Transformer efficiency is basically the measure of how much is the power transformed from the primary to the secondary of the transformer. Transformer with a lower losses is more efficient. Transformer losses can be segregated into two components, the core loss or the no load losses, and the copper loss or the load losses. Core loss is, approximately, constant what ever is the amount of load is while the copper loss is the opposite.

Below is the mathematical proof why transformer has the highest efficiency if the core loss or the no load losses is equal to the copper loss or load losses;

Let,
Plt is the total losses
Pnl is the no load losses, and
Pll is the load losses.
%Plt is the percentage of the total losses with respect to the input load.
Pin is the total input power

Then,

Plt = Pnl + Pll

And,

%Plt = (Pnl + Pll)/Pin = (Pnl/Pin) + (Pll/Pin)

Note that,

Pll = I^2 x R

Thus,

%Plt = (Pnl/Pin) + [(I^2 x R)/Pin]

On the other hand,

I = [(Pin/(3xpf))/(Vxsqrt(3)] = [(sqrt(3) x Pin)/(3 x V x pf)] , this is assuming that the voltage is line to line and the transformer is connected Wye to Wye. (Note: pf is the power factor and sqrt(3) means “square root of 3″)

Expanding the equation for the total losses including the equation for I,

Plt = Pnl + [(sqrt(3) x Pin)/(3 x V x pf)] ^2 x R

Let k = [sqrt(3) /(3 x V x pf)] ^2 x R

Thus,

Plt = Pnl + (k x Pin^2)

And

%Plt = (Pnl/Pin) + (k x Pin^2)/Pin = [Pnl + (k x Pin^2)]/Pin

For optimum value of %Plt, the derivative of %Plt with respect to input power Pin must be equal to zero.

D(%Plt)/dPin = [Pin x (2 x k x Pin) - ((k x Pin^2) +Pnl)]/Pin^2 = 0

D(%Plt)/dPin = (2 x k x Pin^2) - (k x Pin^2) - Pnl = 0

Simplifying the above equation will give the value of Pin where the amount of transformer losses is at minimum.

Pin = sqrt(Pnl/k) , the value of Pin where the amount of transformer losses is at minimum.

Therefore, the minimum value of the percentage of losses will be,

%Plt = {Pnl + [k x (sqrt(Pnl/k))^2]} / sqrt(Pnl/k)

After simplifying,

%Plt = [Pnl/(srt(Pnl/k)] + [Pnl/(srt(Pnl/k)]

That’s it, the percentage of transformer losses is at minimum if the no load losses and load losses are equal.