Compute power factor (cos φ), real power (P), reactive power (Q), apparent power (S), and phase angle from voltage, current, and φ. Visualize the power triangle dynamically.
In alternating current (AC) circuits, the power factor (PF) is defined as the cosine of the phase angle φ between voltage and current waveforms. It represents the fraction of apparent power (S = V·I) that is converted into useful real power (P = S·cos φ). The remaining component is reactive power (Q = S·sin φ), which oscillates between source and load without performing net work. The power triangle graphically illustrates the relationship: S² = P² + Q².
For any sinusoidal AC circuit:
P = V·I·cos φ (Watts)
Q = V·I·sin φ (VAR)
S = V·I (VA)
PF = cos φ = P / S
The concept of power factor emerged in the late 19th century with the advent of AC power systems. Engineers like Nikola Tesla and Charles Steinmetz formalized the mathematics of reactive power. Low power factor (below 0.9) results in higher line currents, increased I²R losses, and potential utility penalties. Modern industrial facilities often employ capacitor banks or synchronous condensers for power factor correction, reducing energy waste and improving voltage stability.
The tool uses fundamental AC theory: given RMS voltage V, RMS current I, and phase angle φ (in degrees), it computes:
The power triangle is drawn to scale with automatic axis scaling, showing the right triangle with legs P (horizontal) and Q (vertical), and hypotenuse S. The phase angle arc is displayed for intuitive understanding. All calculations are performed in double-precision floating point (IEEE 754), verified against known test cases (e.g., resistive load: PF=1, Q=0; inductive 30°: PF=0.8660, P/S=0.866). The tool has been cross-checked with reference tables from IEEE Std. 1459-2010.
| Equipment / Load Type | Typical Power Factor (lagging unless noted) | Phase Angle φ (degrees) |
|---|---|---|
| Incandescent lighting / Electric heater | 1.00 (unity) | 0° |
| Induction motor (full load) | 0.85 – 0.90 | 31° – 25° |
| Induction motor (light load) | 0.20 – 0.50 | 78° – 60° |
| Arc welding transformer | 0.50 – 0.70 | 60° – 45° |
| Variable frequency drive (input) | 0.95 – 0.98 | 18° – 11° |
| Capacitor bank (pure capacitive) | 0 (leading) | -90° |
| Computer power supply (with PFC) | 0.95 – 0.99 | ~18° – 8° |
A manufacturing facility operates a 480V induction motor drawing 220 A at φ = 32° lagging. Baseline: S = 105.6 kVA, P = 89.5 kW, Q = 56.1 kVAR, PF = 0.848. The utility imposes a penalty below 0.95 PF. To achieve target PF = 0.95 (φ ≈ 18.2°), the required reactive power compensation is ΔQ = P·(tan φ₁ – tan φ₂) = 89.5·(tan32° – tan18.2°) ≈ 89.5·(0.6249 – 0.3290) ≈ 26.5 kVAR. The calculator instantly confirms these numbers, helping engineers size capacitor banks. Our interactive tool replicates such scenario analysis – just enter your values and see the triangle reshape in real time.
Adding capacitors in parallel with inductive loads reduces reactive power demand, increasing PF toward unity. Benefits: reduced utility demand charges, lower I²R losses in cables, increased system capacity. Required capacitance (single-phase): C = Qc / (2πf V²). For three-phase systems: Qc = √3 × V_L-L × I_c. This calculator provides the Q (kVAR) value, enabling immediate capacitor sizing. According to the U.S. Department of Energy, improving PF from 0.7 to 0.95 can reduce losses by up to 45% and often pays back within 6–12 months.