--- title: "OC/SC test of 2200/110 V, 40 kVA transformer" subject: "Electrical Machines I" chapter: 2 key: "semester_4/electrical_machines_i:transformer-oc-sc-2200-110-40kva" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/transformer-oc-sc-2200-110-40kva site: BCT Engineering --- # OC/SC test of 2200/110 V, 40 kVA transformer Electrical Machines I · Chapter 2: Transformer - Asked in: 2082 Kartik (new course), 6 marks - Video: [35. OPEN CIRCUIT TEST, SHORT CIRCUIT TEST NUMERICAL PROBLEM, FIND TRANSFORMER CIRCUIT CONSTANT](https://www.youtube.com/watch?v=PCpJF99BoZs) — Electrical Tutorial - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/transformer-oc-sc-2200-110-40kva - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2 ## Question During open circuit and short circuit test of a 2200/110 V, 40 kVA transformer, following results were obtained. The meters for open circuit test were placed on LV side and for short circuit test on HV side. Open circuit: P = 320 W; I = 8.4 A; V = 110 V Short circuit: P = 650 W; I = 18.2 A; V = 85 V Determine: (i) R0 and X0 (ii) The efficiency at full load with pf 0.85 lag (iii) The approx. voltage regulation ## Short answer R₀ = 37.81 Ω, X₀ = 13.96 Ω (LV side). Full-load efficiency at 0.85 pf lag is 97.23%. Approximate voltage regulation is 3.22%. ## Answer The OC test (LV side) gives the shunt branch; the SC test (HV side) gives the series impedance referred to HV. Rated HV current $= 40000/2200 = 18.18$ A, so the SC test (18.2 A) is practically at full load. #### (i) $R_0$ and $X_0$ (LV side) $$ \begin{aligned} \cos\phi_0 &= \frac{320}{110\times8.4} = 0.3463 \\ I_c &= 8.4\times0.3463 = 2.909\ \text{A}, \quad I_m = 8.4\times\sin(69.74^\circ) = 7.880\ \text{A} \\ R_0 &= \frac{110}{2.909} = 37.81\ \Omega, \quad X_0 = \frac{110}{7.880} = 13.96\ \Omega \end{aligned} $$ #### SC test values (HV side) $$ \begin{aligned} R_{eq} &= \frac{650}{18.2^2} = 1.962\ \Omega, \quad Z_{eq} = \frac{85}{18.2} = 4.670\ \Omega \\ X_{eq} &= \sqrt{4.670^2 - 1.962^2} = 4.238\ \Omega \end{aligned} $$ #### (ii) Full-load efficiency at 0.85 pf lag Full-load copper loss (corrected to rated current 18.18 A): $$ \begin{aligned} P_{cu,fl} &= 650\left(\frac{18.18}{18.2}\right)^2 = 648.7\ \text{W} \\ P_{out} &= 40000\times0.85 = 34000\ \text{W} \\ \eta &= \frac{34000}{34000 + 320 + 648.7} = \frac{34000}{34968.7} = 97.23\% \end{aligned} $$ #### (iii) Approximate voltage regulation (0.85 pf lag, $\sin\phi = 0.527$) $$ \begin{aligned} \%R &= \frac{I(R_{eq}\cos\phi + X_{eq}\sin\phi)}{V}\times100 \\ &= \frac{18.18(1.962\times0.85 + 4.238\times0.527)}{2200}\times100 \\ &= \frac{30.33 + 40.59}{2200}\times100 = 3.22\% \end{aligned} $$ **Answer:** $R_0 = 37.81\ \Omega$, $X_0 = 13.96\ \Omega$ (LV side); full-load efficiency = **97.23%**; regulation ≈ **3.22%**. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are R0 and X0 found? A. R0 = P/I0² = 4.5 Ω, X0 from the SC test B. Ic = I0 cos φ0 and Im = I0 sin φ0, then R0 = V/Ic, X0 = V/Im, giving 37.81 Ω and 13.96 Ω C. R0 = V/I0 = 13.1 Ω and X0 = 0 D. R0 = 650/18.2² and X0 = 4.24 Ω Answer: B. cos φ0 = 320/(110×8.4) = 0.3463, so Ic = 2.909 A and Im = 7.88 A. (Key point: Shunt branch) 2. Why is the SC copper loss corrected to 650×(18.18/18.2)²? A. Because the SC test was at half load B. Rated HV current is 40000/2200 = 18.18 A, and Cu loss varies as I² C. Because iron loss adds to the SC reading D. Because rated HV current is 8.4 A Answer: B. Copper loss at rated current is 648.7 W. (Key point: Full-load copper loss) 3. What is the full-load efficiency at 0.85 pf lag? A. 97.9%, from 40000/(40000 + 320 + 648.7) B. 99.1%, because iron loss is ignored C. 97.23%, from 34000/(34000 + 320 + 648.7) D. 95.2%, because losses are counted twice Answer: C. Output = 40000 × 0.85 = 34 kW, and losses are iron 320 W plus full-load copper 648.7 W. (Key point: Efficiency) 4. What is the approximate regulation formula and value? A. I(Req cos φ − Xeq sin φ)/V ×100 ≈ 0.4% B. I(Req cos φ + Xeq sin φ)/V ×100 ≈ 3.22% C. I Zeq/V ×100 = 3.86% always D. I(Req sin φ + Xeq cos φ)/V ×100 ≈ 3.3% Answer: B. For lagging pf the two drops add: 30.33 + 40.59 = 70.92 V of 2200 V. 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