Question C.3: A 10 MVA, 13.8–4.16 kV transformer has a per-unit resistance......

Convert Z_{1} and Z_{2} on any common MVA base and apply Equations C.14 and C.15. The results are as follows:

I_{1}= \frac{IZ_{2}}{Z_{1} \ + \ Z_{2}}
I_{2}= \frac{IZ_{1}}{Z_{1} \ + \ Z_{2}}               (C.14)

S_{1}= \frac{SZ_{2}}{Z_{1} \ + \ Z_{2}}
S_{2}= \frac{SZ_{1}}{Z_{1} \ + \ Z_{2}}               (C.15)

10 MVA transformer: S_{1} = 9.255 <-37.93°
5 MVA transformer: S_{2} = 5.749 <-35.20°

The loads do not sum to 15 MVA because of different power factors. The MW and Mvar components of the transformer loads should sum to the total load components:
10 MVA transformer: 7.30 MW and 5.689 Mvar
5 MVA transformer: 4.698 MW and 3.314 Mvar
The total load is equal to an approximate load MW of 12 MW and 9.0 Mvar. The transformers do not share the load proportional to their ratings.
If the terminal voltages differ, there will be circulating current at no load. With reference to Figure C.5, the load sharing is given by the following equations:
I_{1} = \frac{ E_{1} \ – \ V}{Z_{1}} \ \ \ I_{2} = \frac{E_{2} \ – \ V}{Z_{2}}            (C.16)
where V is the load voltage. This is given by
V = (I_{1} + I_{2})Z_{L} = (\frac{E_{1} \ – \ V}{Z_{1}}+ \frac{E_{2} \ – \ V}{Z_{2}})Z_{L}              (C.17)
This can be written as
V (\frac{1}{Z_{1}}+ \frac{1}{Z_{2}} +  \frac{1}{Z_{L}}) = (\frac{E_{1}}{Z_{1}}+ \frac{E_{2}}{Z_{2}})              (C.18)
For a given load, the calculation is iterative in nature, as shown in Example C.4.