--- title: "10 kVA 250/1000 V tests, equivalent circuit" subject: "Electrical Machines" chapter: 2 key: "semester_4/electrical_machines:tf-10kva-250-1000-parameters" source: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-10kva-250-1000-parameters site: BCT Engineering --- # 10 kVA 250/1000 V tests, equivalent circuit Electrical Machines · Chapter 2: Transformer - Asked in: 2073 Bhadra (old course), 8 marks - Video: [EE213 - 09b - Determining equivalent circuit parameters for a transformer - numerical example](https://www.youtube.com/watch?v=b_6NbUb7U1o) — MAFarooqi - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-10kva-250-1000-parameters - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2 ## Question The data obtained from the test of a 10 kVA, 250 V/1000 V single phase transformer are given below: No-load test (on L.V side): 250 V, 0.8 A, 80 watt Short circuit test (on H.V side): 80 V, 10 A, 120 watt Calculate the equivalent circuit parameters referred to primary side and draw the equivalent circuit. ## Short answer Open-circuit test yields shunt branch R0 = 781.25 Ω, X0 = 341 Ω. Short-circuit test gives equivalent series parameters referred to primary: Req = 0.075 Ω, Xeq = 0.4943 Ω (Zeq ≈ 0.5 Ω). ## Answer **Given:** 10 kVA, 250 V/1000 V. Primary = LV (250 V), $K = 1000/250 = 4$, $K^2 = 16$. Rated HV current $= 10\,000/1000 = 10$ 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. $$ \begin{aligned} \cos\phi_0 &= \frac{P_0}{V_0I_0} = \frac{80}{250\times0.8} = 0.4 \\ I_w &= I_0\cos\phi_0 = 0.32\ \text{A},\qquad I_\mu = \sqrt{I_0^2 - I_w^2} = 0.7332\ \text{A} \\ R_0 &= \frac{V_0}{I_w} = 781.25\ \Omega,\qquad X_0 = \frac{V_0}{I_\mu} = 341\ \Omega\quad(\text{on the LV side}) \end{aligned} $$ #### 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. $$ \begin{aligned} Z_{sc} &= \frac{V_{sc}}{I_{sc}} = \frac{80}{10} = 8\ \Omega \\ R_{sc} &= \frac{P_{sc}}{I_{sc}^2} = \frac{120}{10^2} = 1.2\ \Omega \\ X_{sc} &= \sqrt{Z_{sc}^2 - R_{sc}^2} = 7.909\ \Omega\quad(\text{on the HV side}) \end{aligned} $$ #### Referring to the LV side The turns ratio is $K = 1000/250 = 4$, so $K^2 = 16$. Impedances are multiplied by $K^2$ when moved from LV to HV, and divided by $K^2$ when moved from HV to LV. $$ \begin{aligned} R_{eq} &= 1.2 \div 16 = 0.075\ \Omega \\ X_{eq} &= 7.909 \div 16 = 0.4943\ \Omega \end{aligned} $$ #### Equivalent circuit referred to the LV side ```text Req=0.075 ohm Xeq=0.494 ohm o--+-----/\/\/\-----UUUU------o | [R0=781.25] [X0=341] Load | (in parallel) o--+---------------------------o ``` $R_{eq} = 0.075\ \Omega$, $X_{eq} = 0.4943\ \Omega$ ($Z_{eq} = 0.5\ \Omega$), shunt branch $R_0 = 781.25\ \Omega$ in parallel with $X_0 = 341\ \Omega$. **Answer (referred to primary):** $R_0 = 781.25\ \Omega$, $X_0 = 341.0\ \Omega$, $R_{01} = 0.075\ \Omega$, $X_{01} = 0.4943\ \Omega$ ($Z_{01} = 0.5\ \Omega$) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the 10 kVA, 250/1000 V transformer, why is the SC test current of 10 A a full-load test? A. Because 10 kVA / 250 V = 10 A on the LV side B. Because rated HV current is 10000/1000 = 10 A C. Because 10 A is always used in SC tests D. Because the SC voltage is 80 V Answer: B. The SC test is on the HV side, whose rated current is S/V = 10000/1000 = 10 A, so the copper loss of 120 W is the full-load value. (Key point: Rated HV current check) 2. From the OC test (250 V, 0.8 A, 80 W) on the LV side, what are cos(phi0) and Iw? A. cos(phi0) = 0.4, Iw = 0.32 A B. cos(phi0) = 0.4, Iw = 0.7332 A C. cos(phi0) = 0.32, Iw = 0.8 A D. cos(phi0) = 0.9165, Iw = 0.7332 A Answer: A. cos(phi0) = 80/(250x0.8) = 0.4 and Iw = I0 cos(phi0) = 0.32 A, giving R0 = 250/0.32 = 781.2 Ω. (Key point: Shunt branch parameters) 3. The SC values Rsc = 1.2 Ω and Zsc = 8 Ω are on the HV side with K = 4. Referred to the primary (LV) side, Zeq is A. 128 Ω B. 2 Ω C. 8 Ω D. 0.5 Ω Answer: D. HV to LV means dividing by K^2 = 16: 8/16 = 0.5 Ω. Likewise R01 = 1.2/16 = 0.075 Ω. (Key point: Referring to the primary (LV)) 4. How is the SC reactance obtained? A. Xsc = Vsc / Isc B. Xsc = Psc / Isc C. Xsc = sqrt(Zsc^2 - Rsc^2) D. Xsc = Zsc + Rsc Answer: C. Zsc is the vector sum of R and X, so X = sqrt(Z^2 - R^2) = sqrt(64 - 1.44) = 7.909 Ω. 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