Apparent Power vs Reactive Power vs Real Power
Draw two arrows spinning together sixty times a second, one for voltage and one for current. When they stay locked together, pointing the same direction at every instant, every watt drawn from the source does useful work: heat in an element, torque in a shaft, light from a filament. Give the current a head start or a delay relative to voltage, and something changes: part of what's flowing through the wire stops doing work altogether. It surges into the load and back out again every half-cycle without ever converting into heat, light or motion.
That split between working power and non-working power is what separates three quantities that show up on a nameplate, a utility bill and a switchgear rating almost interchangeably: real power (kW), the rate energy is actually converted to work; apparent power (kVA), the full volt-amp product the supply has to be sized to carry regardless of how much of it works; and reactive power (kVAR), the fraction that oscillates between source and load without ever leaving as usable energy. All three describe the same load at the same instant. They just answer three different questions about it.
Where the phase angle comes from
An AC voltage and an AC current are each sine waves oscillating at the same frequency, but a coil, a motor winding or a bank of capacitors can shift the current's peak away from the voltage's peak by some angle θ. A purely resistive load (a heating element, an incandescent bulb) draws current that peaks at exactly the same instant as the voltage, so θ is 0°. A winding with inductance delays the current's peak, producing a lagging angle; capacitance advances it instead, producing a leading angle. Either way, θ is the number that determines how much of the delivered volt-amps actually does work.
Multiplying instantaneous voltage and current together and averaging over a full cycle produces three related figures. Real power keeps only the in-phase component of current, the part that lines up with voltage and transfers net energy:
Apparent power ignores the phase angle completely and simply multiplies RMS voltage by RMS current: it's what the wiring, the transformer and the source actually have to carry, whether or not that current is doing work:
Reactive power keeps only the out-of-phase component, the part that flows into the load's magnetic or electric field and back out again every cycle without any net transfer:
Because cos²θ + sin²θ = 1 for any angle, those three quantities always sit on a right triangle: apparent power is the hypotenuse, real power is one leg, reactive power is the other.
Power factor is simply the ratio PF = kW / kVA = cos θ, so once real power and power factor are known, apparent power, reactive power and the angle between voltage and current all follow directly from that pair of numbers. That's the calculation the tool below runs on every keystroke.
Notice what the triangle does and doesn't do. Real power and reactive power can't just be added arithmetically to get apparent power, because they're separated by 90° rather than pointing the same direction. That's precisely why the relationship is Pythagorean rather than a simple sum, and why a nameplate that quotes kVA can't be read as "kW plus a bit more." It also means reactive power alone never tells you how big a load is; it only tells you how far that load's current has drifted out of step with its voltage.
Enter a real-power figure and a power factor below to see the other two corners of the triangle, and the angle θ between voltage and current, update together as a supporting reference for the formulas above.
- Apparent
- 53.33 kVA
- Reactive
- 35.28 kVAR
- Angle θ
- 41.4°
Seeing the angle: phasor and power-triangle diagrams
The left half of the figure below is a phasor diagram: it freezes the spinning voltage and current arrows at one instant and shows the angle θ between them, the same θ used in every formula above. The right half redraws that same angle as a power triangle, with the two legs now standing for real power and reactive power in kilowatts and kilovars, and the hypotenuse standing for apparent power in kilovolt-amps. Both panels are drawn at the worked example's angle, 41.41°, and the triangle's leg lengths are scaled to that example's actual kW, kVAR and kVA figures, so the geometry below matches the arithmetic in the next section exactly.
Worked example: a 40 kW load at 0.75 power factor
A 40 kW load (a motor-driven pump, say) running at a 0.75 power factor. Three steps get from the real-power nameplate figure to the full power triangle.
- 1
Apparent power: divide real power by power factor
kVA = kW / PF = 40 / 0.75 = 53.3333 kVA
- 2
Reactive power: the other leg of the triangle
kVAR = √(kVA² − kW²) = √(53.3333² − 40²) = 35.2767 kVAR
- 3
Phase angle: the angle whose cosine is the power factor
θ = cos⁻¹(PF) = cos⁻¹(0.75) = 41.4096°
The supply feeding this pump has to be sized for 53.33 kVA of current-carrying capacity, even though only 40 kW of it converts to shaft work. The remaining 35.28 kVAR is the magnetizing current the motor's windings pull back and forth on every cycle.
Real power, apparent power and reactive power compared
The same three quantities, side by side: what each one measures, how it's derived, and whether an ordinary meter reads it directly.
| Quantity | Symbol | Unit | What it measures | Formula | Read directly by a meter? |
|---|---|---|---|---|---|
| Real power | P | kW | Rate energy is actually converted to work: heat, light, torque | P = V × I × cos θ | Yes, a standard kWh meter integrates this over time |
| Apparent power | S | kVA | Total volt-amps the supply, wiring and transformer must be sized to carry | S = V × I | Sometimes, dedicated kVA-demand meters on larger commercial services |
| Reactive power | Q | kVAR | Power that oscillates into and out of a magnetic or electric field, doing no net work | Q = V × I × sin θ | Rarely on residential meters; needs a dedicated VAR meter |
Why the supply is sized in kVA, not kW
A cable, a breaker and a transformer all heat up according to the current actually flowing through them, and current is set by apparent power, not real power: I = S / V, with no phase angle in that relationship at all. Two loads with the same real-power rating but different power factors draw different currents, and the one with the lower power factor needs the larger conductor and the larger transformer, even though it "uses" the same number of kilowatts. That's why a service, a feeder or a distribution transformer is rated in kVA on its nameplate rather than kW, the rating describes the current-carrying job it has to do, and reactive current takes up exactly as much of that capacity as real current does.
