I have a question considering the transformer's short-circuit test and the impedance (or short-circuit voltage). The transformer's general data is: Rated power: 50kVA Primary voltage: 10000V (10kV) Secondary voltage: 420V (0,42kV) Primary current: 2.89A Secondary current: 68.73A Vector group: Yzn5 No-Load Losses: 190W Load-Losses: 1100W The two transformers were tested by all types of routine tests: Voltage turns-ratio test Windings resistance test Open-circuit test Short-circuit test Dielectric tests All tests were clear to me, but the short-circuit test. The winding resistance test, at some temperature (13°C) gave us the low-voltage and high-voltage transformer winding resistance. The first transformer had RABC = 41.74Ohm's (RABC = (RAB + RAC + RBC) / 3) on the primary windings, and the Rabc = 54.03miliOhm's (Rabc = (Rab + Rac + Rbc) / 3) on the secondary windings. The secondary transformer had RABC = 41.15Ohm's (RABC = (RAB + RAC + RBC) / 3) on the primary windings, and the Rabc = 58.58miliOhm's (Rabc = (Rab + Rac + Rbc) / 3) on the secondary windings. We used these values to calculate the Copper-Losses for both transformers, by using the formula PCu = 1.5*(RABC*I1N2 + Rabc*I2N2). The I1N and the I2N are the rated currents on the primary and the secondary side. The calculated Copper-Losses for the first transformer were 906W. The calculated Copper-Losses for the second transformer were 931W. The short-circuit test is done by short-circuiting the low-voltage side of the transformer and applying the test voltage on the high-voltage side, and slowly increasing this voltage, until the rated current (or the short-circuit current) is not measured on the high-voltage side. We record then this voltage and call it short-circuit voltage. We record this current and call it short-circuit current (this should be the rated secondary current). We record the losses and call them the load losses. On the first transformer we reached the rated secondary current of 68.73A at 396V, and recorded losses of 1060W. On the secondary transformer we reached the rated secondary current of 68.73A at 365V, and recorded losses of 800W. Here is the issue !!! We calculated the Copper-Losses for the first transformer at 906W and we measured the 1060W. The difference from the 906W up to the 1060W are the extra losses of the 154W, which we find OK. The calculated impedance (or the short-circuit voltage) for the first transformer is 3,96%. We calculated the Copper-Losses for the second transformer at 931W and we measured the 800W. The difference from the 931W up to the 800W are the extra losses of the -131W, which we find NOT OK. The calculated impedance (or the short-circuit voltage) for the second transformer is 3,65%. What does it mean? How is this possible? The only technical difference between these two transformers is the cross-section of the low-voltage Copper wire. The cross-section of the low-voltage Copper wire for the first transformer is 24mm2. The cross-section of the low-voltage Copper wire for the second transformer is 21mm2. The cross-section of the high-voltage Copper wire is the same for both transformers 0.78mm2. So, can this difference in the low-voltage Copper wire cross-section be responsible for the issue with the calculated and the measured data, for the second transformer? Thank You for Your answers !!!