# Calculating copper losses of transformer given iron loss and maximum efficiency

I wonder if anyone can help me understand copper losses in transformers a bit better.

I'm having problems understanding the highlighted part on the attached picture half way down the page. I sure its a simple maths thing which escapes me but why the $\left( \frac{1}{0.80} \right)^2$ ? Specifically whats inside the brackets?

• Load power could be seen as a function of current, because the voltage doesn't change. But the copper loss has the variable voltage too, with the current, with the same changing proportion. You know, power is voltage * current. Then the power proportion would be its square because of same proportions. – Ayhan Aug 25 '16 at 12:07

## 1 Answer

The iron losses are 3 kW regardless of load. The copper losses are equal to the iron losses when the load (current) is 80% of full load. Since copper losses are proportional to the square of current, to scale up from 80% load to 100% load multiply the losses at 80% load by the square of the 100/80 increase in load.

• Thank you Charles, I am nearly there. So if for example if the question asked for copper losses @30% load then what would go inside the brackets? (3/8)? – leckytech Aug 26 '16 at 12:07
• Yes. That is correct. – Charles Cowie Aug 26 '16 at 13:39