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.
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