Circulating Current – load switch open
I circulating
E A  EB

Z A  ZB
Circulating Current load switch closed
Icirculating adds to the load current in one
transformer and subtracts from the load current
in the other transformer. At rated load, the
transformer with the higher secondary voltage
will be overloaded!
Example 3.4
• Two 100 kVA single-phase transformers A
and B are to be operated in parallel. The
respective no-load voltage ratios and
respective impedances are obtained from
the nameplates as
– Transformer A
• 2300 – 460 V
%R = 1.36
%X = 3.50
%R = 1.40
%X = 3.32
– Transformer B
• 2300 – 450
Example 3.4 continued
• Determine
– the circulating current in the paralleled
secondaries
– the circulating current as a percent of the
rated current in transformer A
– the percent difference in secondary voltage
that caused the circulating current.
The rated low-side currents are
100kVA
IA 
 217.39 A
460V
100kVA
IB 
 222.22 A
450V
The equivalent resistance and reactance of each transformer
referred to the low side
RPU 
I rated  Req
Vrated
0.0136 
0.0140 
X PU 
217.39  RAeq
460
222.22  RBeq
450
I rated  X eq
0.0350 
0.0332 
 RAeq  0.0288
 RBeq  0.0284
Vrated
217.39  X Aeq
460
222.22  X Beq
450
 X Aeq  0.0741
 X Beq  0.0672
The impedance of the closed loop formed by the two secondaries
Z loop  Z A  Z B
Z loop  0.0288  j 0.0741  0.0284  j 0.0672
Z loop  0.0572  j 0.1413
Z loop  0.152467.97
The circulating current
I circulating 
I circulating
I circulating
E A  EB
Z A  ZB
4600  4500

0.15467.97
 65.6  68 A
The circulating current as a percent of the rated current in transformer A
I circulating
I Arated
65.62

 0.302  30.2%
217.39
The percent difference in secondary voltage that caused the circulating current
VA  VB 460  450

 0.022  2.2%
VB
450