--- title: "Transformer losses and efficiency expression" subject: "Electrical Machine" chapter: 2 key: "semester_4/electrical_machine:transformer-losses-efficiency-derivation" source: https://www.bctengineer.com/q/semester_4/electrical_machine/transformer-losses-efficiency-derivation site: BCT Engineering --- # Transformer losses and efficiency expression Electrical Machine · Chapter 2: Transformer - Asked in: 2071 Bhadra, 8 marks - Video: [13 Losses and Efficiency of Transformer](https://www.youtube.com/watch?v=jShXjmkMyB8) — SP Engg. ACADEMY - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machine/transformer-losses-efficiency-derivation - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-2 ## Question What are different types of losses in transformer? Derive the expression of efficiency of transformer. ## Short answer Core (iron) losses are constant; copper loss varies as I². Efficiency η = x S cos φ / (x S cos φ + Pi + x² Pcu), where x is load fraction, S rating, Pi core loss, Pcu full-load copper loss. Full-load η = S cos φ / (S cos φ + Pi + Pcu). ## Answer #### Types of losses **1. Core (iron) losses** occur in the core and are practically constant from no-load to full load (since V1 and f are constant). - *Hysteresis loss:* Pₕ = ηₕ Bm^1.6 f V, reduced by using silicon steel / CRGO. - *Eddy current loss:* Pₑ = kₑ Bm² f² t² V, reduced by thin insulated laminations. **2. Copper (I²R) losses** occur in the primary and secondary windings and vary as the square of the load current: Pcu = I1² R1 + I2² R2 = I1² R01. **3. Stray and dielectric losses (small):** leakage-flux losses in the tank and structure, and insulation losses (insignificant at normal voltages). Since a transformer has no moving part, there are no friction or windage losses, so efficiency is very high (95–99 %). #### Efficiency expression Efficiency η = output power / input power = output / (output + losses). Let the transformer have rating S (VA), secondary voltage V2, and let a fraction *x* of full load be supplied at a load power factor cos φ. ```text Output power = x S cos φ Iron loss = Pi (constant) Copper loss at x load = x² Pcu (Pcu = full-load copper loss) η = x S cos φ / ( x S cos φ + Pi + x² Pcu ) ``` At full load (x = 1): **η = S cos φ / (S cos φ + Pi + Pcu)** Equivalent form using currents: η = V2 I2 cos φ2 / (V2 I2 cos φ2 + Pi + I2² R02). Observations: - Efficiency depends on the power factor; it is lower for lower power factor. - For a given power factor, efficiency is highest at the load where Pi = x² Pcu. - All-day efficiency matters for distribution transformers, which are on 24 hours but loaded only part of the day. **Example:** 100 kVA, Pi = 1 kW, Pcu = 1.5 kW at unity pf, full load: η = 100/(100 + 1 + 1.5) = 97.6 %. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the core (iron) loss of a transformer practically constant from no-load to full load? A. Because the primary voltage and frequency are constant, so the core flux is constant B. Because the load current does not flow in the core C. Because the winding resistance is constant D. Because the core temperature stays constant Answer: A. Core loss depends on Bm and f, which are fixed by the constant supply voltage V1 and frequency f. (Key point: Core loss behaviour) 2. How does the copper loss of a transformer vary with load current? A. It is constant B. It varies directly with load current C. It varies as the square of the load current D. It varies as the square root of the load current Answer: C. Copper loss is I^2 R, so it varies as the square of the current. (Key point: Copper loss variation) 3. For a fraction x of full load at power factor cos φ, which expression gives the efficiency? A. x S cos φ / (x S cos φ + x Pi + Pcu) B. x S cos φ / (x S cos φ + Pi + x² Pcu) C. S cos φ / (S cos φ + x² Pi + Pcu) D. x S / (x S + Pi + x Pcu) Answer: B. Output is x S cos φ, iron loss stays Pi, and copper loss becomes x² Pcu. (Key point: Efficiency at fraction x of load) 4. How does the transformer efficiency change with load power factor (same kVA)? A. It is lower for lower power factor B. It is higher for lower power factor C. It is independent of power factor D. It is highest at zero power factor Answer: A. Output power is S cos φ, while losses stay the same, so a lower pf gives a lower efficiency. 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