Transformers and power distribution

[Anonymous user]

I need some clarification on transformers and power distribution. The transformer is marked 75 4800, a second unit 167 4800 and a third unit 25 4800.

I am assuming this is 167 kVA (or 75 or 25) and the primary voltage is 4800. Am I correct?

Since our secondary voltage is 120VAC and there is 3-wires coming from the transformer, am I correct to assume this is a center tap and we have 1 phase 2 poles? If this is correct and the transformer is in fact 25kVA, does that mean each pole has 25kVA or the whole transformer is 25kVA and each pole effectively delivers 1/2 the rated value. Thus at 120VAC does each pole have 208 A or 104 A.

[mjk68 · Accepted solution]

There are some unknowns here, so the below are assumptions with the given values. With the secondary voltage of 120 and the markings of 25, 75, and 167 - I would assume as you have stated that these are the kVA ratings and the 4800 is more than likely the primary voltage. Additionally, the 3-wires are either a center tapped winding, or a 2-winding transformer connected internally and only 1 connection lead brought out of the transformer. In either case the secondary voltage would be 120/240 based on your ampacity values stated above. The formula to calculate Full-Load Current with Single-Phase Transformer is = kVA x 1000 / V

Using the 25kVA transformer as an example:

25 x 1000 / 120 = 208 secondary amps or
25 x 1000 / 240 = 104 secondary amps

[Anonymous user]

Thanks for the response, and your assumptions where as mine.

However, the confusing part for me is the panel box.  There is a 200A 2-pole main breaker.  So does each pole have 200A or is it 100A?

[mjk68 · Accepted solution]

200A would be the overall circuit ampacity and not a 100A per pole additive equation statement. The measured voltage of 240 volts would be between each pole or phase and the measured voltage of 120 volts would be from either phase to the center-tap wire. If all the connected loads were balanced the measured amps would be equal.