230/2300 V transformer with load impedance

Refer the load to the primary and use the approximate equivalent circuit, with the exciting branch ($R_0$, $X_0$) connected across the supply.

        I'2   R01=0.1   X01=0.4
  o----->-----[ R ]-----[ X ]----+
                                 |
  V1 = 230 V   Io <- (R0 || X0)  [ Z'L = 4 + j6 ]
                  (across supply)|
  o------------------------------+

(i) Primary current and input power factor

Turns ratio: $a = 2300/230 = 10$ (the transformer steps up).

Load referred to primary:

$$Z_L' = \frac{Z_L}{a^2} = \frac{400 + j600}{100} = 4 + j6\ \Omega$$

Load branch impedance seen by the supply:

$$Z = (R_{01} + jX_{01}) + Z_L' = (0.1 + j0.4) + (4 + j6) = 4.1 + j6.4 = 7.601\angle57.36^\circ\ \Omega$$

Load current (referred to primary), taking $V_1 = 230\angle 0^\circ$:

$$I_2' = \frac{230}{4.1 + j6.4} = 30.261\angle -57.36^\circ\ \text{A} = 16.323 - j25.480\ \text{A}$$

No-load (exciting) current:

$$\begin{aligned} I_c &= \frac{230}{500} = 0.46\ \text{A}, \qquad I_m = \frac{230}{200} = 1.15\ \text{A} \\ I_0 &= 0.46 - j1.15\ \text{A} \end{aligned}$$

Primary current:

$$I_1 = I_2' + I_0 = (16.323 - j25.480) + (0.46 - j1.15) = 16.783 - j26.630\ \text{A} = 31.478\angle -57.78^\circ\ \text{A}$$

Input power factor:

$$\cos\phi_1 = \cos(57.78^\circ) = 0.533\ \text{lagging}$$

(ii) Secondary terminal voltage

Voltage across the referred load:

$$V_2' = I_2' Z_L' = 30.261 \times |4 + j6| = 30.261 \times 7.211 = 218.21\ \text{V}$$ $$V_2 = a\,V_2' = 10 \times 218.21 = 2182.1\ \text{V}$$

Answer: $I_1 = 31.478\ \text{A}$ at power factor $0.533$ lagging; secondary terminal voltage $= 2182.1\ \text{V}$.