--- title: "OC and SC tests to build transformer equivalent circuit" subject: "Electrical Machines I" chapter: 2 key: "semester_4/electrical_machines_i:oc-sc-test-10kva-6600-220" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/oc-sc-test-10kva-6600-220 site: BCT Engineering --- # OC and SC tests to build transformer equivalent circuit Electrical Machines I · Chapter 2: Transformer - Asked in: 2069 Bhadra, 8 marks - Video: [open circuit test and short circuit test | open circuit test of transformer | transformer efficiency](https://www.youtube.com/watch?v=bHd9plIEhNM) — Mind of Engineer - 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-test-10kva-6600-220 - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2 ## Question Describe no load test and short circuit test of a single phase transformer having following ratings: 10 kVA, 6600 V/220 V. How the results obtained from the tests can be utilized to develop the equivalent circuit of the transformer referred to primary side? ## Short answer Perform OC test on LV side at rated voltage to obtain R0 and X0 (shunt branch). Perform SC test on HV side at rated current to obtain R01 and X01 (series branch). Refer shunt parameters to primary by multiplying by (6600/220)^2 = 900. The equivalent circuit: shunt branch (R0' || X0') across primary, then series R01+jX01. ## Answer The **no-load (open-circuit) test** and **short-circuit test** find the equivalent-circuit parameters and losses of the 10 kVA, 6600/220 V transformer without loading it. Rated currents: HV $= 10000/6600 = 1.515$ A; LV $= 10000/220 = 45.45$ A. #### No-load (OC) test ```text 0-220 V +--(A)--[W]--+------+ +-----o variac | | LV | || | HV open 50 Hz (V) | 220V | || | 6600 V +------------+------+ +-----o ``` - Done on the **LV (220 V) side**, HV open. The LV side is used because 220 V is safe and easy to supply and measure; the HV terminals still have 6600 V on them and must be guarded. - Apply **rated voltage, 220 V**, at rated frequency. Read $V_0$, $I_0$, $P_0$. - $I_0$ is small (about 2-5% of 45.45 A, i.e. 1-2 A), so copper loss is negligible and $P_0$ = **iron loss**. Calculations: $$ \cos\phi_0 = \frac{P_0}{V_0I_0}, \quad I_w = I_0\cos\phi_0, \quad I_\mu = I_0\sin\phi_0, \quad R_0 = \frac{V_0}{I_w}, \quad X_0 = \frac{V_0}{I_\mu} $$ These are on the LV side. #### Short-circuit test ```text low V +--(A)--[W]--+------+ +-----+ variac | | HV | || | LV | thick 50 Hz (V) | 6600V| || | 220V| short +------------+------+ +-----+ ``` - Done on the **HV (6600 V) side**, LV shorted by a thick link. The HV side is used because rated current there is only 1.515 A and the needed voltage (5-10% of 6600 V, i.e. about 330-660 V) is easy to measure; on the LV side we would need 45 A at only 11-22 V. - Raise voltage slowly from zero until **rated current 1.515 A** flows. Read $V_{sc}$, $I_{sc}$, $P_{sc}$. - Flux is very low, so iron loss is negligible and $P_{sc}$ = **full-load copper loss**. Calculations (HV side, which is the primary here): $$ Z_{01} = \frac{V_{sc}}{I_{sc}}, \quad R_{01} = \frac{P_{sc}}{I_{sc}^2}, \quad X_{01} = \sqrt{Z_{01}^2 - R_{01}^2} $$ #### Developing the equivalent circuit referred to primary (6600 V) 1. $R_{01}$ and $X_{01}$ come directly from the SC test because it was on the HV (primary) side. 2. $R_0$ and $X_0$ from the OC test are on the LV side, so refer them to the primary by multiplying by $$ \left(\frac{6600}{220}\right)^2 = 30^2 = 900 $$ i.e. $R_0' = 900R_0$, $X_0' = 900X_0$. 3. Draw the approximate equivalent circuit: shunt branch ($R_0' \parallel X_0'$) across the 6600 V input, followed by $R_{01} + jX_{01}$ in series, and load referred to primary $Z_L' = 900Z_L$. ```text o---+---------+---[ R01 ]---[ jX01 ]---o | | + [ R0' ] [ jX0' ] V2' = 30 V2 | | - o---+---------+------------------------o ``` **Example:** if the OC test gives 220 V, 1.2 A, 90 W, then $\cos\phi_0 = 0.341$, $I_w = 0.409$ A, $R_0 = 538\ \Omega$ (LV), so $R_0' = 484\ \text{k}\Omega$ on the 6600 V side. From the same results: iron loss and full-load copper loss give efficiency at any load, and $R_{01}$, $X_{01}$ give voltage regulation. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the OC test done on the LV side? A. Iron loss is larger on LV B. 220 V is safe and easy to supply and measure C. Copper loss is zero on LV D. LV has more turns Answer: B. Rated LV voltage is low and convenient, so the test is done on LV with HV open. (Key point: Side for OC test) 2. Why is the SC test done on the HV side? A. Rated current there is only 1.515 A and the required voltage is easy to measure B. It needs 6600 V applied C. It avoids copper loss D. It measures the iron loss Answer: A. On HV the rated current is small and the needed voltage is 5-10% of 6600 V; on LV 45 A at 11-22 V would be hard. (Key point: Side for SC test) 3. The wattmeter reading in the SC test at rated current gives: A. Iron loss B. Full-load copper loss C. No-load loss D. Output power Answer: B. Flux is very low in SC test, so iron loss is negligible and Psc is the full-load copper loss. (Key point: What the wattmeters read) 4. R0 and X0 from the OC test (LV side) are referred to the 6600 V primary by multiplying by: A. 30 B. 900 C. 1/900 D. 1/30 Answer: B. Impedance is multiplied by the square of the turns ratio: (6600/220)^2 = 900. 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