Three-Phase Power Calculator
Three-Phase System & Phasors
Three line conductors feed a balanced load, connected in star or delta (set in Advanced). On the right, the three phase voltages are 120° apart, the line voltage is √3 times the phase voltage, and the current is shifted from its voltage by the angle θ. The drawing updates with your inputs.
Quick Output: Real-World Examples
Typical balanced three-phase loads, calculated live with P = √3 · VL · IL · cos θ. The currents and power factors are illustrative planning values, not measurements. Press Use to load a row.
| Load | VL (V) | IL (A) | cos θ | P (kW) | S (kVA) | Q (kvar) |
|---|
How to Use
1. Pick the unknown
Solve for the real power, the line voltage, the line current or the power factor.
2. Enter the other three
Use line-to-line volts (V or kV) and line amps (A, kA or mA). Power factor is a number from 0 to 1.
3. Calculate
The page jumps to the result, with kVA, kvar, working and a power versus power-factor chart.
4. Go Advanced
Choose star or delta, find motor output, cable voltage drop, and monthly energy cost.
Frequently Asked Questions
The line voltage between two conductors is √3 times the voltage across one phase, because the phase voltages are 120° apart. Total power is 3 · Vphase · Iphase · cos θ, which equals √3 · VL · IL · cos θ for either star or delta.
Use the line-to-line voltage (such as 400 V) and the current in a single line conductor. If you only know phase values, use Advanced mode and the star or delta relations.
It is the power factor, the cosine of the angle between a phase voltage and its current. It tells you what share of the apparent power √3 VL IL is real power doing useful work.
Not exactly. The √3 formula assumes equal currents and power factors in all three phases. For an unbalanced load, add the power of each phase separately.
Calculating...
Variables
| Symbol | Quantity | Unit |
|---|---|---|
| P | Real power (active power) | kW |
| V_L | Line-to-line RMS voltage | V |
| I_L | Line RMS current | A |
| cos θ | Power factor | dimensionless |
| θ | Phase angle between phase voltage and phase current | ° |
| √3 | Square root of 3 (1.732) — constant for balanced three-phase systems | dimensionless |
What it means
The real power in a balanced three‑phase circuit is P(kW) = (√3 × V_L × I_L × cosθ) / 1000, where V_L is the line‑to‑line voltage, I_L is the line current, and cosθ is the power factor. The √3 factor arises from the relationship between line and phase quantities. This formula is essential for calculating power in industrial motors, generators, and transformers. Example: A three‑phase motor with V_L=400V, I_L=20A, and power factor 0.85 consumes P = √3 × 400 × 20 × 0.85 / 1000 = 1.732 × 400 × 20 × 0.85 / 1000 = 11776 / 1000 ≈ 11.78 kW. This is the actual power drawn by the motor, which determines the electricity cost and required supply capacity.
Worked example
Three‑Phase Power – Practical Example
Real‑World| Parameter | Value |
|---|---|
| VL | 400 V |
| IL | 50 A |
| cos θ | 0.86 |
| Formula | P(kW) = (√3 × VL × IL × cos θ) / 1000 |
Common mistakes
Watch for unit mismatches (W vs kW, single- vs three-phase) and remember to include power factor or efficiency where the formula requires it.Applications
Power in kW (three‑phase): P(kW) = (√3 × V_L × I_L × cosθ)/1000 calculates real power in balanced three‑phase systems. This is used for industrial motors, distribution, and large loads. Engineers use it to design three‑phase systems, to size equipment, and to manage power factor.
- Industrial motor and machinery power calculation
- Three‑phase distribution and transformer sizing
- Power factor correction in industrial plants
- Load flow analysis and system planning
- Educational understanding of three‑phase power
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