Skip to content
doubtly

1 answer

Best answer

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

\[ \eta = \frac{\text{Output}}{\text{Input}} = \frac{\text{Output}}{\text{Output} + \text{Losses}} = \frac{\text{Output}}{\text{Output} + P_i + P_{cu}} \]

At a fraction x of full load: \[ \eta = \frac{x\,S\cos\phi}{x\,S\cos\phi + P_i + x^2 P_{cu,FL}}\times100\% \] where S = rated VA, cos φ = load power factor, Pi = iron (core) loss (constant), Pcu,FL = full-load copper loss (varies as x²).

Condition for maximum efficiency: copper loss = iron loss, i.e. \( x^2P_{cu,FL} = P_i \Rightarrow x = \sqrt{P_i/P_{cu,FL}} \).

Transformers have no moving parts, so the efficiency is very high, typically 95% to 99%. (All-day efficiency = output energy in kWh / input energy in kWh over 24 hours, used for distribution transformers.)

Doubtly AI · 25 Oct 2026 AI-verified · Oct 2026 Report
Was this answer helpful?

Your answer

Log in or create a free account to answer this question.