--- title: "8 kVA 400/120 V tests, parameters on HV side" subject: "Electrical Machines" chapter: 2 key: "semester_4/electrical_machines:tf-8kva-400-120-efficiency-hv" source: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-8kva-400-120-efficiency-hv site: BCT Engineering --- # 8 kVA 400/120 V tests, parameters on HV side Electrical Machines · Chapter 2: Transformer - Asked in: 2071 Magh, 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-8kva-400-120-efficiency-hv - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2 ## Question The following test results were obtained for open circuit and short circuit tests on a 8 kVA, 400/120 V, 50 Hz transformer: Open-circuit Test (LV Side): 120 V, 4 A, 75 W Short-circuit Test (HV Side): 9.5 V, 20 A, 110 W Calculate the equivalent circuit parameters referred to high voltage side. Also calculate the efficiency at half full load and 0.8 power factor lagging load. ## Short answer Referred to HV side: shunt branch R0=2133 Ω, X0=337.5 Ω; series branch Req=0.275 Ω, Xeq=0.3873 Ω. Efficiency at half load, 0.8 pf lag: η=96.90%. ## Answer **Given:** 8 kVA, 400/120 V. HV = 400 V, LV = 120 V, $K = 400/120 = 3.333$, $K^2 = 11.11$. Rated HV current $= 8000/400 = 20$ A, so the SC test is at full load. #### Open-circuit test (on the LV side) The no-load power is almost entirely core loss, so the shunt branch is found from it. $$ \begin{aligned} \cos\phi_0 &= \frac{P_0}{V_0I_0} = \frac{75}{120\times4} = 0.1562 \\ I_w &= I_0\cos\phi_0 = 0.625\ \text{A},\qquad I_\mu = \sqrt{I_0^2 - I_w^2} = 3.951\ \text{A} \\ R_0 &= \frac{V_0}{I_w} = 192\ \Omega,\qquad X_0 = \frac{V_0}{I_\mu} = 30.37\ \Omega\quad(\text{on the LV side}) \end{aligned} $$ #### Short-circuit test (on the HV side) At the reduced test voltage the core loss is negligible, so the wattmeter reading is the copper loss. $$ \begin{aligned} Z_{sc} &= \frac{V_{sc}}{I_{sc}} = \frac{9.5}{20} = 0.475\ \Omega \\ R_{sc} &= \frac{P_{sc}}{I_{sc}^2} = \frac{110}{20^2} = 0.275\ \Omega \\ X_{sc} &= \sqrt{Z_{sc}^2 - R_{sc}^2} = 0.3873\ \Omega\quad(\text{on the HV side}) \end{aligned} $$ #### Referring to the HV side The turns ratio is $K = 400/120 = 3.333$, so $K^2 = 11.11$. Impedances are multiplied by $K^2$ when moved from LV to HV, and divided by $K^2$ when moved from HV to LV. $$ \begin{aligned} R_0' &= 192 \times 11.11 = 2133\ \Omega \\ X_0' &= 30.37 \times 11.11 = 337.5\ \Omega \end{aligned} $$ #### Equivalent circuit referred to the HV side ```text Req=0.275 ohm Xeq=0.387 ohm o--+-----/\/\/\-----UUUU------o | [R0=2133] [X0=337.5] Load | (in parallel) o--+---------------------------o ``` $R_{eq} = 0.275\ \Omega$, $X_{eq} = 0.3873\ \Omega$ ($Z_{eq} = 0.475\ \Omega$), shunt branch $R_0 = 2133\ \Omega$ in parallel with $X_0 = 337.5\ \Omega$. #### Efficiency at half load, 0.8 pf lagging $$ \begin{aligned} P_{out} &= 0.5\times8000\times0.8 = 3200\ \text{W} \\ P_{cu} &= 0.5^2\times110 = 27.5\ \text{W},\qquad P_i = 75\ \text{W} \\ \eta &= \frac{3200}{3200 + 27.5 + 75}\times100 = 96.90\% \end{aligned} $$ **Answer (referred to HV):** $R_0 = 2133\ \Omega$, $X_0 = 337.5\ \Omega$, $R_{eq} = 0.275\ \Omega$, $X_{eq} = 0.3873\ \Omega$; efficiency at half load $= 96.90\%$ ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are K and K^2 for 400/120 V? A. 3.333 and 11.11 B. 0.3 and 0.09 C. 120 and 14400 D. 3.333 and 3.333 Answer: A. K = 400/120. (Key point: Turns ratio) 2. How is R0 = 192 ohm (LV) referred to HV? A. Multiply by K^2 to give 2133 ohm B. Divide by K^2 to give 17.3 ohm C. Multiply by K to give 640 ohm D. Leave unchanged Answer: A. LV to HV multiplies by K^2. (Key point: Shunt to HV) 3. What is the efficiency at half load and 0.8 pf? A. 97.5% B. 96.90% C. 95.6% D. 98.1% Answer: B. Output 3200 W; Pcu = 27.5 W; Pi = 75 W. 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