--- title: "OC and SC tests for efficiency and regulation" subject: "Electrical Machines I" chapter: 2 key: "semester_4/electrical_machines_i:oc-sc-test-procedure-efficiency-regulation" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/oc-sc-test-procedure-efficiency-regulation site: BCT Engineering --- # OC and SC tests for efficiency and regulation Electrical Machines I · Chapter 2: Transformer - Asked in: 2073 Magh, 8 marks - Video: [Expt-5: Open and Short circuit Test of a single phase Transformer : ET lab (part of EN19003)](https://www.youtube.com/watch?v=fpPzgJGhKAA) — Dipankar Debnath, IIT Kharagpur - 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-procedure-efficiency-regulation - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2 ## Question Discuss how to conduct open-circuit and short-circuit tests on a single phase transformer in the laboratory. From the test results how the efficiency and voltage regulation of the transformer can be determined? ## Short answer OC test on LV side at rated voltage gives iron loss. SC test on HV side with LV shorted at rated current gives full-load copper loss and equivalent impedance. Efficiency = output/(output + iron loss + x^2 copper loss). Regulation = I(R cosφ ± X sinφ)/V × 100%. ## Answer The **open-circuit (OC)** and **short-circuit (SC)** tests find the equivalent-circuit parameters, iron loss and copper loss of a transformer without actually loading it, so very little energy is used. #### Open-circuit (no-load) test ```text AC +--(A)--+--[W]--+--------+ +------o supply| | | | LV | HV open (var.)| (V) | || | wdg | wdg +-------+-------+--------+ +------o ``` - Done on the **LV side** with HV open, at **rated voltage and frequency** (using an autotransformer/variac). - Readings: $V_0$, $I_0$, $P_0$. - Since $I_0$ is only 2-5% of rated current, copper loss is negligible, so **$P_0$ = iron loss $P_i$**. - From readings: $\cos\phi_0 = \frac{P_0}{V_0I_0}$, $I_w = I_0\cos\phi_0$, $I_\mu = I_0\sin\phi_0$, $R_0 = \frac{V_0}{I_w}$, $X_0 = \frac{V_0}{I_\mu}$. #### Short-circuit test ```text AC +--(A)--+--[W]--+--------+ +------+ supply| | | | HV | LV | short (low) | (V) | || | wdg | wdg | (thick +-------+-------+--------+ +------+ link) ``` - Done on the **HV side** with LV shorted. Voltage is raised slowly from zero until **rated current** flows; this needs only 5-10% of rated voltage. - Readings: $V_{sc}$, $I_{sc}$, $P_{sc}$. - Flux is very small at this low voltage, so iron loss is negligible and **$P_{sc}$ = full-load copper loss**. - From readings: $Z_{01} = \frac{V_{sc}}{I_{sc}}$, $R_{01} = \frac{P_{sc}}{I_{sc}^2}$, $X_{01} = \sqrt{Z_{01}^2 - R_{01}^2}$ (referred to HV). #### Finding efficiency At a fraction $x$ of full load and power factor $\cos\phi$: $$ \eta = \frac{x\,S\cos\phi}{x\,S\cos\phi + P_i + x^2P_{cu,FL}}\times 100 $$ where $P_i$ is from the OC test and $P_{cu,FL}$ from the SC test (scaled by $(I_{rated}/I_{sc})^2$ if the test was not at rated current). Maximum efficiency occurs at $x = \sqrt{P_i/P_{cu,FL}}$. #### Finding voltage regulation Using $R_{01}$ and $X_{01}$ (or $R_{02}$, $X_{02}$) from the SC test: $$ \%\text{Reg} = \frac{I(R\cos\phi \pm X\sin\phi)}{V}\times 100 $$ (+ for lagging, − for leading pf). Equivalently, $\%\text{Reg} = \%R\cos\phi \pm \%X\sin\phi$ where $\%R = \frac{P_{sc}}{S}\times100$ and $\%Z = \frac{V_{sc}}{V_{rated}}\times100$ when the test is at rated current. **Example:** if a 10 kVA transformer has $P_i = 100$ W and $P_{cu,FL} = 200$ W, full-load efficiency at upf = $\frac{10000}{10300} = 97.1\%$. **Advantages:** low power consumption, no need for a real load, and both losses found separately so efficiency at any load and pf can be predicted. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The open-circuit test is performed with: A. Rated voltage on the LV side, HV open B. Rated current on the HV side, LV shorted C. Reduced voltage on the HV side, LV open D. Rated voltage on the HV side, LV shorted Answer: A. OC test applies rated voltage to the LV side (convenient voltage) with HV open, so wattmeter reads iron loss. (Key point: OC test side) 2. In the SC test, the wattmeter reading is taken as: A. Iron loss at rated flux B. Full-load copper loss, since flux is very small C. Total loss at no load D. Core loss plus load loss equally Answer: B. Applied voltage is only 5-10% of rated, so flux and iron loss are negligible; Psc is full-load copper loss. (Key point: SC test meaning) 3. At a fraction x of full load, the copper loss to use in efficiency is: A. x Pcu,FL B. Pcu,FL C. x^2 Pcu,FL D. Pcu,FL / x Answer: C. Copper loss varies with the square of load current, so it is x^2 Pcu,FL; iron loss stays constant. (Key point: Efficiency formula) 4. Maximum efficiency occurs at load fraction: A. x = sqrt(Pcu,FL / Pi) B. x = Pi / Pcu,FL C. x = Pcu,FL / Pi D. x = sqrt(Pi / Pcu,FL) Answer: D. Max efficiency is when copper loss equals iron loss: x^2 Pcu,FL = Pi. (Key point: Maximum efficiency) ## Related in Electrical Machines I - [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 - [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 - [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 - [Efficiency and regulation of 5 kVA transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Asc-5kva-220-400-efficiency-regulation) — Electrical Machines I, chapter 2 - [Loss assumptions in OC and SC tests](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aoc-sc-test-loss-assumptions) — Electrical Machines I, chapter 2 - [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 other subjects - [OC and SC tests, efficiency and regulation](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Aoc-sc-test-efficiency-regulation) — Electrical Machine, chapter 2 - [20 kVA 250/2500 V tests, regulation, efficiency](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atf-20kva-250-2500-regulation-efficiency) — Electrical Machines, chapter 2 - [5 kVA 220/400 V efficiency and regulation](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atf-5kva-220-400-efficiency-regulation) — Electrical Machines, chapter 2 - [Open circuit and short circuit tests](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aoc-sc-test-describe) — Electrical Machines, chapter 2