Transformer efficiency is the share of input power that reaches the load. The rest is lost as heat in the core and the windings. The basic formula is simple, but the loss-based formula is more useful, because it shows how efficiency changes with load.
The basic formula
Efficiency = output power / input power x 100%. Example: a transformer receives 52.5 kW and delivers 50 kW. Efficiency = 50 / 52.5 x 100 = 95.24%. Total losses = 52.5 – 50 = 2.5 kW. That is all you need when both powers are measured.
Power transformers are very efficient, typically 97 to 99% at full load for larger units. Small transformers are less efficient, and efficiency falls at light loads, which is where the loss method matters.
The loss method
Input = output + iron loss + copper loss. Iron (core) loss comes from hysteresis and eddy currents in the core, and it is roughly constant whenever the transformer is energised, whatever the load. Copper (winding) loss comes from resistance in the windings and grows with the square of the load current.
Efficiency = output / (output + iron loss + copper loss) x 100%. Copper loss at load fraction x is: copper loss at full load x x squared. At half load, copper loss is one quarter of full-load copper loss.
Worked example at partial load
A 100 kVA transformer, power factor 0.8, iron loss 800 watts, full-load copper loss 1,800 watts, running at 50% load. Output = 0.5 x 100 x 0.8 = 40 kW. Copper loss at half load = 1,800 x 0.25 = 450 watts. Total losses = 800 + 450 = 1,250 watts = 1.25 kW. Efficiency = 40 / (40 + 1.25) x 100 = 96.97%.
At full load the same transformer delivers 80 kW with losses of 800 + 1,800 = 2,600 watts, giving 80 / 82.6 = 96.85%. Efficiency is slightly higher at half load here, because the copper loss falls faster than the output.
Maximum efficiency and all-day efficiency
Maximum efficiency happens at the load where iron loss equals copper loss: load fraction = square root of (iron loss / full-load copper loss). For our transformer: square root of (800 / 1,800) = 0.667, so peak efficiency is at about 67% load. Designers pick the core and windings to place that peak near the expected average load.
All-day efficiency is different: it compares total energy out over 24 hours with total energy in, including the iron loss during idle hours. A distribution transformer energised all day but loaded only part of the day is judged on all-day efficiency, not on its full-load figure. That is why low iron loss matters more than peak efficiency for such transformers.
Ratings, cooling, and overloading
A transformer’s kVA rating is its apparent power capacity, not its real power output. A 100 kVA transformer at 0.8 power factor delivers 80 kW of real power. Loading beyond the kVA rating overheats the windings and shortens the transformer’s life, even if the efficiency figure still looks fine.
Cooling sets the real limit: oil-natural-air-natural (ONAN) transformers rely on natural circulation, while forced-air or forced-oil cooling raises capacity. Short-term overloading is sometimes allowed when the transformer has been running cool, but sustained overloading is a failure risk, not a free capacity boost. Size from the rating plate and the load profile, not from the efficiency curve.
Try it faster with the tool
You can follow every step above by hand, but the free Transformer Efficiency Calculator does the arithmetic for you. Enter your numbers and it gives the answer right away, so you can check your hand calculation or skip straight to the result.
Quick tips
- Efficiency = output / input x 100%.
- Iron loss is constant; copper loss grows with load squared.
- Peak efficiency is where iron loss equals copper loss.