Equivalent circuit of 20 kVA 250/2500 V transformer from tests

Shunt branch parameters come from the OC test (done on LV side, so they are directly LV values); series parameters come from the SC test (done on HV side, so they are HV values). Turns ratio $a = 2500/250 = 10$.

Rated currents: LV = 20000/250 = 80 A, HV = 20000/2500 = 8 A, so the SC test was at full load.

From the OC test (LV side)

$$\begin{aligned} \cos\phi_0 &= \frac{P_0}{V_0 I_0} = \frac{105}{250\times1.4} = 0.30 \\ I_c &= I_0\cos\phi_0 = 1.4\times0.30 = 0.42\ \text{A} \\ I_m &= I_0\sin\phi_0 = 1.4\times0.954 = 1.336\ \text{A} \\ R_0 &= \frac{V_0}{I_c} = \frac{250}{0.42} = 595.2\ \Omega \\ X_0 &= \frac{V_0}{I_m} = \frac{250}{1.336} = 187.2\ \Omega \end{aligned}$$

From the SC test (HV side)

$$\begin{aligned} R_{eq,HV} &= \frac{P_{sc}}{I_{sc}^2} = \frac{320}{8^2} = 5\ \Omega \\ Z_{eq,HV} &= \frac{V_{sc}}{I_{sc}} = \frac{120}{8} = 15\ \Omega \\ X_{eq,HV} &= \sqrt{15^2 - 5^2} = 14.14\ \Omega \end{aligned}$$

Referred to LV side (divide HV impedances by $a^2 = 100$)

Approximate equivalent circuit referred to HV side

        Req = 5 ohm    Xeq = 14.14 ohm
  o---+----/\/\/----mmmmmm-----o
      |                         +
  +---+---+
  |       |                    V2'
 [R0]    [X0]
 59.52k  18.72k
  |       |                     -
  +---+---+
      |
  o---+-------------------------o
  V1 = 2500 V

The shunt branch ($R_0 \parallel X_0$) is placed across the supply terminals, and the total series impedance $5 + j14.14\ \Omega$ is in the line.

Answer: referred to LV: $R_0 = 595.2\ \Omega$, $X_0 = 187.2\ \Omega$, $R_{eq} = 0.05\ \Omega$, $X_{eq} = 0.1414\ \Omega$; referred to HV: $R_0 = 59.52$ kΩ, $X_0 = 18.72$ kΩ, $R_{eq} = 5\ \Omega$, $X_{eq} = 14.14\ \Omega$.