--- title: "Open circuit and short circuit tests" subject: "Electrical Machines" chapter: 2 key: "semester_4/electrical_machines:oc-sc-test-describe" source: https://www.bctengineer.com/q/semester_4/electrical_machines/oc-sc-test-describe site: BCT Engineering --- # Open circuit and short circuit tests Electrical Machines · Chapter 2: Transformer - Asked in: 2066 Magh (old course), 8 marks - Video: [Open Circuit Test & Short Circuit Test of Transformer - Simple Explanation](https://www.youtube.com/watch?v=_tHyjbEdCa4) — Electrical lectures - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines/oc-sc-test-describe - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2 ## Question Describe the open circuit test and short circuit test for a single phase transformer. ## Short answer Open circuit test: apply rated voltage to LV with HV open, measures core loss and shunt parameters R0, X0. Short circuit test: short LV, apply reduced voltage to HV to get rated current, gives full‑load copper loss and series parameters Req, Xeq. Both tests determine efficiency and regulation without full loading. ## Answer The open circuit (OC) and short circuit (SC) tests find the equivalent circuit parameters, the losses and the efficiency of a transformer without loading it fully. They need very little power, which is only the loss of the transformer. #### Open circuit test (no-load test) **Purpose:** to find the core loss $P_i$, no-load current $I_0$ and the shunt branch $R_0$, $X_0$. ```text W A AC o---(W)--(A)----+----------+ supply (V) HV winding open o-----------------+----------+ LV winding (rated V) ``` **Procedure:** the HV winding is left open, and rated voltage $V_0$ at rated frequency is applied to the LV winding (this is safer and needs lower-range instruments). A wattmeter, ammeter and voltmeter are connected on the LV side. **Calculations:** $$ P_0 = V_0I_0\cos\phi_0,\quad I_w = I_0\cos\phi_0,\quad I_\mu = I_0\sin\phi_0 $$ $$ R_0 = \frac{V_0}{I_w},\qquad X_0 = \frac{V_0}{I_\mu} $$ Since $I_0$ is small (2 to 5% of rated), the copper loss $I_0^2R_1$ is negligible. So the wattmeter reading $P_0$ is the **core (iron) loss**, constant at all loads. #### Short circuit test **Purpose:** to find the full-load copper loss $P_{cu}$ and the equivalent resistance, reactance and impedance $R_{eq}$, $X_{eq}$, $Z_{eq}$. ```text W A AC o---(W)--(A)----+----------+ supply (low V, (V) LV winding via variac) short-circuited o-----------------+----------+ HV winding ``` **Procedure:** the LV winding is short-circuited with a thick conductor. A low voltage $V_{sc}$ (about 5 to 10% of rated) is applied to the HV winding and slowly increased by a variac until the rated full-load current $I_{sc}$ flows. Wattmeter, ammeter and voltmeter are on the HV side. **Calculations:** $$ Z_{eq} = \frac{V_{sc}}{I_{sc}},\qquad R_{eq} = \frac{P_{sc}}{I_{sc}^2},\qquad X_{eq} = \sqrt{Z_{eq}^2 - R_{eq}^2} $$ Because the applied voltage is low, the flux and hence the core loss are very small (core loss varies as $B^2$). So the wattmeter reading $P_{sc}$ is the **full-load copper loss**. #### Uses of the results - Equivalent circuit parameters, from which regulation is found: $\text{VR} \approx \dfrac{I(R_{eq}\cos\phi \pm X_{eq}\sin\phi)}{V}$. - Efficiency at any load: $\eta = \dfrac{xS\cos\phi}{xS\cos\phi + P_i + x^2P_{cu}}$. - The maximum efficiency load $x = \sqrt{P_i/P_{cu}}$. - Together, the two tests give the transformer's full performance while drawing only the loss power from the supply. This is their advantage over a direct load test. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the OC test, which winding is supplied and at what voltage? A. The LV winding at rated voltage with HV open B. The HV winding at rated voltage with LV open C. The LV winding at 10% voltage with HV shorted D. The HV winding at 10% voltage with LV open Answer: A. Rated voltage is applied to the LV side for safety and easier instruments; HV is left open. (Key point: OC test side and supply) 2. What does the wattmeter read in the OC test? A. Full-load copper loss B. Core loss, since I0 is small and copper loss is negligible C. Sum of core and full-load copper loss D. Only the magnetising power Answer: B. I0 is only 2 to 5% of rated, so I0^2R1 is negligible. (Key point: What OC wattmeter reads) 3. Why is only 5 to 10% of rated voltage applied in the SC test? A. To keep the core saturated B. To measure no-load current C. Just enough to circulate rated current in the shorted winding D. Because the HV winding is open Answer: C. The low impedance of the shorted transformer needs only a small voltage for full-load current. (Key point: SC test voltage) 4. In the SC test, how is R_eq found? A. R_eq = V_sc / I_sc B. R_eq = P_sc / V_sc C. R_eq = V_sc^2 / P_sc D. R_eq = P_sc / I_sc^2 Answer: D. P_sc is the copper loss at I_sc, so R_eq = P_sc/I_sc^2; Z_eq = V_sc/I_sc. 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