--- title: "Equivalent circuit of 100 kVA 2200/220 V transformer" subject: "Electrical Machines I" chapter: 2 key: "semester_4/electrical_machines_i:oc-sc-100kva-2200-220-hv" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/oc-sc-100kva-2200-220-hv site: BCT Engineering --- # Equivalent circuit of 100 kVA 2200/220 V transformer Electrical Machines I · Chapter 2: Transformer - Asked in: 2078 Baisakh, 8 marks - Video: [Equivalent Circuit of Transformer : Problem 2 - Single Phase Transformer](https://www.youtube.com/watch?v=cRoHiuvO_A4) — Ekeeda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/oc-sc-100kva-2200-220-hv - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2 ## Question A 100 kVA, 2200/220 V transformer when tested gave the following results: OC test: 400 W, 10 A, 220 V SC test: 808 W, 30 A, 90 V Compute all the parameters of the equivalent circuit referred to HV side of the transformer. Draw the equivalent circuits also. ## Short answer OC test on LV gives R0=12100 ohm and X0=2237.3 ohm referred to HV. SC test on HV gives R01=0.898 ohm, X01=2.863 ohm, and Z01=3 ohm. ## Answer **Method.** The no-load (OC) test gives the shunt branch: $\cos\phi_0 = \frac{P_0}{V_0 I_0}$, $I_w = I_0\cos\phi_0$, $I_\mu = I_0\sin\phi_0$, $R_0 = \frac{V_0}{I_w}$, $X_0 = \frac{V_0}{I_\mu}$. The SC test gives the series branch: $Z = \frac{V_{sc}}{I_{sc}}$, $R = \frac{P_{sc}}{I_{sc}^2}$, $X = \sqrt{Z^2 - R^2}$. Values found on one side are moved to the other side by multiplying by the square of the voltage ratio. The OC test voltage (220 V) shows it was done on the LV side; the SC test voltage (90 V) and current (30 A, below the rated HV current of $100000/2200 = 45.45$ A) show it was done on the HV side. Ratio factor LV to HV: $\left(\frac{2200}{220}\right)^2 = 100$. #### Shunt branch (OC test: 220 V, 10 A, 400 W on LV) $$ \begin{aligned} \cos\phi_0 &= \frac{400}{220 \times 10} = 0.1818, \quad \sin\phi_0 = 0.9833 \\ I_w &= 10 \times 0.1818 = 1.818\ \text{A}, \quad I_\mu = 10 \times 0.9833 = 9.833\ \text{A} \\ R_0 (\text{LV}) &= \frac{220}{1.818} = 121\ \Omega, \quad X_0 (\text{LV}) = \frac{220}{9.833} = 22.37\ \Omega \\ R_0 (\text{HV}) &= 121 \times 100 = 12100\ \Omega \\ X_0 (\text{HV}) &= 22.37 \times 100 = 2237.3\ \Omega \end{aligned} $$ #### Series branch (SC test: 90 V, 30 A, 808 W on HV) $$ \begin{aligned} Z_{01} &= \frac{90}{30} = 3\ \Omega \\ R_{01} &= \frac{808}{30^2} = 0.898\ \Omega \\ X_{01} &= \sqrt{3^2 - 0.898^2} = 2.863\ \Omega \end{aligned} $$ #### Equivalent circuits Approximate equivalent circuit referred to HV (shunt branch moved to the input terminals): ```text I1 -> I0 I2' o-----+--------+---[ R01 ]---[ jX01 ]---o + | | + [ R0 ] [ jX0 ] load V1 | | V2' - | | - o-----+--------+------------------------o All values referred to HV side: R0 = 12100 ohm, X0 = 2237.3 ohm R01 = 0.898 ohm, X01 = 2.863 ohm ``` The exact circuit has the series impedance split as $R_1 + jX_1$ before the shunt branch and $R_2' + jX_2'$ after it (often taken as $R_1 = R_2' = R_{01}/2$, $X_1 = X_2' = X_{01}/2$) and an ideal transformer at the output; the approximate form above is accurate enough because $I_0$ is small. **Answer (HV side):** $R_0 = 12100\ \Omega$, $X_0 = 2237.3\ \Omega$, $R_{01} = 0.898\ \Omega$, $X_{01} = 2.863\ \Omega$, $Z_{01} = 