--- title: "5 kVA 220/400 V efficiency and regulation" subject: "Electrical Machines" chapter: 2 key: "semester_4/electrical_machines:tf-5kva-220-400-efficiency-regulation" source: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-5kva-220-400-efficiency-regulation site: BCT Engineering --- # 5 kVA 220/400 V efficiency and regulation Electrical Machines · Chapter 2: Transformer - Asked in: 2074 Bhadra (old course) - 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-5kva-220-400-efficiency-regulation - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2 ## Question Open circuit and Short circuit test on 5 kVA, 220/400 V, 50 Hz, single phase transformer gave the following results. Short circuit test (on H.V. side): 40 V, 11.4 A, 200 watts [Open circuit test data is not printed in the paper] Determine the efficiency and the voltage regulation of the transformer at full load at 0.9 pf lagging. ## Short answer Voltage regulation at full load, 0.9 pf lagging, is approximately 8.85%. Efficiency = 4500 / (4500 + Pi + 240.5), which gives 92.97% for an assumed iron loss Pi = 100 W. ## Answer **Given:** 5 kVA, 220/400 V. The SC test is on the HV side: 40 V, 11.4 A, 200 W. The open circuit data is not printed, so the core loss $P_i$ cannot be calculated from the question. I derive the voltage regulation (which needs only the SC test) and give the efficiency with $P_i$ left as a symbol, followed by a worked value for an assumed $P_i$. #### Equivalent impedance from the SC test (HV side) $$ \begin{aligned} Z_{eq} &= \frac{40}{11.4} = 3.509\ \Omega \\ R_{eq} &= \frac{200}{11.4^2} = 1.539\ \Omega \\ X_{eq} &= \sqrt{3.509^2 - 1.539^2} = 3.153\ \Omega \end{aligned} $$ #### Voltage regulation at full load, 0.9 pf lagging Rated HV current: $I = \dfrac{5000}{400} = 12.5$ A, $\cos\phi = 0.9$, $\sin\phi = 0.4359$. $$ \begin{aligned} I(R_{eq}\cos\phi + X_{eq}\sin\phi) &= 12.5(1.385 + 1.374) = 34.49\ \text{V} \\ I(X_{eq}\cos\phi - R_{eq}\sin\phi) &= 12.5(2.838 - 0.671) = 27.09\ \text{V} \end{aligned} $$ Taking the no-load voltage as 400 V, the full-load terminal voltage $V_2$ follows from $400^2 = (V_2 + 34.49)^2 + 27.09^2$, which gives $V_2 = 364.6$ V, so $$ \text{VR} = \frac{400 - 364.6}{400}\times100 = 8.85\% $$ (The approximate formula gives $34.49/400 = 8.62\%$.) #### Efficiency at full load, 0.9 pf lagging The SC test current (11.4 A) is slightly below the rated 12.5 A, so the full-load copper loss is $$ P_{cu,fl} = 12.5^2\times1.539 = 240.5\ \text{W} $$ $$ P_{out} = 5000\times0.9 = 4500\ \text{W},\qquad \eta = \frac{4500}{4500 + P_i + 240.5} $$ For example, if the OC test had given $P_i = 100$ W, then $\eta = \dfrac{4500}{4840.5}\times100 = 92.97\%$. For any other core loss, substitute it in the formula above. **Answer: voltage regulation $\approx 8.85\%$ (lagging); efficiency $= \dfrac{4500}{4740.5 + P_i}$, which is 92.97% for $P_i = 100$ W** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the SC test on the HV side (40 V, 11.4 A, 200 W), Req is found as A. 40/11.4 = 3.51 Ω B. 200/11.4^2 = 1.539 Ω C. 200/11.4 = 17.5 Ω D. sqrt(3.509^2 + 1.539^2) = 3.83 Ω Answer: B. Wattmeter power is the copper loss I^2R, so Req = P/I^2 = 200/129.96 = 1.539 Ω. 40/11.4 is Zeq. (Key point: Equivalent impedance from SC data) 2. The SC test current (11.4 A) is below rated HV current. How is the full-load copper loss obtained? A. Use 200 W directly B. Rated current 12.5 A: 12.5^2 x 1.539 = 240.5 W C. Divide 200 W by 12.5 D. Multiply 200 W by 12.5/11.4 Answer: B. Rated HV current is 5000/400 = 12.5 A. Copper loss scales as I^2, so Pcu = 12.5^2 x 1.539 = 240.5 W. (Key point: Full-load copper loss) 3. Why is the efficiency expressed with the core loss Pi as a symbol? A. Core loss is zero at full load B. The OC test data is not given, so Pi cannot be calculated C. Core loss equals copper loss D. Efficiency does not depend on core loss Answer: B. Core loss comes from the OC test, which is not printed, so efficiency = 4500/(4500 + Pi + 240.5) can only be evaluated for an assumed Pi. (Key point: Why efficiency is left with Pi) 4. The answer finds regulation for 0.9 lagging using V2 from 400^2 = (V2 + 34.49)^2 + 27.09^2. What does 34.49 V represent? A. I(Req cos(phi) + Xeq sin(phi)), the in-phase drop component B. The no-load voltage drop in the core C. The primary voltage D. The load voltage Answer: A. For lagging load the component of drop in phase with V2 is I(R cos + X sin) = 12.5 x 2.759 = 34.49 V; 27.09 V is the quadrature component. 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