What are the losses in a single-phase transformer? Define its efficiency and state the condition for maximum efficiency.

Answer: Iron (core) losses, which are constant, and copper losses, which vary with load; \(\eta=\dfrac{\text{output}}{\text{output}+\text{losses}}\), maximum when copper loss = iron loss.

1. Iron or core loss \(P_i\) (constant at all loads, since flux is nearly constant):

• Hysteresis loss \(\propto B_m^{1.6}f\), due to repeated magnetisation reversal; reduced by using silicon steel.
• Eddy current loss \(\propto B_m^2f^2t^2\), due to circulating currents in the core; reduced by thin, insulated laminations.

2. Copper loss \(P_{cu}\) \(=I_1^2R_1+I_2^2R_2\) in the windings; varies as the square of the load current (at fraction \(x\) of full load, \(P_{cu}=x^2P_{cu,FL}\)).

Efficiency: \[\eta=\frac{\text{Output}}{\text{Output}+P_i+P_{cu}}\times100\%=\frac{x\,S\cos\phi}{x\,S\cos\phi+P_i+x^2P_{cu,FL}}\times100\%.\]

Maximum efficiency occurs when variable (copper) loss = constant (iron) loss, i.e. at load fraction \(x=\sqrt{P_i/P_{cu,FL}}\).