Complex Power Calculator
Combine real and reactive power into complex power: S = P + jQ, with magnitude |S| = √(P² + Q²).
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. Q is positive for an inductive load and negative for a capacitive load. Pick any unit; conversion is automatic.
Results
Enter values and press Calculate. Your answer will appear here.
Power triangle in the complex plane
P lies along the real axis and Q along the imaginary axis. The amber arrow is S, and its length is |S|. With a target power factor, the dashed line shows the triangle after correction and the teal arrow is the capacitor.
What is being calculated
Power factor, angle and Q/P
| cos θ | θ | Q / P = tan θ | Q / |S| = sin θ |
|---|
Multiply P by tan θ to get Q, or |S| by sin θ. Angles are for a lagging load; use −θ for a leading load.
Unit conversions
| Quantity | Conversion |
|---|---|
| Real power | 1 kW = 1,000 W · 1 MW = 1,000 kW |
| Reactive power | 1 kvar = 1,000 var · 1 Mvar = 1,000 kvar |
| Apparent power | 1 kVA = 1,000 VA · 1 MVA = 1,000 kVA |
| Angle | 180° = π rad · 1 rad = 57.2958° |
Formula
- S: complex power (VA); its magnitude |S| is the apparent power
- P: real (active) power (W)
- Q: reactive power (var); positive for inductive, negative for capacitive loads
- θ = atan2(Q, P): angle of S, and cos θ = P / |S| is the power factor
Example: P = 8 kW and Q = 6 kvar. |S| = √(8² + 6²) = 10 kVA, θ = atan(6/8) ≈ 36.87°, and cos θ = 8/10 = 0.8 lagging. In polar form S = 10 kVA ∠ 36.87°.
Complex power comes from phasors as S = V·I*, where I* is the complex conjugate of the current, which makes Q positive when the current lags the voltage (inductive). Other relations: P = |S| cos θ, Q = |S| sin θ. Current is I = |S| / V for single-phase, or |S| / (√3·VL) for balanced three-phase with line-to-line voltage. To raise a lagging power factor from cos θ₁ to cos θ₂ at constant P, add capacitors of Qc = P·(tan θ₁ − tan θ₂); the new reactive power is P·tan θ₂ and the new apparent power is P / cos θ₂. Some textbooks and meters use a different sign convention for Q, so check which one yours uses.
References
- Wikipedia, AC power: complex, real, reactive and apparent power
- Wikipedia, Power factor: power factor and its correction
- Wikipedia, Phasor: phasor representation behind S = V·I*
How to Use This Tool
- Pick a preset, or enter the real power P and reactive power Q with their units.
- Use a negative Q for a capacitive (leading) load.
- In Advanced, start from |S| with a power factor or an angle instead, add a voltage for the current, or a target power factor for capacitor sizing.
- Click Calculate.
- Read |S|, the power factor and the rectangular and polar forms, then check the power triangle.
What This Tool Offers
- Basic and Advanced modes on one calculation engine
- W, kW, MW, var, kvar, Mvar, VA, kVA, MVA and angle unit conversion
- Rectangular and polar forms, power factor with lagging or leading label
- Three input routes, current for single or three-phase, and power-factor correction capacitor sizing (Advanced)
- Power triangle chart and labelled complex-plane diagram
- Reorderable inputs, presets and a mobile-friendly layout
Variables
| Symbol | Quantity | Unit |
|---|---|---|
| S | Complex (apparent) power | VA |
| P | Real (active) power | W |
| Q | Reactive power | VAR |
| j | Imaginary unit (√-1) | — |
| |S| | Apparent power magnitude (√(P²+Q²)) | VA |
| θ | Power factor angle (arg(S)) | ° |
What it means
Complex power S is a mathematical representation of power in AC circuits. It is defined as S = P + jQ, where P is the real power (watts), Q is the reactive power (VAR), and j is the imaginary unit. The magnitude of S is the apparent power |S| = √(P²+Q²) measured in volt‑amperes (VA). The angle of S is the power factor angle, and cosθ = P/|S|. Complex power simplifies power calculations, especially when dealing with multiple loads and power factor correction. It is used in power flow analysis, transformer design, and motor selection. The product of voltage and current phasors (V·I*) gives S. Understanding complex power is essential for analysing AC circuits and for designing efficient power systems. Example: A load with P=1000W and Q=600VAR has S = 1000 + j600 VA. The apparent power magnitude is √(1000²+600²) = √1,360,000 = 1166 VA, and the power factor is 1000/1166 = 0.857.
Worked example
Complex Power – Practical Example
Real‑World| Parameter | Value |
|---|---|
| P | 10 kW |
| Q | 6 kVAR (inductive) |
| Formula | S = P + jQ |
Common mistakes
- Complex power: S = P + jQ – where P is real power (W) and Q is reactive power (VAR).
- Magnitude: |S| = √(P²+Q²) = apparent power (VA).
- Angle: The phase angle θ = arctan(Q/P) – the power factor angle.
- Sign: Q is positive for inductive loads, negative for capacitive.
- Phasor notation: S = V·I* – use RMS values.
Applications
Complex power S = P + jQ combines real power (P) and reactive power (Q) into a single complex quantity. This representation simplifies AC power analysis, allowing the use of vector operations. Engineers use it to analyse power factor, to design power factor correction, and to perform load flow studies. The magnitude of S is the apparent power (VA), and the angle is the power factor angle. By using complex power, professionals can compute the total power in systems with multiple loads and account for phase differences. It is also used in the design of transformers and generators. Understanding complex power is essential for power system engineers.
- AC power analysis and power factor correction
- Load flow studies and system planning
- Transformer and generator rating calculations
- Phasor representation of power in circuits
- Educational understanding of complex power
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