Transformer Percent-Impedance Voltage Drop

The same percent impedance that sets fault current also sets the voltage a transformer loses inside itself. By definition, at full load the internal drop is exactly %Z of rated volts, and the drop scales with loading. So a 1000 kVA, 480 V transformer with 5.75% impedance loses $0.0575 \times 480 = 27.6$ V of its own winding at nameplate load, and 13.8 V at half load. That is why the secondary of a heavily loaded transformer sits low and the tap changer exists.

Treat this as an upper bound rather than a prediction, because it adds the resistive and reactive drops arithmetically when they are really at right angles to each other and to the load current. The true drop depends on power factor: at unity PF a mostly reactive impedance costs far less than this suggests, while at 0.8 lagging the resistive and reactive parts line up closer with the load and the answer here is close to right. The exact form is $V_d = I(R\cos\varphi + X\sin\varphi)$.

The counter-intuitive part is that a leading power factor can make the drop go the other way — a lightly loaded feeder with capacitor banks still switched in can push the secondary above nominal, which is a real problem on rural circuits and on distribution feeders with a lot of solar. High impedance is not simply a defect either: utilities sometimes specify a higher %Z deliberately to hold fault current down to what the existing switchgear can interrupt, and accept the poorer regulation as the price.