Condition for maximum transformer efficiency

Efficiency of a transformer at load fraction $x$ (of full-load current), power factor $\cos\phi$:

$$\eta=\frac{x\,S\cos\phi}{x\,S\cos\phi+P_i+x^2P_{cu}}$$

where $S$ is the rated VA, $P_i$ the iron loss (constant) and $P_{cu}$ the full-load copper loss.

For simplicity, write it with load current $I_2$ and secondary voltage $V_2$ (assumed constant):

$$\eta=\frac{V_2I_2\cos\phi}{V_2I_2\cos\phi+P_i+I_2^2R_{2e}}$$

where $R_{2e}$ is the equivalent resistance referred to the secondary.

Condition for maximum. Divide numerator and denominator by $I_2$:

$$\eta=\frac{V_2\cos\phi}{V_2\cos\phi+\dfrac{P_i}{I_2}+I_2R_{2e}}$$

$\eta$ is maximum when the denominator is minimum. Since $V_2\cos\phi$ is constant, differentiate $\left(\frac{P_i}{I_2}+I_2R_{2e}\right)$ with respect to $I_2$ and equate to zero:

$$-\frac{P_i}{I_2^2}+R_{2e}=0 \ \Rightarrow\ P_i=I_2^2R_{2e}$$

Result: the efficiency is maximum when the iron loss equals the copper loss (variable loss).

Load at maximum efficiency:

$$x=\sqrt{\frac{P_i}{P_{cu}}}, \qquad \text{kVA at } \eta_{max}=\text{kVA}_{rated}\times\sqrt{\frac{P_i}{P_{cu}}}$$

and

$$\eta_{max}=\frac{xS\cos\phi}{xS\cos\phi+2P_i}$$