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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 \$ ?

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  • \$\begingroup\$ 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}\$ \$\endgroup\$
    – skvery
    Mar 27 '20 at 14:07
  • \$\begingroup\$ Are you sure the model answer is correct? \$\endgroup\$
    – skvery
    Mar 27 '20 at 14:15
  • \$\begingroup\$ yes the question said that \$\endgroup\$
    – shineele
    Mar 27 '20 at 14:41
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Because \$P=I^2R\$ since Copper loss is resistive loss so it is proportional to current squared.

Half the loading means half the wattage which means half the current IF your voltage stays the same. After all, you changed the loading to reduce the wattage right? You didn't reduce the input voltage?

If you reduced the wattage such that the voltage was less but current is the same, then copper loss stays the same.

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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}\$.

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