Electrical power calculation: P = U × I, kW vs kVA, single- and three-phase
Calculator for active, reactive and apparent power in 230 V single-phase, 400 V three-phase and DC. Enter voltage, current, cosine phi and efficiency to obtain the useful power, the rated current or the kW ⇄ kVA conversion. Power triangle updated live.
What is the electrical power calculation used for?
The electrical power calculation is the first step of any installation sizing. Before choosing a circuit-breaker rating, a cable cross-section or an RCD rating, the installer or designer must know the active power (in kilowatts, kW) consumed by the load, and the rated current (in amperes, A) that results from it. The basic formula, attributed to James Prescott Joule, is P = U × I in direct current, extended to P = U × I × cosφ × η in single-phase AC and P = √3 × U × I × cosφ × η in balanced three-phase.
In Switzerland, distribution networks are standardised at 230 V single-phase and 400 V three-phase at 50 Hz, with a TN-S earthing system in residential settings according to the NIBT 2020. The efficiency η distinguishes the power drawn from the grid from the useful mechanical power at the shaft output — unavoidable as soon as a motor or a converter enters the chain.
The practical stakes are threefold: protection (the circuit breaker must trip before any heating of the cable), savings (an oversized cable is expensive in copper), and compliance (the OIBT safety report requires consistency between declared power, protection rating and cable cross-section).
Distinguishing kW, kVA and kVAR: the power triangle
The active power P, in watts (W) or kilowatts (kW), corresponds to the energy actually converted into useful work: heat, light, mechanical motion. It is the power billed at the energy tariff.
The reactive power Q, in vars (VAR) or kilovars (kVAR), flows between the source and inductive (coils, motors) or capacitive loads without producing work. It is necessary for the magnetic field of motors but does not contribute to the useful energy balance.
The apparent power S, in volt-amperes (VA) or kilovolt-amperes (kVA), is the vector combination: S = √(P² + Q²). It is what determines the sizing of cables, transformers and UPS units.
The power triangle above visualises this relationship: P forms the base, Q the height, S the hypotenuse, and the angle phi (φ) between P and S has as its cosine the famous cosφ. The correction of the cosine phi by capacitor bank brings the cosφ back towards 0.95 and reduces line Joule losses.
Single-phase vs three-phase: why the √3 factor
En balanced three-phase, three phases L1, L2, L3 are shifted by 120°. The voltage line-to-line (between two phases) is 400 V; the voltage simple (phase-to-neutral) is 230 V. The ratio between these two values is precisely √3 ≈ 1,732. It is the trigonometric projection of the equilateral triangle formed by the three voltage vectors.
At equal power, a three-phase load draws a current 1,73 times lower than a single-phase load: a 5 kW water heater on single-phase draws 21.7 A; the same on three-phase 400 V draws 7.2 A per phase. That is why loads above 4 kW are systematically wired in three-phase in Switzerland.
Common kW ⇄ kVA conversions according to cosine phi
The following table gives the apparent power S (kVA) corresponding to an active power P (kW) for the most commonly encountered cosine phi values. Useful for sizing photovoltaic inverters, generator sets and the UPS which are specified in kVA.
| P (kW) | cosφ = 1,00 | cosφ = 0,95 | cosφ = 0,85 | cosφ = 0,80 | cosφ = 0,70 |
|---|---|---|---|---|---|
| 1 kW | 1,00 kVA | 1,05 kVA | 1,18 kVA | 1,25 kVA | 1,43 kVA |
| 3 kW | 3,00 kVA | 3,16 kVA | 3,53 kVA | 3,75 kVA | 4,29 kVA |
| 5 kW | 5,00 kVA | 5,26 kVA | 5,88 kVA | 6,25 kVA | 7,14 kVA |
| 10 kW | 10,0 kVA | 10,5 kVA | 11,8 kVA | 12,5 kVA | 14,3 kVA |
| 15 kW | 15,0 kVA | 15,8 kVA | 17,6 kVA | 18,8 kVA | 21,4 kVA |
| 22 kW | 22,0 kVA | 23,2 kVA | 25,9 kVA | 27,5 kVA | 31,4 kVA |
| 30 kW | 30,0 kVA | 31,6 kVA | 35,3 kVA | 37,5 kVA | 42,9 kVA |
| 50 kW | 50,0 kVA | 52,6 kVA | 58,8 kVA | 62,5 kVA | 71,4 kVA |
Practical case 1: 3 kW resistive single-phase water heater
A resistive electric water heater of 3 kW is supplied single-phase at 230 V. The sheathed heating element is a purely ohmic : cosφ = 1.00 and η ≈ 1.00. I = P / (U × cosφ × η) = 3000 / (230 × 1 × 1) = 13,04 A. Selected rating: 16 A curve C, cross-section 2,5 mm² copper in installation method B1, compliant with NIBT 5.2.
