--- title: "Efficiency of 25 kVA transformer at half load and overload" subject: "Electrical Machines I" chapter: 2 key: "semester_4/electrical_machines_i:25kva-2200-220-efficiency-half-overload" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/25kva-2200-220-efficiency-half-overload site: BCT Engineering --- # Efficiency of 25 kVA transformer at half load and overload Electrical Machines I · Chapter 2: Transformer - Asked in: 2071 Bhadra, 8 marks - Video: [Transformer efficiency: at Full and Half Load problem.](https://www.youtube.com/watch?v=ek2AQfUmAKo) — Dr Bharat Singh Choudhary - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/25kva-2200-220-efficiency-half-overload - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2 ## Question A 25 kVA single phase 2200/220 V transformer has primary winding resistance of 1 Ω, secondary winding resistance 0.01 Ω, primary leakage reactance of 1.5 Ω and secondary leakage reactance of 0.015 Ω. The iron loss of the transformer is 206 W. Calculate the efficiency of the transformer at: (i) Half load (ii) 50% Overload. ## Short answer Assuming unity power factor, the equivalent resistance referred to primary is 2 Ω, full-load copper loss is 258.26 W, and iron loss is 206 W. Efficiency at half load is 97.88% and at 50% overload is 97.94%. ## Answer Transformation ratio $K = \frac{220}{2200} = 0.1$. Power factor is not given, so **unity power factor** is assumed. #### Equivalent resistance and full-load copper loss $$ \begin{aligned} R_{01} &= R_1 + \frac{R_2}{K^2} = 1 + \frac{0.01}{0.01} = 2\ \Omega \\ I_{1,FL} &= \frac{25000}{2200} = 11.364\ \text{A} \\ P_{cu,FL} &= I_{1,FL}^2R_{01} = 11.364^2 \times 2 = 258.26\ \text{W} \end{aligned} $$ Iron loss $P_i = 206$ W (constant). Leakage reactances cause no power loss and are not needed. #### (i) Half load ($x = 0.5$) $$ \begin{aligned} \text{Output} &= 0.5 \times 25000 \times 1 = 12500\ \text{W} \\ P_{cu} &= 0.5^2 \times 258.26 = 64.57\ \text{W} \\ \eta &= \frac{12500}{12500 + 206 + 64.57}\times 100 = 97.88\% \end{aligned} $$ #### (ii) 50% overload ($x = 1.5$) $$ \begin{aligned} \text{Output} &= 1.5 \times 25000 \times 1 = 37500\ \text{W} \\ P_{cu} &= 1.5^2 \times 258.26 = 581.10\ \text{W} \\ \eta &= \frac{37500}{37500 + 206 + 581.10}\times 100 = 97.94\% \end{aligned} $$ | Load | Output | Iron loss | Cu loss | Efficiency | |---|---|---|---|---| | Half load | 12.5 kW | 206 W | 64.57 W | 97.88% | | 50% overload | 37.5 kW | 206 W | 581.10 W | 97.94% | **Answer (upf):** (i) **97.88%**, (ii) **97.94%**. Maximum efficiency would occur at $x = \sqrt{206/258.26} = 0.893$ of full load, which is why the two values on either side are close. At another pf (say 0.8), multiply the outputs by 0.8 and recompute. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. R01 for this transformer (K = 0.1) is: A. 1.01 ohm B. 2 ohm C. 1.1 ohm D. 11 ohm Answer: B. R01 = R1 + R2/K^2 = 1 + 0.01/0.01 = 2 ohm. (Key point: Equivalent resistance) 2. Full-load primary current is 11.364 A, so full-load copper loss is: A. 258.26 W B. 206 W C. 129 W D. 516 W Answer: A. I1^2 R01 = 11.364^2 x 2 = 258.26 W. (Key point: Full-load copper loss) 3. At 50% overload (x = 1.5), the copper loss is: A. 387.4 W B. 258.26 W C. 1.5^2 x 258.26 = 581.10 W D. 64.57 W Answer: C. Copper loss is proportional to x^2. (Key point: Overload copper loss) 4. Why are the half-load and 50% overload efficiencies almost equal? A. Maximum efficiency lies between them, near 0.893 of full load B. Iron loss changes with load C. Copper loss is constant D. Reactance cancels losses Answer: A. Efficiency peaks at x = sqrt(206/258.26) = 0.893, between 0.5 and 1.5. 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