--- title: "Transformer losses and condition for maximum efficiency" subject: "Electrical Machines I" chapter: 2 key: "semester_4/electrical_machines_i:transformer-losses-max-efficiency" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/transformer-losses-max-efficiency site: BCT Engineering --- # Transformer losses and condition for maximum efficiency Electrical Machines I · Chapter 2: Transformer - Asked in: 2076 Baisakh, 8 marks; 2075 Baisakh, 5+3 marks; 2069 Bhadra, 8 marks (asked 3 times) - 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_machines_i/transformer-losses-max-efficiency - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2 ## Question Describe the various losses in a transformer. When the efficiency of a transformer will be maximum? Derive it mathematically. ## Answer A transformer has no moving parts, so it has no friction or windage loss. Its losses are **core (iron) losses** and **copper losses**. #### Losses in a transformer 1. **Core or iron loss ($P_i$)** – occurs in the core due to the alternating flux. It depends on $B_m$ and $f$, which are almost constant, so it is a **constant loss** (the same at all loads). It is measured by the open-circuit test. - **Hysteresis loss:** $P_h = \eta B_m^{1.6} f V$; reduced by using silicon steel / CRGO. - **Eddy current loss:** $P_e = K_e B_m^2 f^2 t^2 V$; reduced by thin insulated laminations. 2. **Copper loss ($P_{cu}$)** – $I_1^2R_1 + I_2^2R_2 = I_2^2 R_{02}$ in the windings. It varies with the square of the load current, so it is a **variable loss**. The full-load value is measured by the short-circuit test. At a fraction x of full load, $P_{cu} = x^2 P_{cu,fl}$. 3. **Stray loss** – from leakage flux causing eddy currents in the tank, clamps and conductors. Usually small and included in copper loss. 4. **Dielectric loss** – in the insulation (oil, paper); important only at very high voltage. #### Efficiency $$ \eta = \frac{\text{output}}{\text{output} + \text{losses}} = \frac{V_2 I_2\cos\phi_2}{V_2 I_2\cos\phi_2 + P_i + I_2^2 R_{02}} $$ #### Condition for maximum efficiency Take $V_2$ and $\cos\phi_2$ as constant. Dividing the numerator and denominator by $I_2$: $$ \eta = \frac{V_2\cos\phi_2}{V_2\cos\phi_2 + \dfrac{P_i}{I_2} + I_2R_{02}} $$ η is maximum when the varying part of the denominator, $D = \dfrac{P_i}{I_2} + I_2R_{02}$, is minimum: $$ \begin{aligned} \frac{dD}{dI_2} &= -\frac{P_i}{I_2^2} + R_{02} = 0 \\ I_2^2 R_{02} &= P_i \end{aligned} $$ ($d^2D/dI_2^2 = 2P_i/I_2^3 > 0$, so it is a minimum of D.) **Efficiency is maximum when variable copper loss equals constant iron loss.** #### Load at maximum efficiency $$ x = \frac{I_2}{I_{2,fl}} = \sqrt{\frac{P_i}{P_{cu,fl}}}, \qquad \text{kVA at } \eta_{max} = \text{rated kVA}\times\sqrt{\frac{P_i}{P_{cu,fl}}} $$ $$ \eta_{max} = \frac{x S\cos\phi}{x S\cos\phi + 2P_i} $$ Also, for a given kVA, η is highest at unity power factor. **Example:** a transformer with $P_i = 1$ kW and $P_{cu,fl} = 4$ kW has maximum efficiency at $x = \sqrt{1/4} = 0.5$, i.e. half load. Distribution transformers are designed with low $P_i$ so that η is maximum at about 50–70% load, as they are lightly loaded for much of the day. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which loss is constant and which varies with load? A. Copper loss is constant; iron loss varies with load B. Both vary as load current C. Both are constant D. Iron loss is constant; copper loss varies as the square of load current Answer: D. Iron loss depends on Bm and f, which are nearly fixed; copper loss is I²R. (Key point: Constant vs variable) 2. Which tests measure these losses? A. Open-circuit gives copper loss; short-circuit gives iron loss B. Open-circuit test gives iron loss; short-circuit test gives full-load copper loss C. Both from the open-circuit test D. Both from the load test only Answer: B. OC is at rated flux with negligible current; SC is at rated current with very low flux. (Key point: Test source) 3. When is efficiency maximum? A. When the copper loss is zero B. When the load is at full load always C. When variable copper loss equals constant iron loss, I2² R02 = Pi D. When iron loss is twice the copper loss Answer: C. Setting dD/dI2 = 0 for D = Pi/I2 + I2 R02 gives I2² R02 = Pi. (Key point: Condition) 4. At what fraction of full load is η maximum, for Pi = 1 kW and Pcu,fl = 4 kW? A. 0.25 B. 2 C. 0.75 D. 0.5, from x = √(Pi/Pcu,fl) Answer: D. √(1/4) = 0.5, i.e. half load. 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