The worked example above makes the gap concrete: a 40 kW load at unity power factor would need wiring and transformer capacity for 40 kVA. Drop the power factor to 0.75 and the same 40 kW of real work now rides on 53.33 kVA of apparent power: about a third more current-carrying capacity for a load that, from the customer's side of the meter, looks unchanged. Nothing about the useful output grew; the extra capacity exists purely to carry the 35.28 kVAR the motor's windings pull back and forth.
It's also why a utility can end up caring about a figure the customer never directly pays for. A residential kWh bill only charges for real power, because that's the energy actually consumed. Many commercial and industrial tariffs go further and add a kVA-demand charge or a low-power-factor penalty, because a low-PF customer forces the utility to size generation, transformers and feeders for more current than the customer's real-power usage alone would require, capacity that has to be paid for somewhere, even though it never shows up as billed energy. Correcting a low power factor, typically with capacitor banks that offset lagging kVAR, is a distinct topic covered on the power-factor page linked below; this page is about what the three quantities in that correction problem physically represent.
As general reference, none of this substitutes for an actual load study or a code-driven conductor and transformer sizing calculation: a licensed electrician or engineer accounts for demand factors, ambient temperature, harmonics and the applicable code sections that a phase-angle diagram alone doesn't capture.
Going deeper on power quality
Questions
Apparent, reactive and real power FAQ
Questions about what these three quantities physically are, distinct from correcting a low power factor once you've measured one.
Why is reactive power measured in kVAR instead of just watts, since the units are dimensionally the same?
Volt-amps and watts are dimensionally identical (both are just volts times amps), but kVAR is used specifically to flag that the quantity does no net work, so it can never be added directly to a kW figure. Labeling it kVAR instead of kW keeps the two figures from being summed as if they were the same kind of power; combining them correctly requires the Pythagorean relationship in the power triangle, not simple addition.
Can reactive power damage electrical equipment?
Not directly: since it transfers no net energy, reactive power by itself doesn't heat anything or do work, but it isn't harmless, because it adds to the current a conductor, winding or transformer has to carry. That extra current produces extra I²R heating everywhere it flows, which over time can accelerate insulation aging or trip thermal protection sized without it in mind. The damage mechanism is the added current, not the reactive power as such.
Why doesn't reactive power show up as usable energy on a utility bill?
Because a standard kWh meter integrates real power over time, and reactive power isn't energy being consumed: it's energy sloshing into the load's magnetic or electric field and back out again every cycle, netting to zero. Billing it as kWh would overstate what the customer actually used. Some commercial and industrial tariffs do add a separate kVA-demand or low-power-factor charge, but that's a distinct line item from the energy charge, reflecting capacity used rather than energy consumed.
Can apparent power ever be smaller than real power?
No. The power triangle makes this geometrically impossible, because apparent power is the hypotenuse of a right triangle whose legs are real power and reactive power. A hypotenuse is never shorter than either leg, so kVA = √(kW² + kVAR²) can only equal kW when kVAR is exactly zero, a perfectly resistive load at unity power factor. Any reactive component at all makes kVA strictly greater than kW.
What would a phase angle of 0° versus 90° actually mean?
A 0° angle means voltage and current peak at exactly the same instant (a purely resistive load, power factor 1.0, and reactive power of zero), so apparent power and real power are identical. A 90° angle means current is a quarter-cycle out of step with voltage: a theoretical purely reactive load, power factor 0, where real power delivered is zero even though current is still flowing and the source still has to carry it as apparent power.
How does reactive power differ between inductive and capacitive loads?
Inductive loads (motors, transformers, fluorescent ballasts) delay the current's peak behind the voltage's, a lagging angle that draws reactive power from the source. Capacitive loads, such as power-factor correction capacitors or a lightly loaded long cable, advance the current ahead of voltage instead, a leading angle that effectively supplies reactive power back toward the source. By convention lagging kVAR is treated as positive and leading kVAR as negative, which is why adding capacitance to an inductive load cancels part of its reactive demand rather than adding to it.
Does apparent power depend on the power factor, or only on voltage and current?
Apparent power depends only on RMS voltage and RMS current: kVA = V × I regardless of the phase angle between them. Power factor doesn't change how much apparent power exists; it changes how that fixed kVA figure splits between the real-power leg and the reactive-power leg of the triangle. A lower power factor means more of the same kVA shows up as kVAR instead of kW, not that kVA itself grows or shrinks.