The Power Triangle
In AC circuits, the apparent power (S) is the hypotenuse of a right triangle. The horizontal leg is real power (P), the power that does useful work. The vertical leg is reactive power (Q), the power that sloshes back and forth between source and load without doing work.
Q (Reactive Power), VAR. Energy stored and returned by inductors/capacitors each cycle. Does no work but increases current.
S (Apparent Power), VA. The total power the supply must deliver: S = V × I. Sizes cables and transformers.
θ (Phase Angle), Angle between voltage and current. PF = cos(θ). Unity PF (θ = 0°) means all power is real.
Understanding the Power Triangle
In AC circuits, power splits three ways. Real power (P) in watts does the useful work, reactive power (Q) in VAR sloshes between source and load, and apparent power (S) in VA is the total the supply must deliver. Together they form a right triangle linked by S² = P² + Q², and the power factor is the ratio P/S = cos(θ).
The power triangle builds on the basic DC relationships. If you are new to voltage, current and resistance, start with the Ohm's Law calculator, then use the power calculator for straight DC or resistive power before moving to AC power factor here.
Each quantity has its own unit: real power is measured in the watt (W), reactive power in the volt-ampere reactive (VAR), and apparent power in the volt-ampere (VA). Only the watt does real work, while the volt-ampere and volt-ampere reactive describe how much current the supply must carry. This is why equipment such as cables, generators and transformers is rated in VA or kVA, not watts.
How to Use the 4 Input Modes
Pick the mode that matches the values you already know, then type them in. The calculator solves the rest of the triangle and shows the step-by-step working.
- P + Q, enter real power and reactive power. Best when you already know the watt and VAR figures and want apparent power, power factor and phase angle.
- S + PF, enter apparent power and power factor, then choose inductive or capacitive to set whether the load lags or leads. Common for motor and transformer nameplate ratings.
- S + Angle, enter apparent power and the phase angle in degrees. Useful when the angle between voltage and current is known directly.
- V + I + Angle, enter RMS voltage, RMS current and the phase angle. The calculator finds apparent power from S = V × I, then splits it into real and reactive power.
Worked Examples
Motor power factor. A 5 kVA motor runs at a power factor of 0.85 (inductive). Real power P = S × PF = 5 × 0.85 = 4.25 kW. The phase angle θ = arccos(0.85) = 31.8°, so reactive power Q = S × sin(θ) = 5 × 0.527 = 2.63 kVAR.
Transformer sizing. A load draws 10 kW at 0.8 power factor. Apparent power S = P / PF = 10 / 0.8 = 12.5 kVA, so a 12.5 kVA transformer (or the next size up) is needed even though the real load is only 10 kW.
From volts and amps. A 230 V supply delivers 10 A at a 30° phase angle. Apparent power S = V × I = 230 × 10 = 2.3 kVA. Real power P = S × cos(30°) = 1.99 kW, and reactive power Q = S × sin(30°) = 1.15 kVAR.
Why Power Factor Matters
A low power factor means more current is needed to deliver the same real power, which increases cable losses, voltage drop and equipment heating. Suppliers often charge commercial sites for poor power factor, so correcting it with capacitors (power factor correction) can cut bills. Better electrical efficiency usually follows once reactive current is reduced.
Power Triangle Calculator FAQ
What is power factor?
The ratio of real power to apparent power. It ranges from 0 to 1. Unity means all current does useful work. A value of 0.5 means half the current is reactive and does nothing useful.
What causes poor power factor?
Inductive loads such as motors, transformers and fluorescent ballasts. The magnetic field stores and returns energy every cycle, creating reactive current. Lightly loaded motors are the worst offenders.
What is the difference between kW, kVA and kVAR?
kW is real power that does work, kVA is apparent power that sizes equipment, and kVAR is reactive power that oscillates without doing work. They relate by S² = P² + Q².
Why are transformers rated in kVA instead of kW?
Transformer windings must carry the total current set by apparent power. A 100 kVA transformer delivers 100 kW at unity power factor but only 80 kW at 0.8 power factor, because it heats based on total current.
Does power factor matter for residential customers?
Residential meters usually measure real energy only, so there is no direct penalty. However, poor power factor increases current in house wiring, which increases heating and voltage drop.
What power factor do I need?
Most suppliers require 0.90 minimum to avoid penalties. Best practice is 0.95 to 0.99. Over-correction to a capacitive (leading) power factor can cause resonance and voltage rise.
Power Triangle Calculator
In AC circuits, current and voltage fall out of phase. Power splits into two components: real power (does useful work) and reactive power (sloshes back and forth doing nothing). Apparent power is the vector sum — what the supply must deliver and what sizes cables, transformers, and switchgear. These three form a right triangle. The angle between apparent and real power is the phase angle θ, and cos(θ) is the power factor. Enter any two known values — the calculator solves for everything else.
