Equivalent circuit of 220/440 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.

Assumption: the OC test (220 V) is on the LV side and the SC test (24 V, about 5% of 440 V) is on the HV side, as is usual. Factor HV to LV: $\left(\frac{220}{440}\right)^2 = 0.25$.

OC test (LV): 220 V, 1.3 A, 160 W

$$\begin{aligned} \cos\phi_0 &= \frac{160}{220 \times 1.3} = 0.5594, \quad \sin\phi_0 = 0.8289 \\ I_w &= 1.3 \times 0.5594 = 0.727\ \text{A}, \quad I_\mu = 1.3 \times 0.8289 = 1.078\ \text{A} \\ R_0 &= \frac{220}{0.727} = 302.5\ \Omega, \quad X_0 = \frac{220}{1.078} = 204.2\ \Omega \end{aligned}$$

SC test (HV): 24 V, 14 A, 185 W

$$\begin{aligned} Z_{02} &= \frac{24}{14} = 1.714\ \Omega, \quad R_{02} = \frac{185}{14^2} = 0.944\ \Omega \\ X_{02} &= \sqrt{1.714^2 - 0.944^2} = 1.431\ \Omega \end{aligned}$$

Referred to LV:

$$\begin{aligned} R_{01} &= 0.944 \times 0.25 = 0.236\ \Omega \\ X_{01} &= 1.431 \times 0.25 = 0.358\ \Omega \\ Z_{01} &= 1.714 \times 0.25 = 0.429\ \Omega \end{aligned}$$

Equivalent circuit referred to LV side

 I1 ->        I0                 I2'
 o-----+--------+---[ R01 ]---[ jX01 ]---o
 +     |        |                        +
     [ R0 ]  [ jX0 ]                   load
 V1    |        |                       V2'
 -     |        |                        -
 o-----+--------+------------------------o
 All values referred to LV side:
 R0 = 302.5 ohm, X0 = 204.2 ohm
 R01 = 0.236 ohm, X01 = 0.358 ohm

Answer (LV side): $R_0 = 302.5\ \Omega$, $X_0 = 204.2\ \Omega$, $R_{01} = 0.236\ \Omega$, $X_{01} = 0.358\ \Omega$, $Z_{01} = 0.429\ \Omega$.