Question 3 of 5: Equivalent Circuit from OC and SC Tests
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exam 22-Elec-A6 Power Systems and Machines (May 2018, code 16-Elec-A6). Closed book; one double-sided aid sheet and an approved calculator permitted. Five questions of equal value constitute a complete paper — all five are worked here as a study resource. All AC quantities are RMS; three-phase voltages are line-to-line and power is total real power unless noted.
Reference texts. S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (magnetic circuits Ch. 1; transformers Ch. 2; synchronous machines Ch. 4; induction machines Ch. 7). J. D. Glover et al., Power System Analysis and Design, 6th ed. (three-phase power and power-factor correction, Ch. 2). IEEE Std 519 (harmonic limits).
Question 3: Equivalent Circuit from OC and SC Tests (20 marks)
Given. A 75 kVA, 2200/220 V transformer. The OC test is taken on the LV (220 V) side (its voltage equals rated LV) and yields the shunt/excitation branch; the SC test is taken on the HV (2200 V) side (its low voltage, 42 V, is a small fraction of 2200 V) and yields the series branch, already referred to HV.
Given data (test measurements)
Test
Voltage
Current
Power
Side
Open circuit
220 V
9.6 A
710 W
LV
Short circuit
42 V
57 A
1030 W
HV
Find. (a)–(b) the approximate (L-model) equivalent circuit referred to the HV side: series \(R_{eq}+jX_{eq}\) and shunt \(R_c \parallel jX_m\).
Approach. From the SC test compute \(R_{eq}=P_{SC}/I_{SC}^2\), \(Z_{eq}=V_{SC}/I_{SC}\), \(X_{eq}=\sqrt{Z_{eq}^2-R_{eq}^2}\) (already HV). From the OC test compute the excitation admittance on the LV side, split into conductance and susceptance, then refer to HV by multiplying resistances/reactances by \(a^2\) (with \(a=2200/220=10\)).
Series branch from the SC test (HV side).
$$R_{eq}=\frac{P_{SC}}{I_{SC}^{2}}=\frac{1030}{57^{2}}=0.317\ \Omega,\qquad
Z_{eq}=\frac{V_{SC}}{I_{SC}}=\frac{42}{57}=0.737\ \Omega.$$
$$X_{eq}=\sqrt{Z_{eq}^{2}-R_{eq}^{2}}=\sqrt{0.737^{2}-0.317^{2}}=\boxed{0.665\ \Omega\ (R_{eq}=0.317\ \Omega).}$$
Excitation branch from the OC test (LV side). Core-loss conductance and total admittance:
$$G_c=\frac{P_{OC}}{V_{OC}^{2}}=\frac{710}{220^{2}}=0.01467\ \text{S},\qquad
Y_{OC}=\frac{I_{OC}}{V_{OC}}=\frac{9.6}{220}=0.04364\ \text{S}.$$
Magnetizing susceptance:
$$B_m=\sqrt{Y_{OC}^{2}-G_c^{2}}=\sqrt{0.04364^{2}-0.01467^{2}}=0.04110\ \text{S}.$$
On the LV side, \(R_{c,LV}=1/G_c=68.2\ \Omega\), \(X_{m,LV}=1/B_m=24.3\ \Omega\).
Refer the excitation branch to the HV side. With \(a=2200/220=10\), impedances scale by \(a^2=100\):
$$R_c=a^2R_{c,LV}=\frac{(aV_{OC})^2}{P_{OC}}=\frac{2200^2}{710}=\boxed{6817\ \Omega,}$$
$$X_m=a^2X_{m,LV}=100\times24.33=\boxed{2433\ \Omega.}$$
(b) Sketch. The approximate circuit places the excitation branch across the HV terminals, then the series impedance to the (referred) load, as drawn below.
Approximate (L-model) equivalent circuit referred to the HV side. Series \(R_{eq}=0.317\ \Omega\), \(X_{eq}=0.665\ \Omega\); shunt \(R_c=6817\ \Omega\) in parallel with \(X_m=2433\ \Omega\) at the HV terminals.
