Load sharing of 50 kVA and 25 kVA transformers in parallel

For two transformers with the same voltage ratio in parallel, the load is shared in inverse ratio of their equivalent impedances:

$$S_A = S\,\frac{Z_B}{Z_A + Z_B}, \qquad S_B = S\,\frac{Z_A}{Z_A + Z_B}$$

Assumption: both impedances are referred to the 220 V (secondary) side; the load power factor is not given, but it is not needed because both impedances have the same angle.

Load sharing

$$\begin{aligned} Z_A &= 0.01 + j0.02 = 0.02236\angle 63.43^\circ\ \Omega \\ Z_B &= 0.02 + j0.04 = 0.04472\angle 63.43^\circ\ \Omega \\ Z_A + Z_B &= 0.03 + j0.06 = 0.06708\angle 63.43^\circ\ \Omega \end{aligned}$$

Since the angles are equal, the ratios are real numbers:

$$\begin{aligned} \frac{Z_B}{Z_A+Z_B} &= \frac{0.04472}{0.06708} = \frac{2}{3}, \qquad \frac{Z_A}{Z_A+Z_B} = \frac{1}{3} \\ S_A &= 60 \times \tfrac{2}{3} = 40\ \text{kVA}, \qquad S_B = 60 \times \tfrac{1}{3} = 20\ \text{kVA} \end{aligned}$$

Both share the load at the same power factor as the load.

(i) Individual transformer currents

Total load current on the 220 V side:

$$I = \frac{60000}{220} = 272.73\ \text{A}$$ $$\begin{aligned} I_A &= \frac{40000}{220} = 181.82\ \text{A} \quad (\text{primary: } 40000/4400 = 9.09\ \text{A}) \\ I_B &= \frac{20000}{220} = 90.91\ \text{A} \quad (\text{primary: } 20000/4400 = 4.55\ \text{A}) \end{aligned}$$

Check: $181.82 + 90.91 = 272.73$ A.

(ii) Percentage of rated capacity used

$$\begin{aligned} \text{Transformer A} &: \frac{40}{50} \times 100 = 80\% \\ \text{Transformer B} &: \frac{20}{25} \times 100 = 80\% \end{aligned}$$

Both are loaded to the same fraction of their rating because the 25 kVA unit has exactly twice the ohmic impedance of the 50 kVA unit, i.e. both have the same per-unit impedance:

Answer: $I_A = 181.82$ A, $I_B = 90.91$ A (LV side); each transformer runs at 80% of its rated capacity.