Practical case 2: 4 kW three-phase induction motor
A three-phase induction motor of 4 kW mechanical at 400 V, cosφ 0,85 and efficiency η = 0,87: the full formula gives I = P_mech / (√3 × U × cosφ × η) = 4000 / (1,732 × 400 × 0,85 × 0,87) = 7,81 A. This is exactly the reference value returned by the calculator above (« Calculate I » mode, three-phase network, P = 4 kW, U = 400 V, cosφ = 0.85, η = 0.87).
At direct-on-line starting, the motor draws 5 to 8 times I_n for 1 to 5 seconds; you therefore choose a circuit breaker curve D (10-20 In) four-pole 10 A. The apparent power is S = P / cosφ ≈ 5,41 kVA, the reactive power Q = P × tanφ ≈ 2,85 kVAR.
FAQ — Electrical power calculation
What is the difference between kW and kVA?
The kilowatt (kW) measures active power, i.e. the power actually converted into useful work (heat, light, motion). The kilovolt-ampere (kVA) measures the apparent power, which includes the active power and the reactive power needed for the magnetic field of motors and transformers. The relationship is S (kVA) = P (kW) / cosφ. A UPS or a generator set is sized in kVA, not in kW.
Why the √3 factor in balanced three-phase?
Le √3 vient de la géométrie du système triphasé : la tension entre deux phases (composée, 400 V) vaut √3 fois la tension entre une phase et le neutre (simple, 230 V). Lorsqu'on exprime la puissance totale absorbée par une charge triphasée équilibrée en fonction de la tension composée U et du courant de ligne I, le calcul vectoriel donne P = √3 × U × I × cosφ. C'est une conséquence trigonométrique du déphasage de 120° entre les trois phases.
Should you use cosφ = 0.8 or 0.95?
It all depends on the nature of the load. For a induction motor nominal, cosφ is around 0.80 to 0.85. For a purely resistive load (heating, oven, conventional water heater) cosφ equals 1,00. For equipment with switch-mode power supplies recent ones (LED, consumer electronics, PV inverters) the cosφ is between 0.90 and 0.98 thanks to active PFC power-factor correctors. For a default sizing of a mixed sub-distribution board, cosφ = 0.9 is a reasonable average.
What is the efficiency η used for in the formula?
The efficiency η distinguishes the power drawn from the grid from the useful mechanical power at the shaft output. For a 4 kW mechanical motor with η = 0.87, the drawn power is P_abs = 4 / 0.87 ≈ 4.60 kW. The drawn power is always used to size the cable, the circuit breaker and the energy bill. For a purely resistive load, η ≈ 1.
How do you calculate power in direct current?
In direct current there is no phase shift: the formula simplifies to P = U × I × η, the cosine phi is implicitly 1. A 400 Wp photovoltaic panel at 36 V draws 400 / 36 = 11.1 A on the DC side. For high-voltage DC installations (1000 V PV strings, 800 V fast EV chargers), the safety rules change: no zero crossing, special DC disconnectors mandatory.
How do you convert amperes to watts?
You need to know the voltage U and the cosφ. In single-phase 230 V with cosφ=1: W = 230 × A. In three-phase 400 V with cosφ=0.85: W = √3 × 400 × A × 0.85 ≈ 588 × A. A single measured ampere cannot be converted to watts without knowing these two parameters.
Go further with ElectroCAD
The power calculation is natively integrated into the tools ElectroCAD Tools and to the ElectroSchema plugin for AutoCAD. The load schedule automatically fills in P, U, cosφ, η and calculates I, the cross-section sizing NIBT 2020-compliant, and the standardized circuit breaker rating.