3\ \Omega$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How can you tell the OC test was done on the LV side and the SC test on the HV side? A. Both tests were done on LV B. OC current is larger than SC current C. OC voltage of 220 V is LV rated; SC current 30 A is below HV rated 45.45 A D. SC voltage equals rated voltage Answer: C. 220 V is the LV rating, and 30 A is a sensible fraction of the HV rated current 100000/2200 = 45.45 A. (Key point: Identifying the sides of the tests) 2. What multiplier converts the LV shunt values to HV? A. (2200/220)^2 = 100 B. (220/2200) C. 10 D. 1/100 Answer: A. Impedance scales with the square of the turns ratio of 10, hence 100. (Key point: Referring shunt branch to HV) 3. With I0 = 10 A and cos phi0 = 0.1818, what is X0 on the LV side? A. 220/9.833 = 22.37 ohm B. 220/10 = 22 ohm exactly C. 220/1.818 = 121 ohm D. 400/9.833 ohm Answer: A. Imu = I0 sin phi0 = 9.833 A, so X0 = V0/Imu = 22.37 ohm; 121 ohm is R0. (Key point: Magnetising reactance) 4. What are R01 and X01 from the SC test (90 V, 30 A, 808 W)? A. R01 = 3 ohm, X01 = 0.898 ohm B. R01 = 0.0898 ohm, X01 = 2.99 ohm C. R01 = 0.898 ohm, X01 = 2.863 ohm D. R01 = 2.863 ohm, X01 = 0.898 ohm Answer: C. R01 = 808/900 = 0.898; Z01 = 3; X01 = sqrt(9 - 0.806) = 2.863 ohm. (Key point: Series branch) ## Close variants set in other exams - [Equivalent circuit of 20 kVA 250/2500 V transformer from tests](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aoc-sc-20kva-250-2500) — Electrical Machines I, chapter 2, asked 3× ## Related in Electrical Machines I - [Equivalent circuit of 20 kVA 220/2200 V transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aoc-sc-20kva-220-2200-primary) — Electrical Machines I, chapter 2 - [OC/SC test of 2200/110 V, 40 kVA transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Atransformer-oc-sc-2200-110-40kva) — Electrical Machines I, chapter 2 - [OC and SC tests to build transformer equivalent circuit](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aoc-sc-test-10kva-6600-220) — Electrical Machines I, chapter 2 - [Equivalent circuit of 8 kVA 400/120 V transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aoc-sc-8kva-400-120-hv) — Electrical Machines I, chapter 2 - [Equivalent circuit, efficiency and regulation of 5 kVA transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aoc-sc-5kva-220-400-lv) — Electrical Machines I, chapter 2 - [Equivalent circuit of 5 kVA 200/400 V transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aoc-sc-5kva-200-400-primary) — Electrical Machines I, chapter 2, asked 2× ## Related in other subjects - [20 kVA 2200/220 V tests, equivalent circuit LV side](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atf-20kva-2200-220-lv-equivalent) — Electrical Machines, chapter 2 - [10 kVA 200/400 V LV-side parameters](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Atransformer-10kva-oc-sc-lv-circuit) — Electrical Machine, chapter 2 - [10 kVA 200/400 V tests, parameters on LV side](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atf-10kva-200-400-lv-params) — Electrical Machines, chapter 2 - [20 kVA 2200/220 V, LV-side circuit](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Atransformer-20kva-2200-220-lv-side) — Electrical Machine, chapter 2