Transformer (Short-Circuit Analysis): 22-Elec-A6 Power Systems and Machines, December 2014, Q3

Question 3 — Transformer (Short-Circuit Analysis)

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, step-down transformer (turns ratio $a=N_1/N_2=3$) with the low-voltage winding short-circuited and 230 V applied to the high-voltage (primary) winding.

Given data
Turns ratio $a$	3	Applied HV voltage	230 V
Primary $R_1,\,X_1$	$1.2\ \Omega,\ 6\ \Omega$	Secondary $R_2,\,X_2$	$0.05\ \Omega,\ 0.3\ \Omega$

Find. The low-voltage-winding current, the total copper loss, and the input power factor under the short-circuit condition.

Approach. Refer the secondary impedance to the primary with $a^2$, add it to the primary impedance to form the equivalent series impedance, and solve the single-loop short-circuit circuit; the magnetizing branch is negligible at reduced short-circuit voltage.

Approximate equivalent circuit referred to the high-voltage side, low-voltage winding short-circuited. $R_{eq}=R_1+a^2R_2$ and $X_{eq}=X_1+a^2X_2$.

Refer the secondary impedance to the primary. $R_2'=a^2R_2=9(0.05)=0.45\ \Omega$ and $X_2'=a^2X_2=9(0.3)=2.70\ \Omega$.
Equivalent series impedance. $R_{eq}=R_1+R_2'=1.65\ \Omega$, $X_{eq}=X_1+X_2'=8.70\ \Omega$, so $Z_{eq}=1.65+j8.70=8.855\angle79.3^\circ\ \Omega$.
Primary and secondary currents. With the LV winding shorted the reflected short appears across $Z_{eq}$: $$I_1=\frac{230}{|Z_{eq}|}=\frac{230}{8.855}=25.97\ \text{A}\quad\Rightarrow\quad I_2=a\,I_1=3(25.97)=\boxed{77.9\ \text{A}}$$
Copper loss. All real power at short circuit is copper loss: $$P_{cu}=I_1^2R_{eq}=(25.97)^2(1.65)=\boxed{1113\ \text{W}}$$ (check by windings: $I_1^2R_1+I_2^2R_2=809+304=1113\ \text{W}$).
Power factor. $\text{pf}=\cos\!\big(\tan^{-1}(X_{eq}/R_{eq})\big)=R_{eq}/|Z_{eq}| =1.65/8.855=\boxed{0.186\ \text{lagging}}$.

Question 3 — short-circuit results
Quantity	Value
(a) LV-winding current $I_2$	77.9 A
(b) Total copper loss $P_{cu}$	1113 W (1.11 kW)
(c) Power factor	0.186 lagging

Topic. Transformer series (leakage) impedance and the short-circuit test.

Key relations. $R_2'=a^2R_2$, $X_2'=a^2X_2$; $Z_{eq}=R_{eq}+jX_{eq}$; $I_2=aI_1$; $P_{cu}=I_1^2R_{eq}$; $\text{pf}=R_{eq}/|Z_{eq}|$.

Why this works. Impedances transfer across an ideal transformer by the turns-ratio squared, so both windings can be combined on one side into a single series impedance. With the output shorted, only that impedance limits current and all input power is dissipated as copper loss — exactly the basis of the short-circuit test used to measure $R_{eq}$ and $X_{eq}$.

Common pitfall. Referring impedances by $a$ instead of $a^2$, or reporting $I_1$ as the low-voltage current — the low-voltage (secondary) side of a step-down unit carries the larger current, $a$ times the primary.

Source. Chapman, Electric Machinery Fundamentals, 5th ed., §2.6–2.7 (per-unit and short-circuit test); Glover, §3.

Question 3 of 6: Transformer (Short-Circuit Analysis)

Question 3 — Transformer (Short-Circuit Analysis)