--- title: "20 kVA 2200/220 V tests, efficiency" subject: "Electrical Machine" chapter: 2 key: "semester_4/electrical_machine:transformer-20kva-2200-220-primary" source: https://www.bctengineer.com/q/semester_4/electrical_machine/transformer-20kva-2200-220-primary site: BCT Engineering --- # 20 kVA 2200/220 V tests, efficiency Electrical Machine · Chapter 2: Transformer - Asked in: 2073 Bhadra, 8 marks - Video: [Transformer OC test and SC test problems | effeciency, voltage regulation calculations](https://www.youtube.com/watch?v=aB2j_Fqd4zo) — The PiSquare Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machine/transformer-20kva-2200-220-primary - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-2 ## Question The following test results were obtained on a 20 kVA, 2200/220 V, 50 Hz single phase transformer Open circuit test: 220 V, 1.1 A, 125 W Short circuit test: 52.7 V, 8.4 A, 287 W i) Calculate the equivalent circuit parameters referred to primary side and draw the equivalent circuits ii) Calculate maximum efficiency at full load iii) Calculate the efficiency at half full load with 0.8 power factor lagging ## Short answer Referred to primary: Ro=38.72 kΩ, Xo=23.36 kΩ, R01=4.07 Ω, X01=4.78 Ω. Maximum efficiency at full load (unity pf) is 98.0% occurring at 61% load. Half-load efficiency at 0.8 pf lagging is 97.45%. ## Answer **Given:** 20 kVA, 2200/220 V. OC (LV side): 220 V, 1.1 A, 125 W. SC (HV side): 52.7 V, 8.4 A, 287 W. Primary = HV side (2200 V); K = 220/2200 = 0.1. #### (i) Parameters referred to the primary **Shunt branch (the OC test is on the LV side, so transfer it to the HV side with 1/K² = 100):** ```text cos φ0 = 125/(220 × 1.1) = 0.5165 Iw = 1.1 × 0.5165 = 0.568 A ; Iμ = √(1.1² − 0.568²) = 0.942 A On LV: Ro = 220/0.568 = 387.2 Ω ; Xo = 220/0.942 = 233.6 Ω On HV: Ro = 387.2 × 100 = 38 720 Ω ; Xo = 233.6 × 100 = 23 357 Ω ``` **Series branch (the SC test is on the HV side, so the values are used directly):** ```text Z01 = 52.7/8.4 = 6.274 Ω ; R01 = 287/8.4² = 4.067 Ω X01 = √(6.274² − 4.067²) = 4.777 Ω ``` **Equivalent circuit (referred to primary)** ```text o──┬──────┬──────/\/\/────(((────────o │ │ 4.067 Ω 4.777 Ω 38 720 23 357 load Ω Ω V2' o──┴──────┴────────────────────────o ``` This is the approximate equivalent circuit: the shunt branch is moved to the input terminals, and the series branch follows it. #### (ii) Maximum efficiency ```text Rated HV current I1 = 20 000/2200 = 9.091 A Full-load Cu loss = 287 × (9.091/8.4)² = 336.2 W Iron loss Pi = 125 W x (max η) = √(Pi/Pcu) = √(125/336.2) = 0.610 Output at max η, unity pf = 0.610 × 20 000 = 12 196 W η_max = 12 196/(12 196 + 2 × 125) = 97.99 % ``` (Full-load efficiency at unity pf: 20 000/(20 000 + 125 + 336.2) = 97.75 %.) #### (iii) Half load, 0.8 lagging ```text Output = 0.5 × 20 000 × 0.8 = 8000 W Cu loss = 0.5² × 336.2 = 84.0 W η = 8000/(8000 + 125 + 84.0) = 97.45 % ``` **Answer: (i) Ro = 38.72 kΩ, Xo = 23.36 kΩ, R01 = 4.07 Ω, X01 = 4.78 Ω; (ii) η_max ≈ 98.0 % (at unity pf, about 61 % load); (iii) η ≈ 97.45 %.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The OC test was on the LV side. How are Ro and Xo referred to the HV (primary) side? A. Multiply by 100 B. Divide by 100 C. Multiply by 10 D. Leave unchanged Answer: A. Impedance on LV is multiplied by 1/K² = 100 to refer it to HV. (Key point: Shunt branch transfer) 2. The SC test was on the HV side (primary). What is R01? A. 4.067 Ω B. 0.04067 Ω C. 6.274 Ω D. 0.0342 Ω Answer: A. R01 = Wsc/Isc² = 287/8.4² = 4.067 Ω, needing no transfer. (Key point: Series branch) 3. Why is the full-load Cu loss 336.2 W rather than 287 W? A. Because iron loss is added B. Because SC test current 8.4 A is below rated 9.091 A, so loss is scaled by (9.091/8.4)² C. Because of temperature rise D. Because the test is on LV side Answer: B. Rated HV current is 20000/2200 = 9.091 A; copper loss scales with current squared. (Key point: Full-load copper loss) 4. At half load and 0.8 pf lagging, what is the efficiency approximately? A. 96.5 % B. 98.0 % C. 97.45 % D. 99.2 % Answer: C. η = 8000/(8000 + 125 + 84) = 97.45 %. 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