--- title: "Why iron loss neglected in SC test; parameters" subject: "Electrical Machines" chapter: 2 key: "semester_4/electrical_machines:sc-test-iron-loss-400-200" source: https://www.bctengineer.com/q/semester_4/electrical_machines/sc-test-iron-loss-400-200 site: BCT Engineering --- # Why iron loss neglected in SC test; parameters Electrical Machines · Chapter 2: Transformer - Asked in: 2079 Jestha, 2+6 marks - 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/sc-test-iron-loss-400-200 - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2 ## Question Why is iron loss neglected in short circuit test of transformer? A single phase 400/200 V, 50 Hz transformer gave the following test results: Open circuit Test: 200 V, 1.5 A, 110 W on LV side Short circuit Test: 30 V, 18 A, 350 W on HV side Calculate the equivalent circuit parameters referred to primary side and draw the equivalent circuit showing calculated parameters. ## Short answer Short circuit test uses only 5–10% rated voltage, so flux and iron loss are negligible. Equivalent circuit parameters referred to primary (400 V): Req = 1.08 Ω, Xeq = 1.269 Ω, R0 = 1455 Ω, X0 = 573.3 Ω. ## Answer #### Why iron loss is neglected in the SC test In the short-circuit test only a small voltage (about 5 to 10% of the rated value) is applied, just enough to circulate full-load current in the shorted secondary. The core flux is proportional to the applied voltage, so it is only about 5 to 10% of normal flux. Core loss varies approximately as the square of the flux, so it becomes about 0.25 to 1% of the rated core loss, which is negligible. The wattmeter therefore reads the copper loss alone. In this problem $30/400 = 7.5\%$ of rated voltage, so the core loss is about $0.075^2\times110 \approx 0.6$ W against 350 W. #### Equivalent circuit parameters (referred to the primary) Primary = HV side (400 V), secondary = LV side (200 V). #### Open-circuit test (on the LV side) The no-load power is almost entirely core loss, so the shunt branch is found from it. $$ \begin{aligned} \cos\phi_0 &= \frac{P_0}{V_0I_0} = \frac{110}{200\times1.5} = 0.3667 \\ I_w &= I_0\cos\phi_0 = 0.55\ \text{A},\qquad I_\mu = \sqrt{I_0^2 - I_w^2} = 1.396\ \text{A} \\ R_0 &= \frac{V_0}{I_w} = 363.6\ \Omega,\qquad X_0 = \frac{V_0}{I_\mu} = 143.3\ \Omega\quad(\text{on the LV side}) \end{aligned} $$ #### 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. $$ \begin{aligned} Z_{sc} &= \frac{V_{sc}}{I_{sc}} = \frac{30}{18} = 1.667\ \Omega \\ R_{sc} &= \frac{P_{sc}}{I_{sc}^2} = \frac{350}{18^2} = 1.08\ \Omega \\ X_{sc} &= \sqrt{Z_{sc}^2 - R_{sc}^2} = 1.269\ \Omega\quad(\text{on the HV side}) \end{aligned} $$ #### Referring to the HV 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. $$ \begin{aligned} R_0' &= 363.6 \times 4 = 1455\ \Omega \\ X_0' &= 143.3 \times 4 = 573.3\ \Omega \end{aligned} $$ #### Equivalent circuit referred to the HV side ```text Req=1.08 ohm Xeq=1.27 ohm o--+-----/\/\/\-----UUUU------o | [R0=1455] [X0=573.3] Load | (in parallel) o--+---------------------------o ``` $R_{eq} = 1.08\ \Omega$, $X_{eq} = 1.269\ \Omega$ ($Z_{eq} = 1.667\ \Omega$), shunt branch $R_0 = 1455\ \Omega$ in parallel with $X_0 = 573.3\ \Omega$. The shunt branch carries $I_0$ referred to the primary $= 1.5/2 = 0.75$ A. **Answer (referred to primary):** $R_{01} = 1.080\ \Omega$, $X_{01} = 1.269\ \Omega$, $R_0 = 1454.5\ \Omega$, $X_0 = 573.3\ \Omega$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is iron loss neglected in the SC test? A. The low applied voltage gives a small flux, and core loss varies as flux squared B. The core is removed in the SC test C. Copper loss cancels the core loss D. The secondary current demagnetises the core Answer: A. At 5 to 10% voltage the core loss is only about 0.25 to 1% of its rated value. (Key point: Why iron loss negligible) 2. What is cos phi0 in the OC test (200 V, 1.5 A, 110 W)? A. 0.5 B. 0.667 C. 0.3667 D. 0.1833 Answer: C. cos phi0 = 110/(200 x 1.5) = 0.3667. (Key point: OC power factor) 3. The OC shunt values measured on the LV side are referred to the HV side by: A. Dividing by K^2 = 4 B. Multiplying by K^2 = 4 C. Multiplying by K = 2 D. Dividing by K = 2 Answer: B. Impedance moved from LV to HV is multiplied by K^2. (Key point: Referring impedance) 4. What is R_sc in the SC test (30 V, 18 A, 350 W)? A. 1.08 ohm B. 1.667 ohm C. 19.4 ohm D. 0.58 ohm Answer: A. R_sc = P/I^2 = 350/324 = 1.08 ohm. 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