How it works
A transformer has two kinds of loss. The iron loss (hysteresis and eddy currents in the core) depends on the voltage and stays the same from no load to full load. The copper loss in the windings depends on the current, so it grows with the square of the load.
Efficiency is output divided by input. Because the copper loss changes with load, the efficiency changes too, and it is highest at the load where the copper loss equals the iron loss. That load is x = √(Pi ÷ Pcu) times the full load.
A distribution transformer that spends most of its day lightly loaded is therefore designed with a smaller iron loss relative to its copper loss, so that the maximum efficiency falls near its usual load.
Formulas used
- Output = x × kVA × 1000 × pf (x = load as a fraction)
- Copper loss at load x = x² × full-load copper loss
- Efficiency η = output ÷ (output + iron loss + copper loss) × 100
- Maximum efficiency at x = √(iron loss ÷ full-load copper loss), where iron loss = copper loss
Worked example
100 kVA at pf 0.8, full load, iron loss 300 W, copper loss 1500 W: output 80 kW, total loss 1.8 kW, efficiency = 80 ÷ 81.8 = 97.8 %. Maximum efficiency occurs at √(300 ÷ 1500) = 44.7 % of full load, about 98.35 %.
The calculator opens with these example values, so you can compare its answer with the text above.
Frequently asked questions
What is the efficiency of a transformer?
It is the output power divided by the input power. Large transformers reach 98 % to 99.5 % because they have no moving parts.
At what load is transformer efficiency maximum?
When the copper loss equals the iron loss, at a fraction √(iron loss ÷ full-load copper loss) of full load.
Does power factor change efficiency?
Yes. The same kVA at a lower power factor delivers fewer kW of output while the losses stay about the same, so the efficiency drops.
Where do I find the iron and copper loss?
On the nameplate or in the manufacturer’s test report (no-load test gives the iron loss, short-circuit test gives the copper loss at full load).