Formula & Calculator

Complex Power

Represents total power in an AC circuit as a complex quantity combining real and reactive power.

Power SystemsAC Power

Complex Power Calculator S = P + jQ

S = P + j Q   |   |S| = √(P² + Q²)
S = complex power (VA)  ·  P = real power (W)  ·  Q = reactive power (VAR)  ·  |S| = apparent power magnitude
⟹ Solve P, Q, |S|
W
VAR
VA
Please fix the errors above.
Solve for:
Presets:
Apparent Power
P: Q: |S|: θ:
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Apparent Power Gauge
Low (< 50 VA) Medium (50–200 VA) High (> 200 VA)
S = P + jQ  ·  |S| = √(P²+Q²)  ·  θ = atan2(Q, P)  ·  Power factor = cos(θ)

Interpretation

Complex power S = P + jQ combines real power (P) and reactive power (Q) into a single complex quantity.
The magnitude of S is the apparent power (VA), and the angle is the power factor angle.
Example: P=1000W, Q=600VAR → S = 1000 + j600 VA.

S = P + jQ
Complex Power

Variables

SymbolQuantityUnit
SComplex (apparent) powerVA
PReal (active) powerW
QReactive powerVAR
jImaginary unit (√-1)
|S|Apparent power magnitude (√(P²+Q²))VA
θPower 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
Scenario: A load consumes 10 kW of real power and 6 kVAR of reactive power. Express the complex power S.
ParameterValue
P10 kW
Q6 kVAR (inductive)
FormulaS = P + jQ
1Write S:S = 10 + j6 kVA
Final Design S = 10 + j6 kVA ✓ Complex power
Why: Complex power combines real (P) and reactive (Q) power – the magnitude |S| = √(P²+Q²) gives the apparent power.

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

Frequently Asked Questions

Q01What is complex power and how is it defined?
A01

Complex power is S = P + jQ, where P is real power (watts) and Q is reactive power (VAR). Its magnitude is apparent power |S| = √(P²+Q²).

Q02What do P and Q represent?
A02

P = real power (watts), Q = reactive power (VAR). The complex power S = P + jQ is the vector sum.

Q03What is the magnitude of complex power?
A03

|S| = √(P² + Q²) = V·I* (apparent power in VA).

Q04What is the relationship between complex power and impedance?
A04

S = V·I* = I²·Z = V² / Z*, where Z is the complex impedance.

Q05What is the significance of the sign of Q?
A05

Positive Q means inductive (lagging power factor), negative Q means capacitive (leading power factor).

Q06How is complex power used in power factor correction?
A06

By adding capacitive reactance to supply negative Q, reducing the total Q and improving power factor.

Q07What is the complex power in a circuit with both series and parallel elements?
A07

Calculate the total impedance, then S = V² / Z*; the real and imaginary parts give P and Q.

Q08What are the units of reactive power?
A08

Volt-amperes reactive (VAR). Commercial units are kVAR and MVAR.

Q09What is the power triangle?
A09

A right triangle where the hypotenuse is apparent power |S|, adjacent side is P, and opposite side is Q.

Q10What are the common mistakes when using complex power?
A10

Common errors include: 1) confusing P and Q, 2) using the wrong sign for Q, 3) using peak instead of RMS values, 4) forgetting the conjugate, and 5) applying to non-sinusoidal signals without correction.