Load Sharing By Two Transformers
Let us consider the following two cases:
- Equal voltage ratios.
- Unequal voltage ratios.
1.39.1 Equal Voltage Ratios
Assume no-load voltages EA and EB are identical and in phase. Under these conditions if the primary and secondary are connected in parallel, there will be no circulating current between them on no load.
Figure 1.48 Equal Voltage Ratios
Figure 1.48 shows two impedances in parallel. Let RA, XA and ZA be the total equivalent resistance, reactance and impedance of transformer A and RB, XB and ZB be the total equivalent resistance, reactance and impedance of transformer B.
From Figure 1.48, we have
EA=V2+IAZA (1.71)
and EB=V2+IBZB (1.72)
∴ IAZA=IBZB
∴ 
Equation (1.73) suggests that if two transformers with different kVA ratings are connected in parallel, the total load will be divided in proportion to their kVA ratings if their equivalent impedances are inversely proportional to their respective ratings.
Since 
i.e., 
i.e., 
Similarly, 
Similarly, load shared by transformer A,
Similarly, 
Total S=SA+SB=V2I×10-3 kVA
∴ 
2 Unequal Voltage Ratios
For unequal voltage turns ratio, if the primary is connected to the supply, a circulating current will flow in the primary even at no load. The circulating current will be superimposed on the currents drawn by the load when the transformers share a load.
Let V1 be the primary supply voltage, a1 be the turns ratio of transformer A, a2 be the turns ratio of transformer B, ZA be the equivalent impedance of transformer A (= RA + jXA) referred to as secondary, ZB be the equivalent impedance of transformer B (= RB + jXB) referred to as secondary, IA be the output current of transformer A and IB be the output current of transformer B.
The induced emf in the secondary of transformer A is
The induced emf in the secondary of transformer B is
Again, V2 = IZL where ZL is the impedance of the load
∴ V2=(IA+IB)ZL (1.80)
From Equations (1.78), (1.79) and (1.80), we have
EA=IAZA+(IA+IB)ZL (1.81)
and EA=IBZB+(IA+IB)ZL (1.82)
∴ EA − EB = IAZA − IBZB
i.e., 
Substituting IA from Equation (1.83) in Equation (1.82), we have
i.e., 
i.e., 
Similarly, 