--- title: "20 kVA 2200/220 V tests, equivalent circuit LV side" subject: "Electrical Machines" chapter: 2 key: "semester_4/electrical_machines:tf-20kva-2200-220-lv-equivalent" source: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-20kva-2200-220-lv-equivalent site: BCT Engineering --- # 20 kVA 2200/220 V tests, equivalent circuit LV side Electrical Machines · Chapter 2: Transformer - Asked in: 2070 Magh, 8 marks - Video: [(E)EM Ex 2.2 || Equivalent Circuit of a Transformer || Refer to Pri & Secy sides ||Open-Short Test](https://www.youtube.com/watch?v=uJP_rFndPtU) — Electrical Engineering 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-20kva-2200-220-lv-equivalent - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2 ## Question The following test results were obtained on a 20 kVA, 2200/220 V, 50 Hz single phase transformer: Open-circuit Test (LV Side): 220 V, 1.1 A, 125 W Short-circuit Test (HV Side): 52.7 V, 8.4 A, 287 W Calculate the equivalent circuit referred to L.V side and draw the equivalent circuit. ## Short answer Referred to LV side: shunt branch R0=387.2 Ω, X0=233.6 Ω; series branch Req=0.04067 Ω, Xeq=0.04777 Ω (Zeq=0.0627 Ω). ## Answer **Given:** 20 kVA, 2200/220 V, so HV = 2200 V, LV = 220 V, $K = 10$, $K^2 = 100$. (Rated HV current is $20\,000/2200 = 9.09$ A; the SC test was at 8.4 A, near 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. $$ \begin{aligned} \cos\phi_0 &= \frac{P_0}{V_0I_0} = \frac{125}{220\times1.1} = 0.5165 \\ I_w &= I_0\cos\phi_0 = 0.5682\ \text{A},\qquad I_\mu = \sqrt{I_0^2 - I_w^2} = 0.9419\ \text{A} \\ R_0 &= \frac{V_0}{I_w} = 387.2\ \Omega,\qquad X_0 = \frac{V_0}{I_\mu} = 233.6\ \Omega\quad(\text{on the LV side}) \end{aligned} $$ #### 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. $$ \begin{aligned} Z_{sc} &= \frac{V_{sc}}{I_{sc}} = \frac{52.7}{8.4} = 6.274\ \Omega \\ R_{sc} &= \frac{P_{sc}}{I_{sc}^2} = \frac{287}{8.4^2} = 4.067\ \Omega \\ X_{sc} &= \sqrt{Z_{sc}^2 - R_{sc}^2} = 4.777\ \Omega\quad(\text{on the HV side}) \end{aligned} $$ #### Referring to the LV side The turns ratio is $K = 2200/220 = 10$, so $K^2 = 100$. Impedances are multiplied by $K^2$ when moved from LV to HV, and divided by $K^2$ when moved from HV to LV. $$ \begin{aligned} R_{eq} &= 4.067 \div 100 = 0.04067\ \Omega \\ X_{eq} &= 4.777 \div 100 = 0.04777\ \Omega \end{aligned} $$ #### Equivalent circuit referred to the LV side ```text Req=0.0407 ohm Xeq=0.0478 ohm o--+-----/\/\/\-----UUUU------o | [R0=387.2] [X0=233.6] Load | (in parallel) o--+---------------------------o ``` $R_{eq} = 0.04067\ \Omega$, $X_{eq} = 0.04777\ \Omega$ ($Z_{eq} = 0.06274\ \Omega$), shunt branch $R_0 = 387.2\ \Omega$ in parallel with $X_0 = 233.6\ \Omega$. **Answer (referred to LV side):** $R_0 = 387.2\ \Omega$, $X_0 = 233.6\ \Omega$, $R_{eq} = 0.04067\ \Omega$, $X_{eq} = 0.04777\ \Omega$ ($Z_{eq} = 0.0627\ \Omega$) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are the HV-side SC results moved to LV? A. Divide by K^2 = 100 B. Multiply by 100 C. Divide by 10 D. Multiply by 10 Answer: A. Req = 4.067/100 = 0.04067 ohm. (Key point: Series referral) 2. What is cos phi0 for OC (220 V, 1.1 A, 125 W)? A. 0.32 B. 0.5165 C. 0.75 D. 0.9 Answer: B. 125/(220 x 1.1) = 0.5165. (Key point: Power factor OC) 3. How is X_sc found from Zsc and Rsc? A. Zsc - Rsc B. Zsc + Rsc C. sqrt(Zsc^2 - Rsc^2) = 4.777 ohm D. Zsc x Rsc Answer: C. R and X are at right angles in Z. 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