Transformer Short-Circuit Test and Efficiency: 22-Elec-A6 Power Systems and Machines, Undated paper, Q2

Question 2: Transformer Short-Circuit Test and Efficiency (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 single-phase transformer whose winding resistances are known, with a short-circuit test that fixes the series impedance.

Given data
Ratings	2300 / 240 V, 48 kVA, 60 Hz
Primary (HV) resistance	$R_1 = 0.6\ \Omega$
Secondary (LV) resistance	$R_2 = 0.025\ \Omega$
SC applied voltage (HV) for rated current	$V_{sc} = 238\ \text{V}$
Load condition (c)	rated kVA, 0.8 p.f. lagging
Loss model (c)	core loss negligible ("losses are too small to account for")

Find. (a) the equivalent series reactance $X_{eq}$; (b) the power $P$ required to drive rated current at the short-circuit test voltage (= the full-load copper loss); (c) the efficiency at rated load, 0.8 p.f. lagging.

Approach. Refer both winding resistances to the HV side to get $R_{eq}$, use the short-circuit voltage and rated current to get $Z_{eq}$, then $X_{eq} = \sqrt{Z_{eq}^2 - R_{eq}^2}$. The short-circuit input power is the copper loss, and — with core loss neglected — the efficiency follows directly.

Turns ratio and rated currents. $$a = \frac{2300}{240} = 9.583,\qquad I_1 = \frac{48\,000}{2300} = 20.87\ \text{A},\qquad I_2 = \frac{48\,000}{240} = 200\ \text{A}.$$
Equivalent resistance referred to HV. Reflect $R_2$ by $a^2$ and add to $R_1$: $$R_{eq} = R_1 + a^2 R_2 = 0.6 + (9.583)^2(0.025) = 0.6 + 2.296 = 2.896\ \Omega.$$
Equivalent impedance from the SC test. The 238 V drives rated HV current through the series branch: $$Z_{eq} = \frac{V_{sc}}{I_1} = \frac{238}{20.87} = 11.40\ \Omega.$$
Equivalent reactance (a). $$X_{eq} = \sqrt{Z_{eq}^2 - R_{eq}^2} = \sqrt{11.40^2 - 2.896^2} = \sqrt{121.7} = 11.03\ \Omega.$$ $$\boxed{X_{eq} = 11.0\ \Omega\ \text{(referred to the HV side)}.}$$
Power required for rated current (b). All of the short-circuit input is dissipated as copper loss (the core is essentially unexcited at 238 V): $$P_{sc} = I_1^2 R_{eq} = (20.87)^2(2.896) = 1261\ \text{W} \approx 1.26\ \text{kW}.$$ This equals the full-load copper loss $P_{cu}$.
Efficiency at rated load (c). Output power at rated kVA and 0.8 p.f.: $$P_{out} = S\cos\theta = (48\,000)(0.8) = 38\,400\ \text{W}.$$ With core loss neglected, only copper loss subtracts: $$\eta = \frac{P_{out}}{P_{out}+P_{cu}} = \frac{38\,400}{38\,400+1261} = 0.9682 = 96.8\%.$$

Check: Part (b), "P required for rated current/voltage", is read as the real power the supply must deliver in the short-circuit test to circulate rated current at $V_{sc}=238$ V. Because that voltage is only ~10% of rated and produces negligible core flux, this power is entirely copper loss and equals the full-load copper loss (1.26 kW). Part (c)'s closing sentence, "assume that losses are too small to account for", must mean the core losses only — the December 2019 re-issue of this same paper spells it out as "the core losses are too small to account for". Copper loss is therefore retained; had every loss been dropped the efficiency would be a meaningless 100% and part (b) would serve no purpose.

Final results — Question 2
Quantity	Value
(a) Equivalent reactance $X_{eq}$ (HV)	11.0 Ω
(b) Power required for rated current = copper loss	1.26 kW
(c) Efficiency at rated load, 0.8 p.f. lag	96.8%

Topic: Transformer equivalent circuit and the short-circuit (impedance) test.

Key relations: $a = V_1/V_2$; $R_{eq} = R_1 + a^2 R_2$; $Z_{eq} = V_{sc}/I_{rated}$; $X_{eq}=\sqrt{Z_{eq}^2 - R_{eq}^2}$; $P_{cu}=I^2R_{eq}$; $\eta = P_{out}/(P_{out}+P_{cu}+P_{core})$.

Why this works: During the short-circuit test the secondary is shorted and only a small voltage (here 238 V) is needed to circulate rated current, so almost all the applied voltage drops across the series leakage impedance and virtually no flux — hence no core loss — is produced. The wattmeter therefore reads copper loss, and the applied-voltage-to-current ratio is the series impedance. Reflecting $R_2$ across the ideal transformer requires the square of the turns ratio because both the reflected current scales by $1/a$ and the reflected voltage by $a$.

Common pitfall: Forgetting the $a^2$ factor when reflecting the secondary resistance, or treating the 238 V as a line-to-line or secondary quantity. Also, reading "losses are too small to account for" as "there are no losses at all" — copper loss must be kept or the efficiency collapses to a meaningless 100%.

Source: Chapman, §2.6–2.7 (transformer tests and efficiency); Sen, ch. 2.

Question 2 of 5: Transformer Short-Circuit Test and Efficiency

Question 2: Transformer Short-Circuit Test and Efficiency (20 marks)