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According to basic electrical machinery theory, all transformers experience power losses due to winding resistances, eddy currents, leakage flux and hysteresis. So when a load is connected to, say, a transformer with a 13800/240 V rating, the output voltage of a transformer is slightly less than 240 V, for example 225 V.

My question is, when a real transformer has a rating of, for example, 13800/240 V etched on its plate, is this rating just the ideal rating based on the turns ratio N1/N2, or is the transformer built with a slightly different turns ratio to compensate for losses, so that at full load it ACTUALLY delivers 240V?

Also, the same thing for the power rating etched on the plate. Is it the ideal rating or does it take into account losses at full load?

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Ideal transformer - transformation ratio $$\alpha = \frac {V_1} {V_2} = \frac {N_1} {N_2} = \frac {I_2} {I_1}$$ Apparent Power: $$S_1 = S_2$$ Real transformer: $$\alpha = \frac {V_1} {V_2} = \frac {N_1} {N_2}$$ $$S_1 > S_2$$

The kVA rating on transformers deals with the output of the secondary side or $S_2$. There has to be a greater input power (and current) on the primary side ($S_1$) to deal with losses. To get full-load 100kVA at 95% efficiency means 105.26kVA must be put in. The nameplate would show 100kVA. This is the full-load output of the transformer. Actual kVA output will depend upon load.

$S_1$ must be greater than $S_2$ (100kVA) to deal with losses inside the transformer. Since $V_1$ cannot increase, greater $S_1$ means $I_1$ must increase.

13800/240 V means that when you apply 13800V at 60Hz to the primary, you get 240V at the secondary. You would measure 240V at the terminals of the secondary.

The secondary feeds a load. Assume voltage at load is 225V. This is external to the losses of the transformer.

15V was lost to the wires. This voltage loss depends upon current (load) and wire size (cross-sectional area) and length of feeder. Larger current, smaller wire and longer feeder means greater voltage loss.

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