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Power Triangle Relation

Relates apparent, real, and reactive power as sides of a right triangle.

Power SystemsAC Power

Power Triangle Calculator S² = P² + Q²

S² = P² + Q²
S = apparent power (VA)  ·  P = real power (W)  ·  Q = reactive power (VAR)
⟹ Solve S, P, Q
VA
W
VAR
Please fix the errors above.
Solve for:
Presets:
Apparent Power (S)
S: P: Q: PF: θ:
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Power Factor (cos θ)
Low (< 0.7) Medium (0.7–0.9) High (> 0.9)
S² = P² + Q²  ·  PF = P/S = cos θ  ·  θ = atan(Q/P)  ·  All powers in VA, W, VAR.

Interpretation

The power triangle relates apparent power (S), real power (P), and reactive power (Q) by Pythagoras: S² = P² + Q².
This relationship holds for sinusoidal AC circuits.
Example: S=500VA, P=400W → Q = √(500² − 400²) = √(250000−160000) = √90000 = 300 VAR.

S² = P² + Q²
Power Triangle Relation

Variables

SymbolQuantityUnit
SApparent powerVA
PReal (active) powerW
QReactive powerVAR
θPower factor angle (θ = arctan(Q/P))°
pfPower factor (pf = cos θ = P/S)dimensionless

What it means

The power triangle is a right triangle that relates apparent power S, real power P, and reactive power Q. The relationship is S² = P² + Q², which is derived from the complex power equation. The triangle is a graphical representation of the power factor, with the angle θ being the phase angle between voltage and current. The power factor is cosθ = P/S, and sinθ = Q/S. The power triangle is useful for visualising the trade‑offs between real and reactive power and for calculating the required power factor correction. It is often used in design and troubleshooting. Understanding the power triangle helps engineers determine the size of capacitors needed for power factor correction and to understand the effects of inductive or capacitive loads on the power system. Example: If S=500VA and P=400W, then Q = √(500²−400²) = √(250000−160000) = √90000 = 300 VAR. The power triangle has S=500 as hypotenuse, P=400 as adjacent, Q=300 as opposite.

Worked example

Power Triangle – Practical Example

Real‑World
Scenario: A load has apparent power S = 12 kVA and real power P = 9.6 kW. Find the reactive power Q.
ParameterValue
S12 kVA
P9.6 kW
FormulaS² = P² + Q² → Q = √(S² − P²)
1Compute S² − P²: 12² − 9.6² = 144 − 92.16 = 51.84
2Square root:Q = √51.84 = 7.2 kVAR
Final Design Q = 7.2 kVAR ✓ Reactive power
Why: The power triangle relates P, Q, and S – useful for power factor correction calculations.

Common mistakes

  • Pythagorean: S² = P² + Q² – not S = P + Q.
  • Apparent power S: The hypotenuse of the power triangle.
  • Units: S in VA, P in W, Q in VAR – all have the same unit dimensions but different names.
  • Power factor: pf = P/S = cosθ.
  • Lagging/leading: If Q>0, the power factor is lagging (inductive); if Q<0, it is leading (capacitive).

Applications

The power triangle relation S² = P² + Q² is derived from Pythagoras and relates apparent power (S), real power (P), and reactive power (Q) in sinusoidal AC circuits. This geometric relationship helps visualise the power factor and the trade‑off between real and reactive power. Engineers use it to determine the required reactive power compensation to achieve a target power factor. By understanding the power triangle, they can size capacitors or inductors for correction. It also assists in interpreting utility bills and assessing system efficiency. The power triangle is a fundamental tool in AC circuit analysis and power system design.

  • Power factor analysis and correction design
  • Assessment of reactive power needs in installations
  • Understanding utility power factor penalties
  • Design of capacitor banks for compensation
  • Educational visualisation of AC power components

Frequently Asked Questions

Q01What is the power triangle relation?
A01

The power triangle relation is S² = P² + Q², where S is apparent power, P is real power, and Q is reactive power.

Q02What is the power factor angle?
A02

The angle θ between S and P; cosθ is the power factor. θ = arctan(Q/P).

Q03How do you find Q if you know S and P?
A03

Q = √(S² − P²). This is useful for power factor correction sizing.

Q04What is the significance of the triangle relation?
A04

It provides a graphical and mathematical way to understand the relationships between real, reactive, and apparent power.

Q05What happens to the triangle when power factor is unity?
A05

Q = 0, so S = P, and the triangle collapses to a line.

Q06How does the triangle change with inductive vs capacitive loads?
A06

For inductive loads, Q is positive; for capacitive, Q is negative. The triangle is symmetric about P.

Q07What is the apparent power in terms of rms voltage and current?
A07

S = V_rms · I_rms. This is valid for sinusoidal signals.

Q08What are the practical uses of the power triangle?
A08

It helps visualize power factor, size equipment, and understand the effect of adding capacitors.

Q09How do harmonics affect the power triangle?
A09

With harmonics, S² ≠ P² + Q²; there is also a distortion power component D. The relation becomes S² = P² + Q² + D².

Q10What are the common mistakes when using the power triangle?
A10

Common errors include: 1) confusing P and Q, 2) using the wrong sign for Q, 3) not considering harmonics, 4) using peak instead of RMS, and 5) applying to non-linear loads without measurement.