--- title: "20 kVA 250/2500 V tests, regulation, efficiency" subject: "Electrical Machines" chapter: 2 key: "semester_4/electrical_machines:tf-20kva-250-2500-regulation-efficiency" source: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-20kva-250-2500-regulation-efficiency site: BCT Engineering --- # 20 kVA 250/2500 V tests, regulation, efficiency Electrical Machines · Chapter 2: Transformer - Asked in: 2076 Baisakh, 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_machines/tf-20kva-250-2500-regulation-efficiency - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2 ## Question A 20 kVA, 250 V/2500 V, 50 Hz single phase transformer gave the following test result: Open circuit test: 250 V, 1.4 A, 105 watts Short circuit test: 120 V, 8 A, 320 watts Calculate the equivalent circuit parameters referred to primary side and draw the equivalent circuit. Also calculate voltage regulation and efficiency at half full load for an 0.8 p.f. lagging. ## Short answer From OC test: R0=595.2 Ω, X0=187.2 Ω. From SC test referred to primary: R01=0.05 Ω, X01=0.1414 Ω. Voltage regulation at half load, 0.8 pf lag: VR=2.0%. Efficiency at half load: 97.74%. ## Answer **Given:** 20 kVA, 250 V/2500 V. Primary = LV (250 V), $K = 2500/250 = 10$, $K^2 = 100$. #### Equivalent circuit parameters (referred to primary) **OC test (LV side):** 250 V, 1.4 A, 105 W $$ \begin{aligned} \cos\phi_0 &= \frac{105}{250\times1.4} = 0.3,\quad I_w = 0.42\ \text{A},\quad I_\mu = \sqrt{1.4^2 - 0.42^2} = 1.3355\ \text{A} \\ R_0 &= \frac{250}{0.42} = 595.2\ \Omega,\qquad X_0 = \frac{250}{1.3355} = 187.2\ \Omega \end{aligned} $$ **SC test (HV side):** 120 V, 8 A, 320 W. (The rated HV current is $20\,000/2500 = 8$ A, so this is a full-load test.) $$ \begin{aligned} Z_{eq,H} &= \frac{120}{8} = 15\ \Omega,\quad R_{eq,H} = \frac{320}{8^2} = 5\ \Omega,\quad X_{eq,H} = \sqrt{15^2 - 5^2} = 14.14\ \Omega \end{aligned} $$ **Referred to primary (LV)** by dividing by $K^2 = 100$: $$ R_{01} = 0.05\ \Omega,\qquad X_{01} = 0.1414\ \Omega,\qquad Z_{01} = 0.15\ \Omega $$ ```text R01=0.05 ohm X01=0.1414 ohm o--+-----/\/\/\-----UUUU------o | [R0=595.2] [X0=187.2] Load | o--+---------------------------o ``` #### Voltage regulation at half load, 0.8 pf lagging Rated primary current $I_1 = 20\,000/250 = 80$ A; half load gives $I_1 = 40$ A. $$ \begin{aligned} I_1(R_{01}\cos\phi + X_{01}\sin\phi) &= 40(0.05\times0.8 + 0.1414\times0.6) = 4.994\ \text{V} \\ \text{VR} &= \frac{4.994}{250}\times100 = 2.0\% \end{aligned} $$ #### Efficiency at half load, 0.8 pf lagging $$ \begin{aligned} P_{out} &= 0.5\times20\,000\times0.8 = 8000\ \text{W} \\ P_{cu} &= 0.5^2\times320 = 80\ \text{W},\qquad P_i = 105\ \text{W} \\ \eta &= \frac{8000}{8000 + 80 + 105}\times100 = 97.74\% \end{aligned} $$ **Answer: $R_{01} = 0.05\ \Omega$, $X_{01} = 0.1414\ \Omega$, $R_0 = 595.2\ \Omega$, $X_0 = 187.2\ \Omega$; regulation $\approx 2.0\%$; efficiency $= 97.74\%$** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the SC test at 8 A a full-load test? A. Rated HV current = 20000/2500 = 8 A B. Rated HV current = 20000/250 = 80 A C. 8 A is rated LV current D. SC tests are always at full load Answer: A. The test was done on the HV side where the full-load current is 8 A. (Key point: Rated HV current) 2. R_eq,H = 5 ohm becomes what on the primary (LV) side? A. 500 ohm B. 0.05 ohm C. 0.5 ohm D. 0.005 ohm Answer: B. Divide by K^2 = 100. (Key point: Referring to primary) 3. What is the efficiency at half load, 0.8 pf lagging? A. 98.5% B. 97.0% C. 97.74% D. 96.3% Answer: C. Output 8000 W; Pcu = 80 W; Pi = 105 W; eta = 8000/8185. 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