--- title: "Efficiency and loading of 50 kVA transformer at 40 kW" subject: "Electrical Machines I" chapter: 2 key: "semester_4/electrical_machines_i:50kva-200-2200-40kw-overload" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/50kva-200-2200-40kw-overload site: BCT Engineering --- # Efficiency and loading of 50 kVA transformer at 40 kW Electrical Machines I · Chapter 2: Transformer - Asked in: 2069 Bhadra, 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/50kva-200-2200-40kw-overload - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2 ## Question A 50 kVA, 200/2200, 50 Hz single phase transformer has the following parameters: R0 = 600 Ohms, X0 = 200 Ohm referred to primary side and R02 = 2 Ohms and X02 = 4 Ohm referred to secondary side. Calculate the efficiency of the transformer when supplying a power of 40 kW to the load with 0.65 pf lagging at rated voltage. Is the transformer over-loaded or under-loaded? Calculate the percentage by which it is over-loaded or under-loaded. ## Short answer Load kVA = 40 kW / 0.65 = 61.54 kVA, I2 = 61.54kVA/2200V = 27.97 A. Iron loss = 200^2/600 = 66.67 W, Cu loss = 27.97^2 *2 = 1564.87 W. Efficiency = 40000/(40000+66.67+1564.87)*100 = 96.08%. The transformer is overloaded by (61.54-50)/50*100 = 23.08%. ## Answer Given: $R_0 = 600\ \Omega$, $X_0 = 200\ \Omega$ (primary, 200 V); $R_{02} = 2\ \Omega$, $X_{02} = 4\ \Omega$ (secondary, 2200 V). Load: 40 kW at 0.65 pf lagging, at rated voltage 2200 V. #### Load kVA and current $$ \begin{aligned} S_{load} &= \frac{P}{\cos\phi} = \frac{40}{0.65} = 61.54\ \text{kVA} \\ I_2 &= \frac{61538}{2200} = 27.97\ \text{A} \end{aligned} $$ Rated secondary current $= 50000/2200 = 22.73$ A. #### Losses $$ \begin{aligned} P_i &= \frac{V_1^2}{R_0} = \frac{200^2}{600} = 66.67\ \text{W} \\ P_{cu} &= I_2^2R_{02} = 27.97^2 \times 2 = 1564.87\ \text{W} \end{aligned} $$ #### Efficiency $$ \begin{aligned} \eta &= \frac{40000}{40000 + 66.67 + 1564.87}\times 100 \\ &= \frac{40000}{41631.54}\times 100 = 96.08\% \end{aligned} $$ #### Over-load or under-load? The transformer is rated 50 kVA but supplies 61.54 kVA. The power (40 kW) is below 50 kW, but what matters for heating is the kVA (current), so it is **over-loaded**: $$ \%\text{Overload} = \frac{61.54 - 50}{50}\times 100 = 23.08\% $$ Same by current: $\frac{27.97 - 22.73}{22.73}\times 100 = 23.08\%$. | Quantity | Value | |---|---| | Load kVA | 61.54 kVA | | Secondary current | 27.97 A | | Iron loss | 66.67 W | | Copper loss | 1564.87 W | | Efficiency | 96.08% | | Loading | 23.08% overload | **Answer:** Efficiency = **96.08%**; the transformer is **over-loaded by 23.08%** because of the low power factor. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For 40 kW at 0.65 pf and 2200 V, the secondary current is: A. 18.18 A B. 22.73 A C. 27.97 A D. 40 A Answer: C. S = 40/0.65 = 61.54 kVA; I2 = 61538/2200 = 27.97 A. (Key point: Load current) 2. Iron loss is found from R0 (referred to primary) as: A. V1^2/R0 = 66.67 W B. I2^2 R0 C. V2^2/R0 with 2200 V D. R0 x X0 Answer: A. R0 is on the 200 V side, so Pi = 200^2/600 = 66.67 W. (Key point: Iron loss from R0) 3. Why is the transformer over-loaded although 40 kW is less than 50 kVA? A. Heating depends on kVA (current), and the load is 61.54 kVA B. Iron loss is too large C. The voltage is too high D. kW must be equal to kVA Answer: A. Rating is in kVA because heating depends on current; 61.54 kVA exceeds 50 kVA. (Key point: Overload decision) 4. The percentage overload is: A. 23.08% B. 18.9% C. 40% D. 50% Answer: A. (61.54 - 50)/50 x 100 = 23.08%, same as by current (27.97 - 22.73)/22.73. 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