--- title: "Equivalent circuit of 20 kVA 220/2200 V transformer" subject: "Electrical Machines I" chapter: 2 key: "semester_4/electrical_machines_i:oc-sc-20kva-220-2200-primary" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/oc-sc-20kva-220-2200-primary site: BCT Engineering --- # Equivalent circuit of 20 kVA 220/2200 V transformer Electrical Machines I · Chapter 2: Transformer - Asked in: 2072 Asoj, 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-20kva-220-2200-primary - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2 ## Question A 20 kVA, 220V/2200V, 50 Hz single phase transformer gave the following test results: No-load test (on L.V. side): 220 V, 1.4 A, 105 watts Short circuit test (on H.V. side): 120 V, 8 A, 320 watts Calculate the equivalent circuit parameters referred to primary side and draw the equivalent circuit showing the values of parameters. ## Short answer From OC test on LV side compute R0 = 460.95 ohm, X0 = 167.16 ohm. From SC test on HV side compute R02 = 5 ohm, X02 = 14.14 ohm, then refer to primary by factor 0.01 giving R01 = 0.05 ohm, X01 = 0.1414 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. Primary = LV side (220 V). Rated HV current $= 20000/2200 = 9.09$ A (SC test at 8 A). Factor HV to LV: $\left(\frac{220}{2200}\right)^2 = 0.01$. #### No-load test (LV): 220 V, 1.4 A, 105 W $$ \begin{aligned} \cos\phi_0 &= \frac{105}{220 \times 1.4} = 0.3409, \quad \sin\phi_0 = 0.9401 \\ I_w &= 1.4 \times 0.3409 = 0.477\ \text{A}, \quad I_\mu = 1.4 \times 0.9401 = 1.316\ \text{A} \\ R_0 &= \frac{220}{0.477} = 460.95\ \Omega, \quad X_0 = \frac{220}{1.316} = 167.16\ \Omega \end{aligned} $$ #### SC test (HV): 120 V, 8 A, 320 W $$ \begin{aligned} Z_{02} &= \frac{120}{8} = 15\ \Omega, \quad R_{02} = \frac{320}{8^2} = 5\ \Omega \\ X_{02} &= \sqrt{15^2 - 5^2} = 14.14\ \Omega \end{aligned} $$ Referred to primary (× 0.01): $$ \begin{aligned} R_{01} &= 5 \times 0.01 = 0.05\ \Omega \\ X_{01} &= 14.14 \times 0.01 = 0.1414\ \Omega \\ Z_{01} &= 15 \times 0.01 = 0.15\ \Omega \end{aligned} $$ #### Equivalent circuit referred to primary ```text I1 -> I0 I2' o-----+--------+---[ R01 ]---[ jX01 ]---o + | | + [ R0 ] [ jX0 ] load V1 | | V2' - | | - o-----+--------+------------------------o All values referred to primary (LV): R0 = 460.95 ohm, X0 = 167.16 ohm R01 = 0.05 ohm, X01 = 0.1414 ohm ``` **Answer (primary side):** $R_0 = 460.95\ \Omega$, $X_0 = 167.16\ \Omega$, $R_{01} = 0.05\ \Omega$, $X_{01} = 0.1414\ \Omega$, $Z_{01} = 0.15\ \Omega$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the OC test (220 V, 1.4 A, 105 W), cos(phi0) equals: A. 0.477 B. 0.3409 C. 0.9401 D. 0.75 Answer: B. 105/(220 x 1.4) = 0.3409. (Key point: Reading cos phi0) 2. Which formula gives the core-loss resistance R0? A. V0 / Imu B. V0 / I0 C. V0 / Iw D. P0 / Iw Answer: C. R0 = V0/Iw with Iw = I0 cos(phi0) = 0.477 A, so R0 = 460.95 ohm. (Key point: Shunt branch) 3. From the SC test (120 V, 8 A, 320 W), the HV-side R02 is: A. 15 ohm B. 5 ohm C. 14.14 ohm D. 40 ohm Answer: B. R02 = Psc/Isc^2 = 320/64 = 5 ohm. (Key point: SC series values) 4. Referring HV-side values to the 220 V primary uses the factor: A. 10 B. 100 C. 0.1 D. 0.01 Answer: D. (220/2200)^2 = 0.01, so R01 = 0.05 ohm and X01 = 0.1414 ohm. 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