Transformer Equivalent Circuit from OC and SC Tests: 22-Elec-A6 Power Systems and Machines, Undated paper, Q3

Question 3: Transformer Equivalent Circuit from OC and SC Tests (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A 75 kVA, 2200/220 V transformer with the standard pair of tests: the open-circuit test performed from the LV (220 V) side and the short-circuit test from the HV side.

Given data
Ratings	75 kVA, 2200 / 220 V, 60 Hz
Open-circuit test (LV side)	$V_{OC}=220\ \text{V},\ I_{OC}=9.6\ \text{A},\ P_{OC}=710\ \text{W}$
Short-circuit test (HV side)	$V_{SC}=42\ \text{V},\ I_{SC}=57\ \text{A},\ P_{SC}=1030\ \text{W}$

Find. The shunt (magnetizing) branch $R_c,\ X_m$ and the series branch $R_{eq},\ X_{eq}$, all referred to the HV side, and a sketch of the approximate equivalent circuit.

Approach. The OC test (rated voltage, tiny current) yields the core-loss and magnetizing branch on the LV side; refer it to HV by $a^2$. The SC test (rated current, small voltage) yields the series impedance directly on the HV side. Turns ratio $a = 2200/220 = 10$.

Magnetizing branch from the OC test (LV side). The core-loss resistance and magnetizing reactance are $$R_{c,\text{LV}} = \frac{V_{OC}^2}{P_{OC}} = \frac{220^2}{710} = 68.17\ \Omega.$$ The reactive component uses the OC apparent power $S_{OC}=V_{OC}I_{OC}=2112$ VA: $$Q_{OC} = \sqrt{S_{OC}^2 - P_{OC}^2} = \sqrt{2112^2 - 710^2} = 1989\ \text{var},\quad X_{m,\text{LV}} = \frac{V_{OC}^2}{Q_{OC}} = \frac{48\,400}{1989} = 24.33\ \Omega.$$
Refer the shunt branch to the HV side. With $a = 10$, multiply by $a^2 = 100$: $$R_c = 100\,(68.17) = 6817\ \Omega,\qquad X_m = 100\,(24.33) = 2433\ \Omega.$$
Series branch from the SC test (HV side). The test is already on the HV side, so $$R_{eq} = \frac{P_{SC}}{I_{SC}^2} = \frac{1030}{57^2} = 0.317\ \Omega,\qquad Z_{eq} = \frac{V_{SC}}{I_{SC}} = \frac{42}{57} = 0.737\ \Omega.$$
Series reactance. $$X_{eq} = \sqrt{Z_{eq}^2 - R_{eq}^2} = \sqrt{0.737^2 - 0.317^2} = 0.665\ \Omega.$$ $$\boxed{R_{eq}=0.317\ \Omega,\ X_{eq}=0.665\ \Omega,\ R_c=6817\ \Omega,\ X_m=2433\ \Omega\ \text{(all referred to HV)}.}$$
Sketch of the approximate equivalent circuit (b). In the approximate model the shunt branch is moved to the input terminals, ahead of the series impedance:

Approximate equivalent circuit referred to the 2200 V (HV) side: magnetizing branch $R_c\parallel X_m$ across the supply, series leakage impedance $R_{eq}+jX_{eq}$ to the load.

Check — which side each test was run on. The open-circuit test is identified as the LV test because $V_{OC}=220$ V is exactly the rated LV winding voltage, and the short-circuit test as the HV test because $V_{SC}=42$ V is only 1.9% of 2200 V (a credible short-circuit voltage), whereas 42 V applied to the 220 V winding would be 19% and not a short-circuit test at all. Note that the quoted $I_{SC}=57$ A is about 1.67 times the rated HV current ($75\,000/2200 = 34.1$ A), so the examiner's test was not run at exactly rated current. This does not affect the answer: both $R_{eq}=P_{SC}/I_{SC}^2$ and $Z_{eq}=V_{SC}/I_{SC}$ are ratios of the measured quantities and are valid at whatever current the test was performed, provided the same reading pair is used throughout — which is what is done above.

Final results — Question 3 (referred to HV)
Branch	Resistance	Reactance
Series (leakage)	$R_{eq}=0.317\ \Omega$	$X_{eq}=0.665\ \Omega$
Shunt (magnetizing)	$R_{c}=6817\ \Omega$	$X_{m}=2433\ \Omega$

Topic: Building the approximate transformer equivalent circuit from open- and short-circuit tests.

Key relations: OC: $R_c=V_{OC}^2/P_{OC}$, $X_m=V_{OC}^2/Q_{OC}$ with $Q_{OC}=\sqrt{(V_{OC}I_{OC})^2-P_{OC}^2}$. SC: $R_{eq}=P_{SC}/I_{SC}^2$, $X_{eq}=\sqrt{(V_{SC}/I_{SC})^2-R_{eq}^2}$. Refer between sides by $a^2$.

Why this works: The two tests isolate the two branches. Open-circuit at rated voltage drives full core flux but almost no series current, so the wattmeter reads core loss and the branch parameters are the shunt $R_c\parallel X_m$. Short-circuit at rated current needs only a few percent of rated voltage, so the flux and core loss are negligible and the measured impedance is purely the series leakage branch. Because each branch is measured on the side that is convenient, referring one of them across the ideal transformer (here the magnetizing branch, LV → HV) requires the square of the turns ratio.

Common pitfall: Referring the magnetizing branch by $a$ instead of $a^2$, or assuming the SC test was on the LV side — here $V_{SC}=42$ V is a small fraction of 2200 V, so the SC test is the HV one. A related slip is to "correct" the SC data to rated current: the measured pair must be used as it stands, since $R_{eq}$ and $Z_{eq}$ are ratios and rescaling only one of $V_{SC}$ and $I_{SC}$ corrupts them.

Source: Chapman, §2.6 (determining transformer model parameters); Guru & Hiziroglu, ch. 4.

Question 3 of 5: Transformer Equivalent Circuit from OC and SC Tests

Question 3: Transformer Equivalent Circuit from OC and SC Tests (20 marks)