--- title: "250/500 V transformer test, equivalent circuit" subject: "Electrical Machines" chapter: 2 key: "semester_4/electrical_machines:tf-250-500-test-equivalent-primary" source: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-250-500-test-equivalent-primary site: BCT Engineering --- # 250/500 V transformer test, equivalent circuit Electrical Machines · Chapter 2: Transformer - Asked in: 2078 Baisakh, 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-250-500-test-equivalent-primary - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2 ## Question Test data on a 1-phase, 250/500 V, 50 Hz transformer are: No-Load Test: 250 V, 1 A, 80 W (carried on LV side) Short circuit Test: 20 V, 12 A, 100 W (carried on HV side) Then draw equivalent circuit referred to primary side. ## Short answer OC test on LV gives R0=781.25 Ω and X0=263.9 Ω; SC test referred from HV to LV gives Req=0.1736 Ω and Xeq=0.3788 Ω. Thus the equivalent circuit referred to the 250 V primary has shunt R0 and X0 in parallel at the input, followed by series Req and Xeq. ## Answer **Given:** 250/500 V, 50 Hz. The primary is the 250 V (LV) winding, the secondary is the 500 V (HV) winding, $K = 500/250 = 2$. #### Open-circuit test (on the LV side) The no-load power is almost entirely core loss, so the shunt branch is found from it. $$ \begin{aligned} \cos\phi_0 &= \frac{P_0}{V_0I_0} = \frac{80}{250\times1} = 0.32 \\ I_w &= I_0\cos\phi_0 = 0.32\ \text{A},\qquad I_\mu = \sqrt{I_0^2 - I_w^2} = 0.9474\ \text{A} \\ R_0 &= \frac{V_0}{I_w} = 781.25\ \Omega,\qquad X_0 = \frac{V_0}{I_\mu} = 263.9\ \Omega\quad(\text{on the LV side}) \end{aligned} $$ #### 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. $$ \begin{aligned} Z_{sc} &= \frac{V_{sc}}{I_{sc}} = \frac{20}{12} = 1.667\ \Omega \\ R_{sc} &= \frac{P_{sc}}{I_{sc}^2} = \frac{100}{12^2} = 0.6944\ \Omega \\ X_{sc} &= \sqrt{Z_{sc}^2 - R_{sc}^2} = 1.515\ \Omega\quad(\text{on the HV side}) \end{aligned} $$ #### Referring to the LV side The turns ratio is $K = 500/250 = 2$, so $K^2 = 4$. Impedances are multiplied by $K^2$ when moved from LV to HV, and divided by $K^2$ when moved from HV to LV. $$ \begin{aligned} R_{eq} &= 0.6944 \div 4 = 0.1736\ \Omega \\ X_{eq} &= 1.515 \div 4 = 0.3788\ \Omega \end{aligned} $$ #### Equivalent circuit referred to the LV side ```text Req=0.174 ohm Xeq=0.379 ohm o--+-----/\/\/\-----UUUU------o | [R0=781.25] [X0=263.9] Load | (in parallel) o--+---------------------------o ``` $R_{eq} = 0.1736\ \Omega$, $X_{eq} = 0.3788\ \Omega$ ($Z_{eq} = 0.4167\ \Omega$), shunt branch $R_0 = 781.25\ \Omega$ in parallel with $X_0 = 263.9\ \Omega$. **Answer (referred to primary):** $R_0 = 781.25\ \Omega$, $X_0 = 263.9\ \Omega$, $R_{01} = 0.1736\ \Omega$, $X_{01} = 0.3788\ \Omega$ ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which winding is the primary here? A. The 500 V HV winding B. Either, with no effect on K C. The 250 V LV winding D. The winding used for SC test Answer: C. The first voltage in 250/500 V is the primary, so K = 2 and the OC data are directly on the primary. (Key point: Primary side) 2. What are R0 and X0 from the OC test? A. 781.2 ohm and 263.9 ohm B. 250 ohm and 250 ohm C. 312.5 ohm and 263.9 ohm D. 781.2 ohm and 1000 ohm Answer: A. Iw = 0.32 A gives R0 = 250/0.32; Imu = 0.9474 A gives X0 = 250/0.9474. (Key point: Shunt parameters) 3. The SC test on the HV side gives Z = 1.667 ohm. To refer it to the primary (LV) side: A. Multiply by 4 B. Divide by K^2 = 4 C. Leave unchanged D. Divide by 2 Answer: B. Req = 0.6944/4 = 0.1736 ohm, Xeq = 1.515/4 = 0.3788 ohm. 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