Equivalent impedance and copper loss of 50 kVA transformer

Transformation ratio $K = \frac{220}{4400} = 0.05$, so $K^2 = 0.0025$.

(i) Equivalent resistance, reactance and impedance

Referred to primary (4400 V side):

$$\begin{aligned} R_{01} &= R_1 + \frac{R_2}{K^2} = 3.45 + \frac{0.009}{0.0025} = 3.45 + 3.6 = 7.05\ \Omega \\ X_{01} &= X_1 + \frac{X_2}{K^2} = 5.2 + \frac{0.015}{0.0025} = 5.2 + 6.0 = 11.2\ \Omega \\ Z_{01} &= \sqrt{7.05^2 + 11.2^2} = 13.23\ \Omega \end{aligned}$$

Referred to secondary (220 V side):

$$\begin{aligned} R_{02} &= R_2 + K^2R_1 = 0.009 + 0.0025 \times 3.45 = 0.01763\ \Omega \\ X_{02} &= X_2 + K^2X_1 = 0.015 + 0.0025 \times 5.2 = 0.028\ \Omega \\ Z_{02} &= \sqrt{0.01763^2 + 0.028^2} = 0.03309\ \Omega \end{aligned}$$

Check: $Z_{02} = K^2 Z_{01} = 0.0025 \times 13.23 = 0.0331\ \Omega$.

(ii) Full-load copper loss

Full-load currents:

$$I_1 = \frac{50000}{4400} = 11.364\ \text{A}, \qquad I_2 = \frac{50000}{220} = 227.27\ \text{A}$$

Using individual winding resistances:

$$\begin{aligned} I_1^2R_1 &= 11.364^2 \times 3.45 = 445.5\ \text{W} \\ I_2^2R_2 &= 227.27^2 \times 0.009 = 464.9\ \text{W} \\ P_{cu} &= 445.5 + 464.9 = 910.4\ \text{W} \end{aligned}$$

Using $R_{01}$:

$$P_{cu} = I_1^2R_{01} = 11.364^2 \times 7.05 = 910.4\ \text{W}$$

Using $R_{02}$:

$$P_{cu} = I_2^2R_{02} = 227.27^2 \times 0.017625 = 910.4\ \text{W}$$

All three methods give the same value, which confirms that referring resistances through $K^2$ keeps the power loss unchanged.

Answer: $R_{01}=7.05\ \Omega$, $X_{01}=11.2\ \Omega$, $Z_{01}=13.23\ \Omega$; $R_{02}=0.01763\ \Omega$, $X_{02}=0.028\ \Omega$, $Z_{02}=0.0331\ \Omega$; full-load copper loss = 910.4 W by every method.