# What is these factor multiplying the current of the copper wires_?

I was reading some solved problems about transformers but I don’t understand how was calculated the losses in the copper.

- Enunciate: A one phase, 10kVA, 2400/240, 60Hz distribution transformer has the following characteristics:
- Core loss at full voltage=100W
- Copper loss at half load=60W
Determine the efficiency of the transformer when it delivers full load at 0.8 power factor lagging.

• Solution:
P=10*0.8=8kW

Pcore=100W

PCU,FL (Full load)=60($**2^2**)=240W$

The Efficiency is then
(8000/(8000+100+240))*100=95.92%

Not a big deal, but why the copper loss at half load was multyplied by $2^2$? What is this?

• Because the copper loss was specified at half load. Full load = 2* the current = 2**2 power. – Brian Drummond May 21 '18 at 19:40
• Because copper losses are from a constant resistance, full load means two times the current, and the power over a resistor is P=I²*R. – Janka May 21 '18 at 19:40
• Ok, I got the idea – riccs_0x May 21 '18 at 19:52