Calcylator
Transformer apparent power

Transformer rating:
why kVA, not kW, is the number on the plate

Learn why transformers are rated in kVA, how to convert measured volts and amps into that rating, and how much margin to keep.

Calcylator Editorial Team

Updated · 5 min read

Why a transformer is rated in kVA

Look at the nameplate of a transformer and you will see kVA, not kW. The reason is practical. The windings and the core heat up because of the current flowing through them and the voltage across them. Neither of those depends on whether the load turns that current into useful work. A motor with a poor power factor and a heater with a perfect one can draw the same current and heat the transformer the same amount, even though one delivers fewer kilowatts.

Apparent power, measured in volt-amperes (VA) or kilovolt-amperes (kVA), is the product of the RMS voltage and the RMS current. It is the quantity that sizes copper, insulation and cooling. Real power in kW comes after, once you apply the power factor of the load.

This also explains why a generator, a UPS or an inverter is rated in the same way. Whenever a device must carry the full current of a load regardless of the useful work that current performs, its limit is expressed in volt-amperes. Real power, in watts, is the part that spins a shaft or warms a room.

For household sizing, the distinction matters most when motors and compressors are involved. Their power factor is often noticeably below one, so a 1 kW motor may demand more than 1 kVA from the supply.

The single-phase calculation

Apparent power (single-phase) =V × I ÷ 1,000
V:
RMS voltage across the load in volts
I:
RMS current in amperes
kVA:
Result in kilovolt-amperes
Do not include power factor here; apparent power is voltage times current by definition.

To go the other way, divide the rating by the voltage. A 5 kVA unit working at 230 V can carry 5,000 ÷ 230 = 21.7 A before it reaches its rated current. That is the figure to compare with a clamp-meter reading.

If you know real power instead, divide it by the power factor to get apparent power. A load drawing 3.68 kW at a power factor of 0.8 needs 3.68 ÷ 0.8 = 4.6 kVA, the same as our measured example.

Primary and secondary sides can be checked separately. A 5 kVA, 230 V to 115 V unit has a rated secondary current of 5,000 ÷ 115 = 43.5 A and a rated primary current of 21.7 A. The larger current on the low-voltage side is the reason its conductors are thicker.

Always use the voltage that is actually present. A nominal 230 V supply can sit a few percent either side, and the computed kVA moves in step with it.

Worked example: 230 V and 20 A against a 5 kVA unit

  • Voltage

    230 V

  • Current

    20 A

  • Transformer rating

    5 kVA

Apparent power

4.6 kVA (92% loaded)

230 × 20 = 4,600 VA = 4.6 kVA. Loading = 4.6 ÷ 5 = 0.92. At a power factor of 0.8, real power is 4.6 × 0.8 = 3.68 kW.

The transformer is within its rating, but only 8% headroom is left. A motor starting current or a new appliance could push it over. Many designers aim to keep continuous load nearer 70 to 80% of the rating so the unit runs cooler and has room for growth.

Measure at the busiest time of day, not at an arbitrary reading. A snapshot at noon on a quiet Sunday can understate the evening peak by a wide margin.

To see how much a future addition would cost, work in current. The spare capacity is 5 kVA − 4.6 kVA = 0.4 kVA, which at 230 V is 0.4 × 1,000 ÷ 230 = 1.74 A. That is less than the current of a small fan heater, so this transformer cannot take any significant new load without upgrading.

kVA, kW and power factor side by side

QuantitySymbol and unitWhat it representsHow it is found
Apparent powerS, in kVATotal burden on the supply and transformerV × I ÷ 1,000
Real powerP, in kWPower turned into useful work or heatS × power factor
Reactive powerQ, in kvarPower swapped back and forth with inductive or capacitive parts√(S² − P²)
Power factorcos φ, 0 to 1Share of apparent power that does real workP ÷ S

For the worked example, S is 4.6 kVA, P is 3.68 kW and Q follows from the relationship above: √(4.6² − 3.68²) = 2.76 kvar. The three quantities form a right-angled triangle, with S as the longest side.

Picking a size with margin

  1. List each load with its voltage, current and power factor, or use VA ratings from the nameplates.
  2. Add the apparent powers of loads that run together. For mixed power factors, add real and reactive components separately, then combine.
  3. Allow for starting surges of motors and for growth you expect over the life of the installation.
  4. Select the next standard size above the result, and check the transformer's temperature and duty class for your conditions.

Standard ratings step up in preset sizes, so you will usually end up with a little more than you strictly need. That spare capacity is not wasted; it lowers winding temperature and extends insulation life, and it leaves room for future loads.

Ambient temperature matters as well. Ratings assume a stated ambient, often 40 °C. A unit in a hot, poorly ventilated enclosure may need derating, which is another reason to leave some room. Ask the manufacturer for the derating curve if the installation is unusual.

Efficiency losses are small by comparison with the rating but not zero. A typical distribution transformer converts a high share of its input to output, and the remainder appears as heat that the enclosure must shed.

Common slips to avoid

  • Using the transformer's primary and secondary voltages interchangeably. The kVA figure is the same on both sides in an ideal transformer, but the currents differ, so use the voltage that matches the current you measured.
  • Applying this single-phase formula to a three-phase supply. A three-phase system carries √3 times the line voltage and line current.
  • Mistaking peak or instantaneous current for RMS current. Meters usually report RMS, and the formula expects RMS.
  • Forgetting harmonic loads. Electronics and drives distort current, and the true RMS current can be higher than a simple reading suggests.

A basic power calculator that multiplies voltage and current is enough for the arithmetic here, as long as you take the voltage and current from the same point in the circuit.

Common questions

How do you calculate transformer kVA?

For a single-phase transformer multiply the voltage by the current and divide by 1,000. At 230 V and 20 A, that is 230 × 20 = 4,600 VA, or 4.6 kVA. Power factor is not needed for this step.

Why are transformers rated in kVA and not kW?

Heating in the windings and core depends on voltage and current, not on how much of the power does useful work. Because the power factor depends on the load, the manufacturer rates the transformer in kVA instead.

How do I convert kVA to kW?

Multiply kVA by the power factor. A 4.6 kVA load at a power factor of 0.8 delivers 4.6 × 0.8 = 3.68 kW. If the power factor is unknown, assume a realistic value rather than 1.

What current can a 5 kVA transformer supply at 230 V?

Divide 5,000 VA by 230 V to get about 21.7 A. That is the rated current. Running continuously at that level leaves no margin, so many designers keep the continuous load lower.

Can I load a transformer to 100% of its kVA rating?

Within its rating it can supply that load, but continuous operation at full load runs it hot and shortens insulation life. Keeping typical load somewhat lower leaves room for surges, harmonics and ambient heat.

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