Equivalent circuit of 20 kVA 220/2200 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 (220 V). Rated HV current $= 20000/2200 = 9.09$ A (SC test at 8 A). Factor HV to LV: $\left(\frac{220}{2200}\right)^2 = 0.01$.

No-load test (LV): 220 V, 1.4 A, 105 W

$$\begin{aligned} \cos\phi_0 &= \frac{105}{220 \times 1.4} = 0.3409, \quad \sin\phi_0 = 0.9401 \\ I_w &= 1.4 \times 0.3409 = 0.477\ \text{A}, \quad I_\mu = 1.4 \times 0.9401 = 1.316\ \text{A} \\ R_0 &= \frac{220}{0.477} = 460.95\ \Omega, \quad X_0 = \frac{220}{1.316} = 167.16\ \Omega \end{aligned}$$

SC test (HV): 120 V, 8 A, 320 W

$$\begin{aligned} Z_{02} &= \frac{120}{8} = 15\ \Omega, \quad R_{02} = \frac{320}{8^2} = 5\ \Omega \\ X_{02} &= \sqrt{15^2 - 5^2} = 14.14\ \Omega \end{aligned}$$

Referred to primary (× 0.01):

$$\begin{aligned} R_{01} &= 5 \times 0.01 = 0.05\ \Omega \\ X_{01} &= 14.14 \times 0.01 = 0.1414\ \Omega \\ Z_{01} &= 15 \times 0.01 = 0.15\ \Omega \end{aligned}$$

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 = 460.95 ohm, X0 = 167.16 ohm
 R01 = 0.05 ohm, X01 = 0.1414 ohm

Answer (primary side): $R_0 = 460.95\ \Omega$, $X_0 = 167.16\ \Omega$, $R_{01} = 0.05\ \Omega$, $X_{01} = 0.1414\ \Omega$, $Z_{01} = 0.15\ \Omega$.