What is the difference between kW and kVA?
What is the difference between kW and kVA?
kW measures active power; kVA measures apparent power. A kilowatt describes the rate at which electrical energy is converted into heat, light or mechanical work; electrical active input power includes losses. A kilovolt-ampere describes apparent power associated with voltage and current. Power factor (PF) links the two.
kW = kVA × PF and kVA = kW ÷ PF. Therefore, “How many kW is 10 kVA?” has no single answer: at PF 0.8 it is 8 kW, and at PF 1 it is 10 kW. The second formula requires PF greater than zero.
The rate of electrical energy conversion. 1 kW = 1000 W.
Used to assess the voltage–current loading of sources and conductors. 1 kVA = 1000 VA.
The ratio of active to apparent power. It is dimensionless and differs from motor efficiency.
kW–kVA conversion calculator
Use the power factor for the load at the same operating point. Decimal points and commas are accepted. 0.8 is an example, not a universal value.
Try an example:
10 kVA × 0.8 = 8 kW
This tool converts power values; it does not complete generator or UPS sizing. Motor efficiency, starting current, harmonic loading and separate equipment kW/kVA limits are not assessed automatically.
The same kVA, different kW: why?
At an apparent power of 10 kVA, changing PF gives the following active power values. These are calculation examples, not a promise that one source can operate at every point. Its own kW limit still applies.
| Apparent power | PF | Active power |
|---|---|---|
| 10 kVA | 1 | 10 kW |
| 10 kVA | 0.9 | 9 kW |
| 10 kVA | 0.8 | 8 kW |
| 10 kVA | 0.6 | 6 kW |
In the reverse calculation, a load drawing 10 kW at PF 0.8 requires 12.5 kVA. This is the VA versus watts distinction expressed in units a thousand times larger.
kW, kVA and kvar: what does the power triangle show?
In circuits containing inductance or capacitance, energy can move back and forth between the source and magnetic or electric fields. Reactive power (Q), measured in var or kvar, describes this exchange. It is not interpreted as consumed active power, but it affects current and equipment loading.
The numerical difference between 10 kVA and 8 kW is not “2 kW of losses”. Reactive power in the sinusoidal example above is 6 kvar. When voltage or current waveforms are distorted, this simple triangle alone does not describe every power component.
Are PF and cos φ always the same?
PF = cos φ for sinusoidal voltage and current. Nonlinear loads such as rectifiers and variable frequency drives can draw harmonic currents. True power factor accounts for waveform distortion as well as phase displacement, so cos φ alone may not give the correct kW–kVA conversion. This distinction also matters when assessing harmonic filtering.
Why do generators, UPS systems and transformers need different checks?
Generators: a kVA rating does not remove the kW limit
A generator labelled 100 kVA / 80 kW may be rated at PF 0.8 under its specified operating conditions. Connecting a PF 1 load does not automatically turn that set into a 100 kW source: its 80 kW active power limit still applies. When assessing power capacity, keep continuous, prime and standby ratings separate. Generator selection also considers load steps, environmental conditions and manufacturer limits.
UPS systems: respect both kW and kVA limits
Consider an example 10 kVA / 9 kW UPS. An 8 kW load at PF 0.7 requires 8 ÷ 0.7 ≈ 11.43 kVA: the kW value looks acceptable, but the kVA limit is exceeded. A 9.5 kW load at PF 1 draws 9.5 kVA; this time the kVA value looks acceptable, but the kW limit is exceeded.
The UPS type and permitted load characteristics require separate consideration. UPS input power factor is not the power factor of its connected output load. Battery runtime cannot be determined from the kVA rating alone either.
Transformers: kVA is not an energy consumption figure
A 100 kVA rating on a transformer does not mean it continuously consumes 100 kW. At a secondary loading of 100 kVA and PF 0.8, the load receives 80 kW of active power; transformer losses add to the input requirement. Harmonics, temperature and cooling conditions can affect usable capacity.
Where does motor efficiency enter the calculation?
An electric motor’s kW nameplate rating usually describes shaft output power. For example, 7.5 kW shaft power, 90% efficiency and PF 0.85 give an electrical active input of 7.5 ÷ 0.90 ≈ 8.33 kW and apparent input of 7.5 ÷ (0.90 × 0.85) ≈ 9.80 kVA.
Efficiency = output kW / input kW; PF = input kW / input kVA. These are different ratios. Do not apply efficiency again if you already have measured electrical input kW. Also, motor starting current requires an assessment beyond the steady-state calculation.
How does kVA relate to current?
For a single-phase circuit, use S(kVA) = U × I ÷ 1000. For balanced, sinusoidal three-phase operation, use S(kVA) = √3 × Uline × Iline ÷ 1000. Voltage is in volts and current in amperes, both RMS values. The three-phase formula uses line-to-line voltage.
For example, 10 kVA on a balanced 400 V three-phase supply corresponds to approximately 14.43 A line current. A 10 kW load at PF 0.8 on the same supply requires 12.5 kVA and approximately 18.04 A. See calculating current from power for more examples.