In one line
Transformer efficiency is output power divided by input power, and it is maximum at the load where variable copper loss equals constant iron loss.
Explained simply
Efficiency answers "of the power I put in, how much comes out usefully?" Because a transformer has no moving parts, it is one of the most efficient machines made, typically 95–99%. At very light loads, the fixed iron loss is a big fraction of the small output; at very heavy loads, copper loss grows rapidly. In between there is a best point.
Technically, \( \eta = \dfrac{\text{output}}{\text{output} + \text{losses}} \). For a load fraction \( x \) of full-load rating \( S \) (VA) at power factor \( \cos\phi \), output \( = xS\cos\phi \), losses \( = P_i + x^2P_{cu} \).
How it works
- Write \( \eta = \dfrac{xS\cos\phi}{xS\cos\phi + P_i + x^2P_{cu}} \).
- Divide top and bottom by \( x \): \( \eta \) is maximum when \( P_i/x + xP_{cu} \) is minimum.
- Differentiate with respect to \( x \) and equate to zero: \( -P_i/x^2 + P_{cu} = 0 \).
- So \( x^2P_{cu} = P_i \): copper loss = iron loss.
- Load for maximum efficiency: \( x = \sqrt{P_i/P_{cu}} \) times full load.
- For the same kVA, efficiency is higher at unity power factor than at lower power factor.
Key formula
\( \eta = \dfrac{xS\cos\phi}{xS\cos\phi + P_i + x^2P_{cu}} \times 100\%, \qquad \eta_{max} \text{ when } x^2P_{cu} = P_i \)
\( S \) = rated VA, \( x \) = fraction of full load, \( \cos\phi \) = load power factor, \( P_i \) = iron loss (W), \( P_{cu} \) = full-load copper loss (W).
Quick example
Distribution transformers are lightly loaded for many hours of the day, so they are designed with low iron loss, which places their maximum efficiency below full load. Large power transformers run near full load, so they are designed for maximum efficiency near rated load.
Common mistakes
- Writing efficiency as input/output.
- Saying maximum efficiency occurs at full load; it occurs where copper loss equals iron loss.
- Using kVA instead of kW as output.
How MU asks it
MU states numericals are not expected in this module, so questions are theory. Typical: "Define efficiency of a transformer and derive the condition for maximum efficiency" (5 marks). For full marks: give the efficiency expression with \( x \), differentiate clearly, state 'copper loss = iron loss', give \( x = \sqrt{P_i/P_{cu}} \), and sketch efficiency versus load showing the peak.