Transformers

Using current division with transformer impedances: $Z_A = 2 + j3 = 3.605 \angle 56.31^\circ\ \Omega$ $Z_B = 2.3 + j5 = 5.462 \angle 66.25^\circ\ \Omega$ $Z_A + Z_B = 4.2 + j8 = 9.035 \angle 62.30^\circ\ \Omega$ $I_A = 600 \angle -36.87^\circ \times \frac{5.462 \angle 66.25^\circ}{9.035 \angle 62.30^\circ} = 362.72 \angle -32.92^\circ$ A $P_A = V_L \times I_A \times \cos(-32.92^\circ) = 304.48 \times V_L$ $I_B = 600 \angle -36.87^\circ \times \frac{3.605 \angle 56.31^\circ}{9.035 \angle 62.30^\circ} = 239.40 \angle -42.86^\circ$ A $P_B = V_L \times I_B \times \cos(-42.86^\circ) = 175.48 \times V_L$ $\text{Ratio} = \frac{P_A}{P_B} = \frac{304.48 \times V_L}{175.48 \times V_L} = 1.735 \approx 1.74$

Total load $S_{\text{load}} = \frac{180}{0.90} \angle -25.84^\circ = 200 \angle -25.84^\circ$ kVA. $Z_A = 0.01 + j0.06 = 0.0608 \angle 80.54^\circ$ $Z_B = 0.02 + j0.05 = 0.0538 \angle 68.20^\circ$ $Z_A + Z_B = 0.03 + j0.11 = 0.1140 \angle 74.74^\circ$ $S_A = 200 \angle -25.84^\circ \times \frac{0.0538 \angle 68.2^\circ}{0.1140 \angle 74.74^\circ} = 94.38 \angle -32.38^\circ$ kVA $P_A = 94.38 \times \cos(-32.38^\circ) = 79.7 \approx 80$ kW $S_B = 200 \angle -25.84^\circ \times \frac{0.0608 \angle 80.54^\circ}{0.1140 \angle 74.74^\circ} = 106.67 \angle -20.04^\circ$ kVA $P_B = 106.67 \times \cos(-20.04^\circ) = 100.2 \approx 100$ kW

Total load $S_{\text{load}} = \frac{280}{0.80} \angle -36.87^\circ = 350 \angle -36.87^\circ$ kVA. Since ratings and bases are equal: $Z_A = 0.02 + j0.08 = 0.08246 \angle 75.96^\circ$ $Z_B = 0.01 + j0.06 = 0.06080 \angle 80.53^\circ$ $Z_A + Z_B = 0.03 + j0.14 = 0.14318 \angle 77.90^\circ$ $S_A = 350 \angle -36.87^\circ \times \frac{0.0608 \angle 80.53^\circ}{0.14318 \angle 77.9^\circ} = 148.81 \angle -34.24^\circ$ kVA $P_A = 148.81 \times \cos(-34.24^\circ) = 123$ kW $S_B = 350 \angle -36.87^\circ \times \frac{0.08246 \angle 75.96^\circ}{0.14318 \angle 77.9^\circ} = 201.83 \angle -38.81^\circ$ kVA $P_B = 201.83 imes \cos(-38.81^\circ) = 157$ kW (Note: This question is printed with a minor typographical error as Problem 789 on page 549 of Romeo Rojas Jr.'s book, but solved under number 689 in the solutions section.)

Convert the percent impedances to a common base kVA of $125$ kVA: $Z_A = 0.04$ $Z_B' = 0.03 \times \frac{125}{75} = 0.05$ Now, calculate the shared load carried by each: $S_A = S_{\text{load}} \times \frac{Z_B'}{Z_A + Z_B'} = 140 \times \frac{0.05}{0.04 + 0.05} = 77.78$ kVA $S_B = S_{\text{load}} \times \frac{Z_A}{Z_A + Z_B'} = 140 \times \frac{0.04}{0.04 + 0.05} = 62.22$ kVA

Assuming both transformers have the same turns ratio and their impedances are referred to the secondary side: $S_A = S_{\text{load}} \times \frac{Z_B}{Z_A + Z_B} = 175 \times \frac{12.5}{9.6 + 12.5} \approx 99$ kVA $S_B = S_{\text{load}} \times \frac{Z_A}{Z_A + Z_B} = 175 \times \frac{9.6}{9.6 + 12.5} \approx 76$ kVA