--- title: "250/500 V transformer, parameters both sides" subject: "Electrical Machines" chapter: 2 key: "semester_4/electrical_machines:tf-250-500-open-hv-test" source: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-250-500-open-hv-test site: BCT Engineering --- # 250/500 V transformer, parameters both sides Electrical Machines · Chapter 2: Transformer - Asked in: 2065 Chaitra (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-250-500-open-hv-test - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2 ## Question A 1 phase 250/500 V, 50 Hz transformer gave the following test results: Open circuit test: 250 V, 1 A, 80 W on H.V. side Short-circuit test: 20 V, 12 A, 100 W on L.V. side Calculate the equivalent circuit parameters and draw the equivalent circuit referred to low voltage side and high voltage side. ## Short answer Using the printed test-side assignment, LV side: Req=0.694 Ω, Xeq=1.515 Ω, R0=195.3 Ω, X0=65.97 Ω; HV side: Req=2.778 Ω, Xeq=6.06 Ω, R0=781.25 Ω, X0=263.9 Ω. If the OC and SC side labels are interchanged, LV values become R0=781 Ω, X0=264 Ω, Req=0.174 Ω, Xeq=0.379 Ω. ## Answer **Data as printed:** the OC test is on the **HV** winding (250 V, 1 A, 80 W) and the SC test is on the **LV** winding (20 V, 12 A, 100 W). The test values are used exactly as given. The transformer is 250/500 V, so the LV winding is 250 V, the HV winding is 500 V, and $K = 2$, $K^2 = 4$. #### Open-circuit test (HV side) $$ \begin{aligned} \cos\phi_0 &= \frac{80}{250\times1} = 0.32,\quad I_w = 0.32\ \text{A},\quad I_\mu = \sqrt{1 - 0.32^2} = 0.9474\ \text{A} \\ R_{0H} &= \frac{250}{0.32} = 781.25\ \Omega,\qquad X_{0H} = \frac{250}{0.9474} = 263.9\ \Omega \end{aligned} $$ #### Short-circuit test (LV side) $$ \begin{aligned} Z_{eq,L} &= \frac{20}{12} = 1.667\ \Omega,\qquad R_{eq,L} = \frac{100}{12^2} = 0.6944\ \Omega \\ X_{eq,L} &= \sqrt{1.667^2 - 0.6944^2} = 1.515\ \Omega \end{aligned} $$ #### Referred to the high-voltage (HV) side Impedance moves from LV to HV by multiplying by $K^2 = 4$: $$ R_{eq,H} = 4\times0.6944 = 2.778\ \Omega,\qquad X_{eq,H} = 4\times1.515 = 6.060\ \Omega $$ with shunt branch $R_{0H} = 781.25\ \Omega$, $X_{0H} = 263.9\ \Omega$. ```text HV side: Req=2.778 ohm Xeq=6.06 ohm o--+-----/\/\/\-----UUUU------o | [R0=781.25] [X0=263.9] Load | o--+---------------------------o ``` #### Referred to the low-voltage (LV) side Shunt branch is divided by $K^2$: $$ R_{0L} = \frac{781.25}{4} = 195.3\ \Omega,\qquad X_{0L} = \frac{263.9}{4} = 65.97\ \Omega $$ with series branch $R_{eq,L} = 0.6944\ \Omega$, $X_{eq,L} = 1.515\ \Omega$. ```text LV side: Req=0.694 ohm Xeq=1.515 ohm o--+-----/\/\/\-----UUUU------o | [R0=195.3] [X0=65.97] Load | o--+---------------------------o ``` **Answer: HV side: $R_0 = 781.25\ \Omega$, $X_0 = 263.9\ \Omega$, $R_{eq} = 2.778\ \Omega$, $X_{eq} = 6.060\ \Omega$. LV side: $R_0 = 195.3\ \Omega$, $X_0 = 65.97\ \Omega$, $R_{eq} = 0.694\ \Omega$, $X_{eq} = 1.515\ \Omega$** **Note:** 250 V is the LV rating, so the side labels in the paper are probably swapped (normally the OC test is done on the LV side and the SC test on the HV side). With that reading the results are: LV side $R_0 = 781.25\ \Omega$, $X_0 = 263.9\ \Omega$, $R_{eq} = 0.1736\ \Omega$, $X_{eq} = 0.3788\ \Omega$; HV side $R_0 = 3125\ \Omega$, $X_0 = 1055.5\ \Omega$, $R_{eq} = 0.6944\ \Omega$, $X_{eq} = 1.515\ \Omega$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The OC data are on the HV winding. On which side is the shunt branch then first found? A. HV side B. LV side C. Neither D. Primary only Answer: A. Parameters are first found on the side where the test was done. (Key point: Where OC data apply) 2. To refer the SC values from the LV side to the HV side: A. Divide by 4 B. Multiply by K^2 = 4 C. Multiply by 2 D. Divide by 2 Answer: B. Req,H = 4 x 0.6944 = 2.778 ohm. (Key point: LV to HV conversion) 3. What are R0 and X0 referred to the LV side? A. 781.3 ohm and 263.9 ohm B. 1562.5 ohm and 527.8 ohm C. 195.3 ohm and 65.97 ohm D. 390.6 ohm and 131.95 ohm Answer: C. Divide the HV values by 4: 781.25/4 and 263.9/4. 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