Efficiency and condition for maximum efficiency

Efficiency is the ratio of output power to input power for a transformer:

$$\eta = \frac{\text{output}}{\text{input}} = \frac{\text{output}}{\text{output} + P_i + P_{cu}}$$

where $P_i$ is the core loss (constant) and $P_{cu} = I_2^2R_{02}$ is the copper loss (varies as the square of the load).

For load current $I_2$, secondary voltage $V_2$ and power factor $\cos\phi_2$:

$$\eta = \frac{V_2I_2\cos\phi_2}{V_2I_2\cos\phi_2 + P_i + I_2^2R_{02}}$$

Condition for maximum efficiency

Dividing by $I_2$ gives

$$\eta = \frac{V_2\cos\phi_2}{V_2\cos\phi_2 + \dfrac{P_i}{I_2} + I_2R_{02}}$$

With $V_2$ and $\cos\phi_2$ constant, $\eta$ is maximum when the denominator is minimum:

$$\frac{d}{dI_2}\left(\frac{P_i}{I_2} + I_2R_{02}\right) = -\frac{P_i}{I_2^2} + R_{02} = 0$$ $$P_i = I_2^2R_{02}\quad\Rightarrow\quad \text{iron loss} = \text{copper loss}$$

Current at maximum efficiency

$$I_{2,\eta max} = \sqrt{\frac{P_i}{R_{02}}}$$

In terms of the full-load current $I_{2,fl}$ and full-load copper loss $P_{cu,fl} = I_{2,fl}^2R_{02}$:

$$I_{2,\eta max} = I_{2,fl}\sqrt{\frac{P_i}{P_{cu,fl}}}$$

The corresponding kVA load is $\text{kVA}_{\eta max} = \text{kVA}_{rated}\sqrt{P_i/P_{cu,fl}}$ and the maximum efficiency is

$$\eta_{max} = \frac{V_2I_2\cos\phi_2}{V_2I_2\cos\phi_2 + 2P_i}$$

Distribution transformers (on load all day) are designed with a small $P_i$ so that maximum efficiency occurs near the usual load, which is below full load.