The three powers, and why they differ
AC circuits have three distinct "power" quantities, and confusing them is the source of most electrical sizing mistakes.
- Real power (P), in watts. What the equipment converts into heat, light, rotation or sound. This is what you pay for, and it is measured in watts.
- Apparent power (S), in volt-amperes. The product of voltage and current. This is what the conductors, the transformer and the meter must be rated for. It is always greater than or equal to real power.
- Reactive power (Q), in volt-amperes reactive. Power that oscillates between the source and the load without being consumed. It exists because inductive loads (motors, transformers) and capacitive loads (capacitor banks) store and release energy each cycle.
The link between them is the power factor:
P = S × cos φ · so S = P ÷ PF
And the current that flows is simply apparent power over voltage:
I = S ÷ V
Worked example
A three-phase 400 V supply powers a 5 kW machine at a power factor of 0.9.
- Real power P = 5,000 W
- Apparent power S = 5,000 ÷ 0.9 = 5,556 VA
- Reactive power Q = √(5,556² − 5,000²) = 2,645 VAR
- Line current I = 5,556 ÷ 400 = 13.9 A
With the standard 125% continuous-load margin, the circuit needs 17.4 A, so a 20 A breaker is the right choice. This is why you size the breaker on current, not on watts — the watts tell you the bill, the current tells you the cable and the protection.
Single-phase versus three-phase
The formulas above are the general AC case. Three-phase changes the relationship between line and phase quantities:
Three-phase real power: P = √3 × VL × IL × PF
The √3 (1.732) factor is the reason a 20 A three-phase supply can deliver 6.9 kW at 400 V while a 20 A single-phase supply at 230 V delivers only 4.6 kW. In practice the calculator computes from the power you supply rather than from current and voltage, so it does not need the √3 — but the same numbers will show a lower current on the three-phase side, and that is correct.
Where the power factor actually goes
A low power factor is not a nuisance; it is billed. Utilities charge for apparent power on commercial and industrial tariffs, or impose reactive power penalties. A 5 kW load at PF 0.6 draws 8,333 VA, so you pay for 8,333 units while using 5,000 — a 67% overhead.
The common causes and fixes:
- Induction motors — the usual offender. Fix with a VFD or a local correction capacitor bank sized at about 0.7 × kW × (tan φ1 − tan φ2).
- Older fluorescent lighting with magnetic ballasts. Replacing the ballast or the fitting fixes it permanently.
- Large UPS and rectifier loads — modern switch-mode supplies usually have active power factor correction and are fine; older ones are not.
The target is PF ≥ 0.95 for most commercial premises, and utilities will often require it as a condition of supply.
What the calculator does and does not cover
It gives you the electrical quantities: real, apparent and reactive power, current, and a suggested breaker size from the 125% continuous-load rule. It deliberately does not size cable — that depends on installation method, ambient temperature, grouping factor and local code, and getting it wrong is a fire risk rather than an inconvenience. Use the current figure as the input to a proper cable calculation, and check the breaker against the local standard (IEC 60898-1 in Europe, NEC 240.4 in North America) since breaker ratings are regional.