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 E_A and E_B 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 R_A, X_A and Z_A be the total equivalent resistance, reactance and impedance of transformer A and R_B, X_B and Z_B be the total equivalent resistance, reactance and impedance of transformer B.

From Figure 1.48, we have

E_A=V₂+I_AZ_A     (1.71)

and          E_B=V₂+I_BZ_B     (1.72)

∴      I_AZ_A=I_BZ_B

∴    

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=S_A+S_B=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, Z_A be the equivalent impedance of transformer A (= R_A + jX_A) referred to as secondary, Z_B be the equivalent impedance of transformer B (= R_B + jX_B) referred to as secondary, I_A be the output current of transformer A and I_B 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 = IZ_L where Z_L is the impedance of the load

∴    V₂=(I_A+I_B)ZL    (1.80)

From Equations (1.78), (1.79) and (1.80), we have

E_A=I_AZ_A+(I_A+I_B)Z_L    (1.81)

and    E_A=I_BZ_B+(I_A+I_B)Z_L    (1.82)

∴ E_A − E_B = I_AZ_A − I_BZ_B

i.e.,    

Substituting I_A from Equation (1.83) in Equation (1.82), we have

i.e.,    

i.e.,    

Similarly,