1000 VA 230/115 V transformer equivalent circuit

Given: 1000 VA, 230/115 V. HV = 230 V, LV = 115 V, so $K = 2$ and $K^2 = 4$.

The OC test was done on the LV side at rated voltage, so the shunt branch is already on the LV side. The SC test was done on the HV side, so the series branch must be divided by $K^2 = 4$ to refer it to the LV side.

(i) Equivalent circuit referred to the LV side

Open-circuit test (LV side): $V_{oc} = 115$ V, $I_{oc} = 0.11$ A, $P_{oc} = 3.9$ W.

$$\begin{aligned} \cos\phi_0 &= \frac{3.9}{115\times0.11} = 0.3083 \\ I_w &= 0.11\times0.3083 = 0.03391\ \text{A},\quad I_\mu = \sqrt{0.11^2 - 0.03391^2} = 0.1046\ \text{A} \\ R_c &= \frac{115}{0.03391} = 3391\ \Omega,\qquad X_m = \frac{115}{0.1046} = 1099\ \Omega \end{aligned}$$

Short-circuit test (HV side): $V_{sc} = 17.1$ V, $I_{sc} = 8.7$ A, $P_{sc} = 38.1$ W.

$$\begin{aligned} Z_{eq,H} &= \frac{17.1}{8.7} = 1.966\ \Omega,\qquad R_{eq,H} = \frac{38.1}{8.7^2} = 0.5034\ \Omega \\ X_{eq,H} &= \sqrt{1.966^2 - 0.5034^2} = 1.900\ \Omega \end{aligned}$$

Referred to the LV side:

$$R_{eq,L} = \frac{0.5034}{4} = 0.1258\ \Omega,\qquad X_{eq,L} = \frac{1.900}{4} = 0.4750\ \Omega$$

   Req=0.1258 ohm   Xeq=0.475 ohm
 o--+-----/\/\/\-----UUUU------o
    |
   [Rc=3391] [Xm=1099]    Load (LV side, 115 V)
    |
 o--+---------------------------o

(ii) Efficiency at rated load, 0.8 pf lagging

Rated LV current: $I_2 = \dfrac{1000}{115} = 8.696$ A.

$$\begin{aligned} P_{cu} &= I_2^2R_{eq,L} = 8.696^2\times0.1258 = 9.515\ \text{W} \\ P_{core} &= 3.9\ \text{W (OC test at rated voltage)} \\ P_{out} &= 1000\times0.8 = 800\ \text{W} \\ \eta &= \frac{800}{800 + 9.515 + 3.9}\times100 = 98.35\% \end{aligned}$$

(The same copper loss follows from the HV side: rated HV current $4.348$ A, $4.348^2\times0.5034 = 9.515$ W.)

Answer: $R_c = 3391\ \Omega$, $X_m = 1099\ \Omega$, $R_{eq} = 0.126\ \Omega$, $X_{eq} = 0.475\ \Omega$ (LV side); $\eta = 98.35\%$