50 kVA transformer: equivalent parameters, copper loss

The equivalent values are found by moving one winding's impedance to the other side using the square of the turns ratio.

Turns ratio: $a = \dfrac{V_1}{V_2} = \dfrac{4400}{220} = 20$

Rated currents:

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

(i) Equivalent resistance, reactance and impedance

Referred to primary:

$$\begin{aligned} R_{01} &= R_1 + a^2R_2 = 3.45 + 400(0.009) = 7.050\ \Omega \\ X_{01} &= X_1 + a^2X_2 = 5.2 + 400(0.015) = 11.200\ \Omega \\ Z_{01} &= \sqrt{R_{01}^2 + X_{01}^2} = \sqrt{7.050^2 + 11.200^2} = 13.234\ \Omega \end{aligned}$$

Referred to secondary:

$$\begin{aligned} R_{02} &= R_2 + \frac{R_1}{a^2} = 0.009 + \frac{3.45}{400} = 0.01762\ \Omega \\ X_{02} &= X_2 + \frac{X_1}{a^2} = 0.015 + \frac{5.2}{400} = 0.02800\ \Omega \\ Z_{02} &= \sqrt{R_{02}^2 + X_{02}^2} = 0.03309\ \Omega \end{aligned}$$

(ii) Total copper loss

Using individual resistances:

$$P_{cu} = I_1^2R_1 + I_2^2R_2 = (11.364)^2(3.45) + (227.27)^2(0.009) = 445.5 + 464.9 = 910.4\ \text{W}$$

Using equivalent resistance referred to primary:

$$P_{cu} = I_1^2R_{01} = (11.364)^2(7.050) = 910.4\ \text{W}$$

Using equivalent resistance referred to secondary:

$$P_{cu} = I_2^2R_{02} = (227.27)^2(0.01762) = 910.4\ \text{W}$$

All three methods give the same loss.

Answer: $R_{01} = 7.050\ \Omega$, $X_{01} = 11.200\ \Omega$, $Z_{01} = 13.234\ \Omega$; $R_{02} = 0.01762\ \Omega$, $X_{02} = 0.02800\ \Omega$, $Z_{02} = 0.03309\ \Omega$; total copper loss $\approx 910\ \text{W}$.