Power Triangle Calculator

Real (P), Reactive (Q) and Apparent (S) power, plus power factor and phase angle.
From P & Q
From S & Power Factor

Apparent Power & PF from P and Q

S = √(P2+Q2)  •  PF = P/S = cosφ  •  φ = atan(Q/P)
Motor: P=8kW, Q=6kVAR
HVAC: P=50kW, Q=20kVAR
Resistive load: Q=0
kW
kVAR
Enter values and press Calculate.

P & Q from Apparent Power and PF

P = S×PF  •  Q = S×sin(φ), φ=acos(PF)  •  PF = cosφ
S=10kVA, PF=0.8 lag
S=25kVA, PF=0.95
S=5kVA, PF=1 (unity)
kVA
Enter values and press Calculate.

Understanding the Power Triangle

In an AC circuit with any reactance (inductive or capacitive load), the power delivered splits into three related quantities that form a right triangle: real power (P), measured in watts, does actual work (heat, motion, light); reactive power (Q), measured in VAR, is exchanged with the source without net work (stored/released in inductors and capacitors); and apparent power (S), measured in VA, is the vector sum the utility actually has to supply. The angle φ between P and S is the same phase angle between voltage and current, and its cosine is the power factor.

QuantityFormulaUnit
Apparent powerS = √(P²+Q²)VA (kVA)
Power factorPF = P/S = cosφ0–1 (or %)
Phase angleφ = atan(Q/P) = acos(PF)degrees
Real power from S, PFP = S×PFW (kW)
Reactive power from S, PFQ = S×sinφVAR (kVAR)

Inductive loads (motors, transformers, ballasts) draw lagging reactive power (current lags voltage); capacitive loads draw leading reactive power. Utilities bill for kVA (what they must supply) even though only kW (P) does useful work, which is why a low power factor costs extra — see our Power Factor Correction Capacitor Calculator to fix it.

Real-World Applications & Examples

Worked examples

1. Small induction motor. P=8kW, Q=6kVAR: S=√(64+36)=10kVA, PF=8/10=0.8, φ=36.9°.
2. HVAC compressor bank. P=50kW, Q=20kVAR: S=√(2500+400)≈53.9kVA, PF≈0.928.
3. Purely resistive load. P=10kW, Q=0: S=10kVA, PF=1.0 (unity) — no reactive component, e.g. a resistive heater.
4. From S and PF (transformer rating). S=10kVA at PF=0.8 lagging: P=10×0.8=8kW, φ=acos(0.8)=36.9°, Q=10×sin(36.9°)=6kVAR.
5. High-efficiency facility. S=25kVA, PF=0.95: P=25×0.95=23.75kW, Q=25×sin(18.2°)≈7.8kVAR — a well-corrected facility with most of its kVA doing real work.
6. Unity power factor. S=5kVA, PF=1: P=5kW, Q=0 — the ideal case where the utility supplies only real power.

Frequently Asked Questions

What is the power triangle?

A right-triangle representation of AC power showing real power (P, the horizontal leg), reactive power (Q, the vertical leg), and apparent power (S, the hypotenuse), related by S=√(P²+Q²).

What is the difference between real, reactive, and apparent power?

Real power (P, watts) does actual work. Reactive power (Q, VAR) is energy that oscillates between source and load (inductors/capacitors) without net work. Apparent power (S, VA) is the total the source must supply, combining both.

What is power factor?

The ratio PF=P/S=cosφ, showing what fraction of the apparent power is actually doing useful work. PF=1 (unity) means no reactive power; lower PF means more of the supply capacity is "wasted" on reactive current.

Why does reactive power matter if it does no work?

It still flows through wires, transformers, and generators, causing I²R heating and consuming capacity, even though it delivers no net energy to the load — that is why utilities often charge extra for low power factor.

What is the difference between lagging and leading power factor?

Lagging PF means current lags voltage (typical of inductive loads like motors); leading PF means current leads voltage (typical of capacitive loads). Both represent reactive power, just in opposite directions.

How is the phase angle φ related to power factor?

φ is the angle between voltage and current (and between P and S in the triangle); PF = cosφ. A larger φ means more reactive power relative to real power and a lower PF.

Why are transformers and generators rated in kVA, not kW?

Because winding and conductor heating depend on current (which relates to kVA), not on how much of that current does useful work. A device must be sized for the full apparent power regardless of the load's PF.

Can Q be negative?

By convention inductive (lagging) loads have positive Q and capacitive (leading) loads have negative Q; a purely resistive load has Q=0.

How do you improve a poor power factor?

Add capacitors (for inductive/lagging loads) to supply local reactive power, reducing the reactive current the source must provide — see our Power Factor Correction Capacitor Calculator.

What is a typical power factor for an induction motor?

Often between 0.7 and 0.9 lagging at full load, and considerably lower (0.3–0.5) at light load, since magnetizing (reactive) current stays roughly constant while real power drops.

Does the power triangle apply to three-phase systems?

Yes — the same P, Q, S relationships hold for total three-phase power; just ensure P, Q, and S are all computed consistently as total (all three phases) or per-phase values, not mixed.

What units are used for P, Q, and S?

P is in watts (W/kW/MW), Q is in reactive volt-amperes (VAR/kVAR), and S is in volt-amperes (VA/kVA/MVA) — all three share the same numeric relationship regardless of the magnitude prefix used.

How do I find S if I only know P and PF?

S = P/PF, which follows directly from PF=P/S rearranged; then Q can be found from S and P using Q=√(S²−P²).

Is a power factor greater than 1 possible?

No, PF is bounded between 0 and 1 (or -1 to 1 with sign convention) because P can never exceed S — real power is always the resistive-projection component of the total apparent power.

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