--- title: "Iron loss and maximum efficiency of 20 kVA 220/2200 V transformer" subject: "Electrical Machines I" chapter: 2 key: "semester_4/electrical_machines_i:20kva-iron-loss-max-efficiency-current" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/20kva-iron-loss-max-efficiency-current site: BCT Engineering --- # Iron loss and maximum efficiency of 20 kVA 220/2200 V transformer Electrical Machines I · Chapter 2: Transformer - Asked in: 2077 Chaitra, 8 marks - Video: [43. Transformer Efficiency, Load KVA, Maximum Efficiency Condition, Maximum Efficiency Calculation.](https://www.youtube.com/watch?v=EsDMulvvKHo) — Electrical Tutorial - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/20kva-iron-loss-max-efficiency-current - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2 ## Question A 20 kVA, 220V/2200V, 50 Hz single phase transformer has the following parameters: R0 = 500 ohm, X0 = 160 ohm, R01 = 0.04 ohm and X01 = 0.1 ohm. (i) Calculate the iron loss of the transformer. (ii) Calculate the primary current at which the efficiency for the transformer will be maximum and also calculate the maximum efficiency. ## Short answer Iron loss is V1^2/R0=96.8 W. Maximum efficiency occurs when copper loss equals iron loss, giving primary current 49.19 A, about 54% of full load, and maximum efficiency 98.24% at unity pf. ## Answer Given (referred to the 220 V primary): $R_0 = 500\ \Omega$, $X_0 = 160\ \Omega$, $R_{01} = 0.04\ \Omega$, $X_{01} = 0.1\ \Omega$. Rated primary voltage $V_1 = 220$ V. #### (i) Iron loss Iron loss is the power taken by the core-loss resistance $R_0$ at rated voltage: $$ P_i = \frac{V_1^2}{R_0} = \frac{220^2}{500} = 96.8\ \text{W} $$ ($X_0$ only carries the magnetising current, which takes no real power.) #### (ii) Current for maximum efficiency Efficiency is maximum when copper loss equals iron loss: $$ \begin{aligned} I_1^2 R_{01} &= P_i \\ I_1 &= \sqrt{\frac{P_i}{R_{01}}} = \sqrt{\frac{96.8}{0.04}} = \sqrt{2420} = 49.19\ \text{A} \end{aligned} $$ Full-load primary current is $20000/220 = 90.91$ A, so maximum efficiency occurs at $49.19/90.91 = 0.541$, i.e. about **54% of full load**. #### Maximum efficiency Assuming unity power factor (pf not given, and maximum efficiency is normally quoted at upf): $$ \begin{aligned} \text{Output} &= V_1 I_1 \cos\phi = 220 \times 49.19 \times 1 = 10822.6\ \text{W} \\ \text{Total loss} &= P_i + P_{cu} = 96.8 + 96.8 = 193.6\ \text{W} \\ \eta_{max} &= \frac{10822.6}{10822.6 + 193.6} \times 100 = 98.24\% \end{aligned} $$ **Answer:** Iron loss = **96.8 W**; maximum efficiency occurs at a primary current of **49.19 A**; $\eta_{max} \approx$ **98.24%** at unity pf. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the iron loss found from the parameters? A. V1^2/R0 = 96.8 W B. I^2 R01 at full load C. V1^2/X0 = 302 W D. V1^2/(R0 + X0) Answer: A. Iron loss is the real power in R0 at rated voltage: 220^2/500 = 96.8 W. (Key point: Iron loss) 2. Why does X0 not enter the iron loss? A. It only matters at full load B. It is in series C. It carries magnetising current that takes no real power D. It is much larger than R0 Answer: C. A pure reactance dissipates no real power. (Key point: Role of X0) 3. What current gives maximum efficiency? A. I1 = sqrt(Pi x R01) B. I1 = Pi/R01 C. I1^2 R01 = Pi, so I1 = sqrt(96.8/0.04) = 49.19 A D. The full-load current 90.91 A Answer: C. Maximum efficiency occurs when copper loss equals iron loss. (Key point: Maximum efficiency condition) 4. Why is total loss 193.6 W at maximum efficiency, and what is eta max? A. Because Pcu = 2Pi; eta = 97% B. Because Pi doubles at this load; eta = 95% C. Because Pcu = Pi = 96.8 W; eta = 10822.6/(10822.6+193.6) = 98.24% D. Because Pcu = 0; eta = 99% Answer: C. With equal losses total loss is 2 x 96.8 = 193.6 W; at unity pf output is 220 x 49.19 W. 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