kVA to kW calculator
Converting apparent power to active power through the power factor.
Fill in the fields and the result will appear here automatically.
Convert apparent power S in kVA and active power P in kW using power factor PF. This models a load, not a generator's or UPS's guaranteed output: equipment may have separate VA and W limits. The additional reactive power Q applies to a sinusoidal steady state, where PF = cos φ. With distorted current, the triangle does not identify all non-active power as Q.
How it works
Formula and logic
P = S·PF; S = P/PF. For sinusoidal voltage and current, Q² = S² − P², so |Q| = S·√((1−PF)(1+PF)). This is not S−P. Inductive or capacitive sign cannot be obtained without phase information. At PF = 1, P = S and Q = 0.
Example
10 kW at PF = 0.8 corresponds to 12.5 kVA and 7.5 kvar. A 5 kVA load at PF = 0.8 gives 4 kW and 3 kvar, not 1 kvar. At PF = 1, the same 5 kVA corresponds to 5 kW.
Fields and units
- What to find — list option
- Active power — kW
- Apparent power — kVA
- Power factor (cos φ) — 1
How to use
- — Choose the unknown kVA or kW and fill the one visible known-power field.
- — Enter nonnegative power and PF greater than 0 and at most 1.
- — Use measured or specified load PF; the example's 0.8 is not a universal household value.
- — Read Q as a magnitude in kvar. Check equipment W and VA limits separately.
Method and limitations
- Calculation method
- Formula and logic
- Data or methodology source
- OpenStax: average power, RMS values and cos φ
- Limitation
- Nonnegative-power sinusoidal load model. Harmonics, Q sign, protection sizing and actual generator or UPS output capability are not calculated.
FAQ
Does 5 kVA always mean less than 5 kW?
Numerically P = S·PF. At PF = 1, kW and kVA values coincide; below 1, active power is smaller. This does not guarantee source capability: separate equipment limits still apply.
What if the power factor is unknown?
Measure it or obtain the actual load specifications. PF depends on operating conditions and waveform. An assumed 0.8 produces a scenario, not a verified conversion for equipment.
Why isn't reactive power the remaining kVA?
In a sinusoidal regime the powers form a triangle: S² = P² + Q². Watts and vars cannot be linearly subtracted from volt-amperes; 5 kVA with 4 kW gives a Q magnitude of 3 kvar.
Is zero PF physically impossible?
No. An ideal purely reactive load can have PF = 0 and P = 0. This page requires PF > 0 so S = P/PF is determinate. Zero power with positive PF is accepted as an algebraic zero case.