θ

Common examples

CaseSP

Illustrative inputs, not manufacturer data.

Effect of power factor at 400 V and 50 A

cos θReal power PReactive power Q

The apparent power stays at 34.64 kVA; only the split between real and reactive power changes.

Common nominal line-to-line voltages

VLPhase voltage (VL/√3)Typical use
208 V120 VNorth American commercial
400 V230 VEurope and many other regions
415 V240 VUK, Australia, India (older nominal)
480 V277 VNorth American industrial
11 kV6.35 kVMedium-voltage distribution

Unit conversions

QuantityConversion
Power1 kW = 1,000 W · 1 MW = 1,000 kW · 1 HP (mechanical) = 745.70 W
Voltage1 kV = 1,000 V · 1 V = 1,000 mV
Current1 kA = 1,000 A · 1 A = 1,000 mA
Angle180° = π rad · 1 rad = 57.2958°
Energy1 kWh = 1,000 Wh = 3.6 MJ

Formula

P = √3 · VL · IL · cos θ
  • P: real (active) power (W)
  • VL: line-to-line voltage (V)
  • IL: line current (A)
  • cos θ: power factor, where θ is the angle between voltage and current

Example: VL = 400 V, IL = 50 A, cos θ = 0.85. S = √3 × 400 × 50 = 34,641 VA, so P = 34,641 × 0.85 ≈ 29,445 W ≈ 29.45 kW.

The formula assumes a balanced load, and it is the same for star and delta connections when written in line quantities. In a star system Vphase = VL/√3 and Iphase = IL; in a delta system Vphase = VL and Iphase = IL/√3. Either way P = 3·Vphase·Iphase·cos θ. Advanced mode also gives apparent power S = √3·VL·IL, reactive power Q = S·sin θ, output power with an efficiency, and energy for a number of hours. Use measured values for VL, IL and the power factor; for unbalanced or distorted loads a power meter is more accurate.

References

How to Use This Tool

  1. Pick a preset, or enter the line-to-line voltage VL and its unit.
  2. Enter the line current IL and the power factor cos θ.
  3. In Advanced, enter the phase voltage or the angle θ instead, and add efficiency or operating hours.
  4. Click Calculate.
  5. Read P in the highlighted result, then check S, Q and the power chart.

What This Tool Offers

  • Basic and Advanced modes on one calculation engine
  • Voltage, current, angle and power (W, kW, MW, HP) unit conversion
  • Apparent, reactive and real power with phase angle
  • Phase quantities, motor output, losses and energy (Advanced)
  • P-and-Q-versus-angle chart and labelled three-phase diagram
  • Reorderable inputs, presets and a mobile-friendly layout

Variables

Symbol Quantity Unit
P Real power (active power) W (or kW)
V_L Line-to-line RMS voltage V
I_L Line RMS current A
cos θ Power factor (pf) dimensionless
θ Phase angle between phase voltage and phase current °
√3 Square root of 3 (1.732) dimensionless

What it means

For a balanced three‑phase system, the total real power is given by P = √3 * V_L * I_L * cosθ, where V_L is the line‑to‑line voltage, I_L is the line current, and cosθ is the power factor. This formula applies to both star (wye) and delta connections when line quantities are used. The √3 factor comes from the relationship between line and phase voltages (V_L = √3 * V_ph for star) and currents (I_L = √3 * I_ph for delta). Three‑phase power is the standard for industrial and large‑scale power distribution because it provides constant power and higher efficiency. This formula is essential for calculating the power in motors, generators, and transformers. It is also used for sizing cables, switchgear, and protective devices. Understanding three‑phase power is crucial for electrical engineers working in power systems. Example: A three‑phase motor with V_L=400V, I_L=10A, and power factor 0.8 draws P = √3 * 400 * 10 * 0.8 = 1.732*400*10*0.8 = 5542 W (about 5.54 kW).

Worked example

Three‑Phase Power – Practical Example

Real‑World
Scenario: A 400 V three‑phase motor draws 50 A with a power factor of 0.85. Calculate the real power.
ParameterValue
VL400 V
IL50 A
cos θ0.85
FormulaP = √3 · VL · IL · cos θ
1Substitute:P = 1.732 × 400 × 50 × 0.85
2Compute: 1.732 × 400 = 692.8; × 50 = 34640; × 0.85 = 29444 W
3Result:P ≈ 29.4 kW
Final Design P ≈ 29.4 kW ✓ Real power
Why: Three‑phase power is √3 times the product of line voltage, line current, and power factor – standard for industrial loads.

Common mistakes

  • V_L and I_L: Line‑to‑line voltage and line current – not phase quantities.
  • Power factor cosθ: The angle between phase voltage and phase current – same as between line quantities.
  • Balanced load: This formula assumes a balanced three‑phase system.
  • Phase voltage: If you have phase voltage, use P = 3·V_ph·I_ph·cosθ.
  • Units: V in volts, I in amperes → P in watts.

Applications

Three‑phase power using line values is P = √3 · V_L · I_L · cosθ, applicable to balanced star and delta systems. This formula is fundamental for industrial power systems, where three‑phase motors and loads are common. Engineers use it to size generators, transformers, and cables, to calculate the total power consumption of factories, and to assess power factor. By using line values, they avoid the need to know whether the load is star or delta. This formula also underpins the design of three‑phase distribution networks and the analysis of power factor correction. Understanding three‑phase power is essential for anyone working with large electrical loads.

  • Power calculation in industrial and commercial installations
  • Sizing of three‑phase transformers and generators
  • Power factor correction in three‑phase systems
  • Motor and load power estimation
  • Educational understanding of three‑phase systems