Transformer Efficiency from Core and Copper Losses

Transformers are the most efficient machines anyone builds — 98 or 99 per cent is ordinary — and the reason this page exists anyway is that the last one or two per cent is worth a great deal of money on a device that is energised every hour of its life. What makes the calculation interesting is that the two losses inside it behave in completely different ways.

Core loss, or iron loss, is hysteresis and eddy currents in the laminations. It depends on the flux, the flux depends on the applied voltage, and the applied voltage does not change with load. So it is there the instant the transformer is energised, it is the same at midnight with nothing connected as at the afternoon peak, and it is measured by the OPEN-CIRCUIT test: rated voltage on one winding, the other open, nothing but magnetising current flowing. Copper loss is $I^2R$ in the windings, so it goes as the square of the load — half load is a quarter of the copper loss, not half — and it is measured by the SHORT-CIRCUIT test: one winding shorted, the applied voltage raised until rated current flows, which happens at a few per cent of rated volts and therefore at negligible flux and negligible core loss. Both figures are on the test report of any transformer worth buying.

Put them together and the efficiency at a load fraction x is $\eta = xS\cos\phi / (xS\cos\phi + P_c + x^2P_{cu})$. A 100 kVA unit with 400 W of core loss and 1500 W of copper loss at full load runs at 98.14 per cent at unity power factor and full load. At half load and 0.8 power factor the copper loss drops to 375 W but the core loss is unchanged, and the efficiency comes out 98.10 per cent — almost identical, from a very different balance.

That near-coincidence is not luck. Differentiate and you find that efficiency peaks where the variable loss equals the constant one, $x^2P_{cu} = P_c$, or $x = \sqrt{P_c/P_{cu}}$ — here $\sqrt{400/1500} = 0.52$, just over half load. Distribution transformers are deliberately designed to peak there, because that is where they spend their lives, and it is why the utility's "all-day efficiency" — energy out over energy in across a full daily load cycle — is the figure that actually decides a purchase, not the number at full load.

Two things trip people up. The first is that the power factor scales the useful output and does not touch either loss: the same transformer carrying the same CURRENT at 0.7 power factor is markedly less efficient than at unity, because the losses are set by the current and the useful power is not. The second is that solving backwards for the load fraction has two answers. Efficiency rises from zero at no load, peaks, and falls again, so every efficiency below the peak is reached at two loadings — one light, one heavy — and which one you meant is a question the arithmetic cannot answer.