# Why should we multiply Copper loss with $(\frac{1}{2})^2$, not $(\frac{1}{2})$ when the transformer is in half loading?

A 10 kVA transformer, its iron loss is 120 W, the copper loss is 240 W (full loading), now when the load factor is 0.9 delay, what is the efficiency when the transformer is in half loading?

The answer said $$\\eta=\frac{\frac{1}{2}*10K*0.9}{\frac{1}{2}*10K*0.9+120+\frac{1}{2}^2*240}\$$.

I want to ask why should we multiply copper loss with $$\ (\frac{1}{2})^2 \$$, not $$\(\frac{1}{2})\$$.

The unit of copper loss is Watt,this means it should have a relation with real power,that is $$\IVcos\theta\$$, and the real power has a relation with resistor, and as we know,the relation between resistance and power is $$\P=IV=I^2R=V^2/R\$$, so I wonder does $$\ (\frac{1}{2})^2 \$$ have relation with $$\P=IV=I^2R=V^2/R\$$? If not,can anyone tell me the reason of multiplying Copper loss with $$\ (\frac{1}{2})^2 \$$ ?

• just a side note about load factor and power factor. The power factor is normally defined as $\cos(\theta)$ and the load factor the $P_{average}/P_{peak}$ Mar 27 '20 at 14:07
• Are you sure the model answer is correct? Mar 27 '20 at 14:15
• yes the question said that Mar 27 '20 at 14:41

Because $$\P=I^2R\$$ since Copper loss is resistive loss so it is proportional to current squared.
The $$\I^2R\$$ loss is $$\240\ W\$$ at defined at full load of $$\10\ k\mathit{VA}\$$.
The $$\I^2R\$$ loss at half load of $$\5\ k\mathit{VA}\$$ is $$\{240\ W\over{2^2}} = 60\ W\$$ irrespective of the power factor and the voltage loss remains at $$\120\ W\$$.
The active power efficiency is therefore $$\{90\ \%\cdot 5\ kW}\over{90\ \%\cdot 5\ kW + 120\ W + 60\ W}\$$.