Core Formulas
PF = P / S = cos(θ) — power factor
θ = arccos(PF) = arctan(Q / P) — phase angle
P = S × cos(θ) — real power from apparent and angle
Q = S × sin(θ) — reactive power from apparent and angle
S = V × I — apparent power from voltage and current
P = real power (W, kW) — does useful work. Q = reactive power (VAR, kVAR) — does no work. S = apparent power (VA, kVA) — total delivered by the supply. +Q = inductive (lagging). −Q = capacitive (leading). For the underlying V = IR relationship, see the Ohm's Law Calculator.
Four Input Modes
Mode 2: S + PF — know apparent power and power factor. Solves P, Q, θ.
Mode 3: S + Angle — know apparent power and phase angle. Solves P, Q, PF.
Mode 4: V + I + Angle — know voltage, current, and phase angle. Solves S, P, Q, PF.
Induction Motor (S = 5 kVA, PF = 0.85)
Mode 2. A typical induction motor nameplate reads 5 kVA at 0.85 power factor, inductive.
θ = arccos(0.85) = 31.8°
Q = 5 × sin(31.8°) = 5 × 0.527 = 2.63 kVAR — reactive, creates magnetic field
The supply delivers 5 kVA, but only 4.25 kW does useful work. The remaining 2.63 kVAR magnetises the motor's stator — necessary for operation but it increases current demand without producing output. At 230 V single-phase: I = 5000/230 = 21.7 A. If the power factor were unity, only 18.5 A would be needed for the same mechanical output.
Power Factor Correction (P = 10 kW, Q = 6 kVAR)
Mode 1. A factory load measured at 10 kW real and 6 kVAR reactive (inductive).
PF = 10 / 11.66 = 0.857
θ = arctan(6/10) = 30.96°
Power factor of 0.857. Most suppliers penalise below 0.90–0.95. To correct to PF = 0.95, the target Q is: Q_target = 10 × tan(arccos(0.95)) = 10 × 0.329 = 3.29 kVAR. The capacitor bank must supply: Q_cap = 6 − 3.29 = 2.71 kVAR of leading reactive power. This reduces apparent power from 11.66 kVA to 10.53 kVA, lowering current by 10% and eliminating the penalty. For the capacitive reactance needed to supply those kVAR, see the Capacitive Reactance Calculator.
Resistive Load (S = 2.4 kVA, θ = 0°)
Mode 3. A heater, kettle, or incandescent bulb — purely resistive.
Q = 2.4 × sin(0°) = 0 kVAR
PF = 1.0 (unity)
S = P and Q = 0. All power does useful work. No reactive component, no power factor penalty, no oversized cables needed. This is the ideal case that power factor correction tries to approach. For the energy cost of running this load, use the Electrical Energy Calculator.
Measured V, I, and Angle (230 V, 10 A, 30°)
Mode 4. UK mains, measured with a clamp meter and power analyser.
P = 2300 × cos(30°) = 2300 × 0.866 = 1992 W ≈ 2.0 kW
Q = 2300 × sin(30°) = 2300 × 0.5 = 1150 VAR = 1.15 kVAR
PF = 0.866
The supply delivers 2.3 kVA but only 2.0 kW is real. The 1.15 kVAR reactive component means 10 A is flowing, but a resistive load doing the same work would only draw 8.66 A. The difference is reactive current that heats cables and transformers without producing useful output.
Why the Power Triangle Matters
Transformer Sizing
Transformers are rated in kVA (apparent power), not kW. A 100 kVA transformer at PF 0.8 delivers only 80 kW of real power. If your load needs 100 kW at PF 0.8, you need a 125 kVA transformer.
Cable Sizing
Cables carry current. Current is determined by apparent power: I = S/V. A load drawing 10 kW at PF 0.7 requires I = (10/0.7)/230 = 62 A. The same load at PF 0.95 requires only 46 A. Lower power factor means bigger cables, bigger breakers, and more copper cost.
Electricity Bills
Residential customers pay for kWh (real energy). Industrial customers often also pay a reactive power penalty or a maximum demand charge based on kVA. A factory at PF 0.7 pays roughly 43% more in demand charges than at PF 1.0 for the same useful output. To calculate the energy cost itself, use the Power Calculator.
Power Factor Correction
Adding capacitors in parallel with an inductive load supplies the reactive power locally instead of drawing it from the grid. The real power stays the same, but Q drops, S drops, current drops, and the power factor improves.
Power Factor Correction — Step by Step
2. Choose target PF (typically 0.95 or 0.99)
3. Q_target = P × tan(arccos(PF_target))
4. Q_capacitor = Q_current − Q_target
5. Install Q_capacitor kVAR of correction capacitors
Example: P = 50 kW, current PF = 0.75 (Q = 44.1 kVAR). Target PF = 0.95 (Q_target = 16.4 kVAR). Capacitor bank needed: 44.1 − 16.4 = 27.7 kVAR. Apparent power drops from 66.7 kVA to 52.6 kVA — a 21% reduction in current demand.
Frequently Asked Questions
Last updated: March 2026