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Loss vs Load (0–125%)
Iron loss Pfe (constant)
Copper loss Pcu(x) (∝ x²)
Total loss
A transformer's total loss has two very different characters. Iron (core) loss Pfe — made of hysteresis and eddy-current loss in the laminated core — depends only on the applied voltage and frequency, so it is present continuously and stays essentially constant regardless of load, even at no load. Copper loss Pcu is I²R heating in the windings, so it is zero at no load and grows with the square of the load current: Pcu(x) = x²·Pcu(rated).
| Quantity | Formula |
|---|---|
| Iron loss (any load) | Pfe = constant (hysteresis + eddy current) |
| Copper loss at load x | Pcu(x) = x² × Pcu(rated) |
| Total loss | Ptotal(x) = Pfe + Pcu(x) |
| Losses equal at | x* = √(Pfe / Pcu(rated)) |
Because copper loss is a parabola (x²) and iron loss is a flat line, they cross at x* = √(Pfe/Pcu) — the same load point where efficiency peaks (see the Efficiency & Regulation calculator). Understanding the two loss types separately helps in diagnosing where energy is wasted: iron loss is a "standing charge" whenever the transformer is energised, while copper loss only appears when current flows.
It is the energy lost in the laminated steel core from hysteresis (repeated magnetisation) and eddy currents. It depends on the applied voltage and frequency, not on the load current, so it is present continuously whenever the transformer is energised.
Copper loss is I²R heating in the primary and secondary windings caused by their resistance. It is zero at no load and grows with the square of the load current, so it dominates at high load.
With an open-circuit test: apply rated voltage to one winding with the other open. The input power measured (with negligible current in the open winding) is essentially the iron loss, since there is almost no copper loss.
With a short-circuit test: short one winding and apply just enough reduced voltage to circulate rated current. The input power measured is essentially the copper loss at that current, since core flux (and hence iron loss) is very low.
Power dissipated in a resistance is P = I²R. Since current scales linearly with load fraction x, copper loss scales as x² — doubling the load current quadruples the copper loss.
It stays essentially unchanged, because iron loss depends on the applied voltage and frequency (usually held constant), not on the load current. A transformer at no load still dissipates its full iron loss.
At load fraction x* = √(Pfe/Pcu(rated)). This is also the load at which the transformer's efficiency is at its maximum.
Use thinner laminations and low-loss (grain-oriented or amorphous) core steel to reduce hysteresis and eddy-current loss. This is a design/material choice, not something the user can change after purchase.
Use larger-gauge (lower-resistance) winding conductors, shorter winding lengths, and better conductor packing. In operation, running the transformer nearer its optimum load (not chronically overloaded) limits copper loss.
Yes. Both core and copper losses ultimately appear as heat in the transformer, which must be removed by natural or forced cooling to keep the windings and insulation within their temperature rating.
Distribution transformers are energised 24/7 even when lightly loaded, so their iron loss is a continuous, unavoidable energy cost across the whole grid, which is why efficiency standards specifically regulate no-load loss.
This tool shows the loss breakdown directly; the Efficiency & Regulation calculator uses the same Pfe and Pcu figures to compute efficiency at each load and find the same maximum-efficiency crossing point.
Efficiency & Regulation • VA / kVA Sizing • Turns (Faraday's Law) • All Calculators