The ideal transformer of the last lesson passes every watt through. A real one keeps a little, as heat, in two places that behave completely differently. Telling them apart is the whole lesson.

The core loss, $P_{c}$, is hysteresis and eddy currents in the iron. It depends on the flux, the flux depends on the voltage, and the voltage does not change with load. So it is there from the moment the transformer is energised, and it stays constant all day whether anything is plugged in or not. The copper loss is $I^{2}R$ heating in the windings, so it follows the SQUARE of the current. Call the load fraction $x$: the load as a fraction of the nameplate rating, a bare number, with 1 for full load and 0.5 for half. If $P_{cu}$ is the copper loss at full load, the copper loss at load $x$ is $x^{2} P_{cu}$. Half load is a quarter of the copper loss, not half. Both losses are in watts, and both come off the test report: the open-circuit test reads $P_{c}$, the short-circuit test reads $P_{cu}$.

Put them under the output and you have the efficiency: $\eta = \dfrac{100\, x S \cos\varphi}{x S \cos\varphi + P_{c} + x^{2} P_{cu}}$, read aloud eta equals one hundred x S cos phi, over x S cos phi plus P-c plus x squared P-c-u. $\eta$ (eta) is the efficiency in percent, and it is what you are solving for. $S$ is the rated apparent power in volt-amperes, so watch the kilo on a kVA nameplate. $\cos\varphi$ is the load's power factor, a bare number written PF on the cards, and $x S \cos\varphi$ is the real power actually delivered, in watts. Notice what the power factor does: it shrinks the useful output without touching either loss, because the losses follow the current and the current is set by the kVA.

One consequence is worth carrying out of the room. Efficiency peaks where the two losses are equal, $x^{2} P_{cu} = P_{c}$, which gives $x = \sqrt{P_{c} / P_{cu}}$. Distribution transformers are designed to peak near half load, because that is where they spend their lives. Units guide, they do not confess: watts over watts is a bare ratio either way, and the units will never notice that you scaled the copper loss by $x$ when it wanted $x^{2}$.