--- title: "Equivalent circuit of 20 kVA 250/2500 V transformer from tests" subject: "Electrical Machines I" chapter: 2 key: "semester_4/electrical_machines_i:oc-sc-20kva-250-2500" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/oc-sc-20kva-250-2500 site: BCT Engineering --- # Equivalent circuit of 20 kVA 250/2500 V transformer from tests Electrical Machines I · Chapter 2: Transformer - Asked in: 2076 Bhadra, 8 marks; 2070 Bhadra, 8 marks (asked 2 times) - 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_i/oc-sc-20kva-250-2500 - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2 ## Question A 20 kVA, 250/2500 V transformer when tested gave the following results: O.C. test (L.V. side): 105 W, 1.4 A, 250 V S.C. test (H.V. side): 320 W, 8 A, 120 V Compute all the parameters of the equivalent circuit as referred to the L.V. side and also draw the resultant circuit as referred to H.V. side. ## Short answer From OC test (LV): R₀ = 595.2 Ω, X₀ = 187.2 Ω. From SC test, parameters referred to LV side: R_eq = 0.05 Ω, X_eq = 0.1414 Ω. Equivalent circuit referred to HV side: R₀ = 59.52 kΩ, X₀ = 18.72 kΩ; R_eq = 5 Ω, X_eq = 14.14 Ω. ## Answer Shunt branch parameters come from the OC test (done on LV side, so they are directly LV values); series parameters come from the SC test (done on HV side, so they are HV values). Turns ratio $a = 2500/250 = 10$. Rated currents: LV = 20000/250 = 80 A, HV = 20000/2500 = 8 A, so the SC test was at full load. #### From the OC test (LV side) $$ \begin{aligned} \cos\phi_0 &= \frac{P_0}{V_0 I_0} = \frac{105}{250\times1.4} = 0.30 \\ I_c &= I_0\cos\phi_0 = 1.4\times0.30 = 0.42\ \text{A} \\ I_m &= I_0\sin\phi_0 = 1.4\times0.954 = 1.336\ \text{A} \\ R_0 &= \frac{V_0}{I_c} = \frac{250}{0.42} = 595.2\ \Omega \\ X_0 &= \frac{V_0}{I_m} = \frac{250}{1.336} = 187.2\ \Omega \end{aligned} $$ #### From the SC test (HV side) $$ \begin{aligned} R_{eq,HV} &= \frac{P_{sc}}{I_{sc}^2} = \frac{320}{8^2} = 5\ \Omega \\ Z_{eq,HV} &= \frac{V_{sc}}{I_{sc}} = \frac{120}{8} = 15\ \Omega \\ X_{eq,HV} &= \sqrt{15^2 - 5^2} = 14.14\ \Omega \end{aligned} $$ #### Referred to LV side (divide HV impedances by $a^2 = 100$) | Parameter | LV side | HV side (×100) | |---|---|---| | $R_0$ | 595.2 Ω | 59.52 kΩ | | $X_0$ | 187.2 Ω | 18.72 kΩ | | $R_{eq}$ | 0.05 Ω | 5 Ω | | $X_{eq}$ | 0.1414 Ω | 14.14 Ω | #### Approximate equivalent circuit referred to HV side ```text Req = 5 ohm Xeq = 14.14 ohm o---+----/\/\/----mmmmmm-----o | + +---+---+ | | V2' [R0] [X0] 59.52k 18.72k | | - +---+---+ | o---+-------------------------o V1 = 2500 V ``` The shunt branch ($R_0 \parallel X_0$) is placed across the supply terminals, and the total series impedance $5 + j14.14\ \Omega$ is in the line. **Answer:** referred to LV: $R_0 = 595.2\ \Omega$, $X_0 = 187.2\ \Omega$, $R_{eq} = 0.05\ \Omega$, $X_{eq} = 0.1414\ \Omega$; referred to HV: $R_0 = 59.52$ kΩ, $X_0 = 18.72$ kΩ, $R_{eq} = 5\ \Omega$, $X_{eq} = 14.14\ \Omega$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which parameters come from each test? A. OC gives series; SC gives shunt B. Both from the OC test C. Both from the SC test D. OC test gives the shunt branch R0, X0; SC test gives series Req, Xeq Answer: D. OC has rated flux, little current; SC has rated current, little flux. (Key point: Which test gives what) 2. What are R0 and X0 (LV)? A. R0 = 187.2 Ω, X0 = 595.2 Ω B. R0 = 595.2 Ω, X0 = 187.2 Ω from Ic = 0.42 A and Im = 1.336 A C. R0 = 250/1.4 = 178.6 Ω for both D. R0 = 5.95 Ω, X0 = 1.87 Ω Answer: B. cos φ0 = 105/(250×1.4) = 0.3; R0 = V/Ic, X0 = V/Im. (Key point: No-load parameters) 3. What are the HV-side Req and Xeq? A. Req = 15 Ω, Xeq = 5 Ω B. Req = 2.5 Ω, Xeq = 7.07 Ω C. Req = 5 Ω, Xeq = 14.14 Ω D. Req = 40 Ω, Xeq = 14.14 Ω Answer: C. Req = 320/8² = 5 Ω; Zeq = 120/8 = 15 Ω; Xeq = √(15² − 5²). (Key point: SC parameters) 4. How are the HV series values referred to LV? A. Multiply by 100 B. Divide by 10 only C. Multiply by 10 D. Divide by a² = 100, giving 0.05 Ω and 0.1414 Ω Answer: D. Impedance referred across scales with the square of the turns ratio; LV is the lower side. 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