--- title: "10 kVA 200/400 V tests, parameters on LV side" subject: "Electrical Machines" chapter: 2 key: "semester_4/electrical_machines:tf-10kva-200-400-lv-params" source: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-10kva-200-400-lv-params site: BCT Engineering --- # 10 kVA 200/400 V tests, parameters on LV side Electrical Machines · Chapter 2: Transformer - Asked in: 2068 Bhadra, 8 marks - Video: [Numerical 7: A 10 kVA, 200/400V, 50Hz transformer has the following test results](https://www.youtube.com/watch?v=nFmiCeoM8Mc) — Progress Edu - 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-200-400-lv-params - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2 ## Question A 10 kVA, 200/400 V, 50 Hz, 1 phase, transformer gave the following test results: OC test (HV open): 200 V, 1.3 A, 120 W SC test (LV short): 22 V, 30 A, 200 W Determine shunt and series branch parameters referred to Low Voltage Side and hence draw equivalent circuit diagram also. ## Short answer Referred to LV side: shunt branch R0=333.3 Ω, X0=173.4 Ω; series branch Req=0.0556 Ω, Xeq=0.1747 Ω. ## Answer **Reading of the tests:** "HV open" means the OC test is done on the LV winding at its rated 200 V. "LV short" means the LV winding is short-circuited while the supply is applied to the HV (400 V) winding, so the SC readings are on the HV side. Hence $K = 400/200 = 2$, $K^2 = 4$. #### Open-circuit test (on the LV side) The no-load power is almost entirely core loss, so the shunt branch is found from it. $$ \begin{aligned} \cos\phi_0 &= \frac{P_0}{V_0I_0} = \frac{120}{200\times1.3} = 0.4615 \\ I_w &= I_0\cos\phi_0 = 0.6\ \text{A},\qquad I_\mu = \sqrt{I_0^2 - I_w^2} = 1.153\ \text{A} \\ R_0 &= \frac{V_0}{I_w} = 333.3\ \Omega,\qquad X_0 = \frac{V_0}{I_\mu} = 173.4\ \Omega\quad(\text{on the LV side}) \end{aligned} $$ #### 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. $$ \begin{aligned} Z_{sc} &= \frac{V_{sc}}{I_{sc}} = \frac{22}{30} = 0.7333\ \Omega \\ R_{sc} &= \frac{P_{sc}}{I_{sc}^2} = \frac{200}{30^2} = 0.2222\ \Omega \\ X_{sc} &= \sqrt{Z_{sc}^2 - R_{sc}^2} = 0.6989\ \Omega\quad(\text{on the HV side}) \end{aligned} $$ #### Referring to the LV side The turns ratio is $K = 400/200 = 2$, 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. $$ \begin{aligned} R_{eq} &= 0.2222 \div 4 = 0.05556\ \Omega \\ X_{eq} &= 0.6989 \div 4 = 0.1747\ \Omega \end{aligned} $$ #### Equivalent circuit referred to the LV side ```text Req=0.0556 ohm Xeq=0.175 ohm o--+-----/\/\/\-----UUUU------o | [R0=333.3] [X0=173.4] Load | (in parallel) o--+---------------------------o ``` $R_{eq} = 0.05556\ \Omega$, $X_{eq} = 0.1747\ \Omega$ ($Z_{eq} = 0.1833\ \Omega$), shunt branch $R_0 = 333.3\ \Omega$ in parallel with $X_0 = 173.4\ \Omega$. **Answer (referred to LV side):** shunt branch $R_0 = 333.3\ \Omega$, $X_0 = 173.4\ \Omega$; series branch $R_{eq} = 0.0556\ \Omega$, $X_{eq} = 0.1747\ \Omega$ ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In this problem, 'OC test (HV open)' and 'SC test (LV short)' mean the readings are taken on which sides? A. OC on the HV side, SC on the LV side B. OC on the LV side at 200 V, SC on the HV side (supply to the 400 V winding) C. Both tests on the LV side D. Both tests on the HV side Answer: B. With the HV winding open, the supply is on the LV winding (200 V). With the LV shorted, the supply must be applied to the HV winding, so the SC readings are HV-side values. (Key point: Which side each test is on) 2. The OC readings are 200 V, 1.3 A, 120 W. What are the working current Iw and R0? A. Iw = 0.6 A, R0 = 333.3 Ω B. Iw = 1.153 A, R0 = 173.4 Ω C. Iw = 0.6 A, R0 = 173.4 Ω D. Iw = 1.3 A, R0 = 153.8 Ω Answer: A. cos(phi0) = 120/(200x1.3) = 0.4615, so Iw = 1.3x0.4615 = 0.6 A and R0 = 200/0.6 = 333.3 Ω. Iw is the core-loss component; Iμ = 1.153 A gives X0. (Key point: Shunt branch from OC test) 3. Why can the SC wattmeter reading (200 W) be taken as the copper loss? A. Because the core flux is at its maximum during the SC test B. Because the voltage applied is small, so the core loss is negligible C. Because iron loss and copper loss are always equal D. Because the LV winding has no resistance Answer: B. Only a reduced voltage (22 V) is applied, so flux and core loss are tiny, and the power is essentially I^2R in the windings. (Key point: Why SC wattmeter gives copper loss) 4. Rsc = 0.2222 Ω and Xsc = 0.6989 Ω are on the HV side (K = 2). What are Req and Xeq referred to the LV side? A. 0.8889 Ω and 2.796 Ω (multiply by 4) B. 0.1111 Ω and 0.3495 Ω (divide by 2) C. 0.0556 Ω and 0.1747 Ω (divide by 4) D. 0.2222 Ω and 0.6989 Ω (unchanged) Answer: C. Moving an impedance from HV to LV divides it by K^2 = 4, giving 0.05556 Ω and 0.1747 Ω. 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