Transformer losses and efficiency expression

Losses in a transformer

The first two together are the iron (core) loss $P_i$. Stray load loss and dielectric loss are small and neglected. Because there is no rotating part, there is no friction or windage loss.

Expression for efficiency

$$\eta = \frac{\text{output power}}{\text{input power}} = \frac{\text{output}}{\text{output} + \text{losses}}$$

For a load that takes a fraction $x$ of the full-load current, at power factor $\cos\phi$ and rated secondary voltage $V_2$:

$$\begin{aligned} \text{Output} &= x\,S\cos\phi\quad (S = V_2I_{2,fl}\ \text{rated VA}) \\ P_{cu} &= x^2P_{cu,fl} \\ \text{Losses} &= P_i + x^2P_{cu,fl} \end{aligned}$$ $$\boxed{\eta = \frac{xS\cos\phi}{xS\cos\phi + P_i + x^2P_{cu,fl}}\times100\%}$$

Remarks:

• Efficiency is zero at no load, rises with load, reaches a maximum when $P_i = x^2P_{cu,fl}$, and then falls.
• For a good power transformer $\eta$ is 95 to 99%.
• The efficiency is higher at a higher power factor, since the output power is larger for the same losses.