Equivalent circuit of 10 kVA 250/1000 V transformer

Method. The no-load (OC) test gives the shunt branch: $\cos\phi_0 = \frac{P_0}{V_0 I_0}$, $I_w = I_0\cos\phi_0$, $I_\mu = I_0\sin\phi_0$, $R_0 = \frac{V_0}{I_w}$, $X_0 = \frac{V_0}{I_\mu}$. The SC test gives the series branch: $Z = \frac{V_{sc}}{I_{sc}}$, $R = \frac{P_{sc}}{I_{sc}^2}$, $X = \sqrt{Z^2 - R^2}$. Values found on one side are moved to the other side by multiplying by the square of the voltage ratio.

Primary = LV side (250 V). Rated HV current $= 10000/1000 = 10$ A, so the SC test is at full load. Factor HV to LV: $\left(\frac{250}{1000}\right)^2 = \frac{1}{16} = 0.0625$.

No-load test (LV): 250 V, 0.8 A, 80 W

$$\begin{aligned} \cos\phi_0 &= \frac{80}{250 \times 0.8} = 0.4, \quad \sin\phi_0 = 0.9165 \\ I_w &= 0.8 \times 0.4 = 0.32\ \text{A}, \quad I_\mu = 0.8 \times 0.9165 = 0.733\ \text{A} \\ R_0 &= \frac{250}{0.32} = 781.25\ \Omega, \quad X_0 = \frac{250}{0.733} = 340.97\ \Omega \end{aligned}$$

SC test (HV): 80 V, 10 A, 120 W

$$\begin{aligned} Z_{02} &= \frac{80}{10} = 8\ \Omega, \quad R_{02} = \frac{120}{10^2} = 1.2\ \Omega \\ X_{02} &= \sqrt{8^2 - 1.2^2} = 7.909\ \Omega \end{aligned}$$

Referred to primary (× 0.0625):

$$\begin{aligned} R_{01} &= 1.2 \times 0.0625 = 0.075\ \Omega \\ X_{01} &= 7.909 \times 0.0625 = 0.494\ \Omega \\ Z_{01} &= 8 \times 0.0625 = 0.5\ \Omega \end{aligned}$$

Equivalent circuit referred to primary

 I1 ->        I0                 I2'
 o-----+--------+---[ R01 ]---[ jX01 ]---o
 +     |        |                        +
     [ R0 ]  [ jX0 ]                   load
 V1    |        |                       V2'
 -     |        |                        -
 o-----+--------+------------------------o
 All values referred to primary (LV):
 R0 = 781.25 ohm, X0 = 340.97 ohm
 R01 = 0.075 ohm, X01 = 0.494 ohm

Answer (primary side): $R_0 = 781.25\ \Omega$, $X_0 = 340.97\ \Omega$, $R_{01} = 0.075\ \Omega$, $X_{01} = 0.494\ \Omega$, $Z_{01} = 0.5\ \Omega$.

Useful extras: iron loss = 80 W, full-load copper loss = 120 W.