--- title: "Transformer coil temperature and power rating" subject: "Basic Electrical Engineering" chapter: 1 key: "semester_1/basic_electrical_engineering:transformer-coil-temperature-rise" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/transformer-coil-temperature-rise site: BCT Engineering --- # Transformer coil temperature and power rating Basic Electrical Engineering · Chapter 1: General Electric System - Asked in: 2074 Chaitra, 6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/transformer-coil-temperature-rise - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-1 ## Question A coil of stranded copper wire having a resistance of 12Ω at 25°C is embedded in the core of a large transformer supplied at 230 V. After the transformer has been in service for several hours, the resistance of the coil is found to be 13.4 Ω. What is the temperature of the core? Also find the power rating of the resistance. Assume temperature coefficient of wire as 0.00125/°C at 15°C. ## Short answer Reference resistance at 15°C is 11.852 Ω. With α=0.00125, the hot resistance yields core temperature ≈ 119.5°C. Power rating at 230 V = 230²/13.4 ≈ 3.95 kW. ## Answer The resistance of a copper coil rises with temperature, so the hot resistance tells us the core temperature. #### Data R₂₅ = 12 Ω, R_hot = 13.4 Ω, α₁₅ = 0.00125 per °C (as given), V = 230 V. #### Step 1: Resistance at the reference temperature 15°C R₂₅ = R₁₅ [1 + α₁₅ (25 − 15)] 12 = R₁₅ × (1 + 0.0125) = R₁₅ × 1.0125 R₁₅ = 11.852 Ω #### Step 2: Core temperature R_t = R₁₅ [1 + α₁₅ (t − 15)] 13.4 = 11.852 [1 + 0.00125 (t − 15)] 1 + 0.00125 (t − 15) = 1.1306 t − 15 = 104.5 t = 119.5°C (Converting α to 25°C, α₂₅ = 0.00125/1.0125 = 0.001235, gives the same result.) #### Step 3: Power rating of the coil In service, P = V²/R = 230²/13.4 = 3947.8 W (At the cold value of 12 Ω it would be 4408 W.) **Answer: Core temperature ≈ 119.5°C; power ≈ 3.95 kW** Note: the usual α for copper at 15°C is about 0.004 per °C; the value 0.00125 is used here exactly as given. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The coefficient 0.00125 per degC is given at 15 degC while the 12 ohm is at 25 degC. What is the first step? A. Convert 12 ohm to R15 using R25 = R15 [1 + alpha15 (25 - 15)] B. Ignore 12 ohm and use only 13.4 ohm with the supply C. Use 13.4/12 as the temperature in degC D. Add 25 to the final answer only Answer: A. The formula needs resistance at the reference temperature of alpha, so R15 = 12/1.0125 = 11.852 ohm. (Key point: Reference temperature) 2. Which relation gives the core temperature? A. 13.4 = 11.852 [1 + 0.00125 (t - 25)] B. 13.4 = 12 [1 + 0.00125 (t - 15)] C. 13.4 = 11.852 [1 + 0.00125 (t - 15)], giving about 119.5 degC D. 13.4 = 11.852 x 0.00125 x t Answer: C. Use R_t = R15 [1 + alpha15 (t - 15)]: 1.1306 - 1 = 0.00125 (t - 15), so t = 119.5 degC. (Key point: Core temperature) 3. The power rating of the coil in service at 230 V is found by: A. V^2/R with R = 13.4 ohm, about 3.95 kW B. V^2/R with R = 12 ohm, about 4.41 kW C. V x R = 3082 W D. I^2 R with I = 230 A Answer: A. After heating, resistance is 13.4 ohm, so P = 230^2/13.4 = 3947.8 W. 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