Question 3 of 6: Single-Phase Transformer — Tests, Efficiency and Regulation
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: Closed-book written exam, 3 hours. Six questions of equal value (20 marks each); five constitute a complete paper. All six are solved here as a study resource. AC quantities are rms; three-phase voltages are line-to-line and power is total power unless noted.
Reference texts: S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (McGraw-Hill); B. S. Guru & H. R. Hiziroglu, Electric Machinery and Transformers, 3rd ed. (Oxford); P. C. Sen, Principles of Electric Machines and Power Electronics; T. Wildi, Electrical Machines, Drives, and Power Systems.
Given. Turns ratio $a=208/120=1.733$; shunt-branch $R_C=416\ \Omega$, $X_m=172\ \Omega$ (HV); series $R_{eq}=0.21\ \Omega$, $X_{eq}=0.98\ \Omega$ (LV); rating 1800 VA.
Find. (a) OC and SC wattmeter readings; (b) efficiency and %VR at the 1200 W, 0.67-lag load; (c) the same for a 12 Ω capacitive load.
Approximate equivalent circuit; the shunt (core-loss / magnetizing) branch and the series impedance are shown referred to their stated sides.
Approach. The OC test measures core loss (rated voltage, negligible copper loss); the SC test measures full-load copper loss (rated current, negligible core loss). For (b) and (c), add copper and core losses to the output for efficiency, and phasor-add the series drop for regulation.
Open-circuit test = core loss. At rated voltage the wattmeter reads the power in $R_C$:
$$P_{OC}=\frac{V_{HV}^2}{R_C}=\frac{208^2}{416}=\boxed{104\ \text{W}.}$$
(The same 104 W results if measured on the LV side, since $R_C$ refers by $a^2$.)
Short-circuit test = full-load copper loss. Rated LV current $I=1800/120=15$ A:
$$P_{SC}=I^2R_{eq}=15^2(0.21)=\boxed{47.25\ \text{W}.}$$
(b) Load current and losses.$I=\dfrac{P}{V\,\text{pf}}=\dfrac{1200}{120(0.67)}=14.93$ A. Copper loss $=I^2R_{eq}=46.8$ W; core loss $=104$ W (rated voltage).
$$\eta=\frac{P_{out}}{P_{out}+P_{cu}+P_{core}}=\frac{1200}{1200+46.8+104}=\boxed{88.8\%.}$$
(b) Voltage regulation. With $I=14.93\angle{-47.9^\circ}$ A and $Z=0.21+j0.98\ \Omega$,
$$E = V + IZ = 120 + 14.93\angle{-47.9^\circ}(1.002\angle 77.9^\circ)=133.2\ \text{V},$$
$$\text{VR}=\frac{|E|-V}{V}\times100=\frac{133.2-120}{120}\times100=\boxed{+11.0\%.}$$
(c) Capacitive 12 Ω load. With $Z_L=-j12\ \Omega$ across 120 V, $I=V/Z_L=10\angle{+90^\circ}$ A (leading). A pure capacitor absorbs no average power, so $P_{out}=0$ and hence $\eta=0$ (only losses are supplied). The leading current makes the series drop subtract:
$$E=120+10\angle 90^\circ(1.002\angle 77.9^\circ)=110.2\ \text{V},\qquad \text{VR}=\frac{110.2-120}{120}\times100=\boxed{-8.2\%.}$$
Check: Part (c) states a purely capacitive impedance, so the delivered real power — and therefore the efficiency — is zero; the instructive result is the negative voltage regulation produced by the leading load. Had the 12 Ω load carried a resistive component, efficiency would follow the same method as part (b).
