Load sharing of two transformers in parallel

Two transformers with equal turns ratio have the same secondary voltage, and they share the load current in inverse proportion to their impedances.

Given: $Z_A = 0.5 + j3\ \Omega$, $Z_B = 0.6 + j10\ \Omega$ (referred to secondary), total load $P = 100$ kW at 0.8 pf lagging.

Total load

$$S = \frac{100}{0.8} = 125\ \text{kVA},\qquad \phi = \cos^{-1}0.8 = 36.87^\circ\ \text{lagging}$$

Current division. Let the total current be $I = I\angle-36.87^\circ$ with $V_2$ as reference. For parallel branches:

$$I_A = I\frac{Z_B}{Z_A + Z_B},\qquad I_B = I\frac{Z_A}{Z_A + Z_B}$$ $$\begin{aligned} Z_A + Z_B &= 1.1 + j13\ \Omega \\ \frac{Z_B}{Z_A + Z_B} &= 0.7676 + j0.0188 = 0.7679\angle1.40^\circ \\ \frac{Z_A}{Z_A + Z_B} &= 0.2324 - j0.0188 = 0.2331\angle-4.63^\circ \end{aligned}$$

Load shared (kVA = fraction of 125 kVA, since $V_2$ is common):

$$\begin{aligned} S_A &= 0.7679\times125 = 95.98\ \text{kVA at angle } 36.87^\circ - 1.40^\circ = 35.47^\circ \\ P_A &= 95.98\cos35.47^\circ = 78.17\ \text{kW},\quad \text{pf}_A = 0.814\ \text{lag} \\ S_B &= 0.2331\times125 = 29.14\ \text{kVA at angle } 36.87^\circ + 4.63^\circ = 41.50^\circ \\ P_B &= 29.14\cos41.50^\circ = 21.83\ \text{kW},\quad \text{pf}_B = 0.749\ \text{lag} \end{aligned}$$

Check: $P_A + P_B = 78.17 + 21.83 = 100$ kW.

Answer: A supplies about 78.2 kW (96.0 kVA) and B about 21.8 kW (29.1 kVA). The transformer of lower impedance (A) carries the greater share.