This document discusses parallel and autotransformer operations. It begins by explaining how circulating currents occur in parallel transformers when their voltage ratios are unequal and how this reduces their load capacity. It then covers methods for calculating circulating current values and load division between parallel transformers. The document next discusses how autotransformers operate using a single tapped coil to change voltages and provides examples for calculating step-down, step-up, and load values. It concludes by showing how an autotransformer can be modeled from a two-winding transformer connection.

1. Lesson 12: Parallel Transformers
and Autotransformers
ET 332b Ac Motors, Generators
and Power Systems
1
Lesson 12_et332b.pptx

2. Learning Objectives
After this presentation you will be able to:
 Explain what causes circulating currents in parallel and
compute its value.
 Compute the load division between parallel
transformers.
 Explain how autotransformers operate
 Make calculation using ideal autotransformer model
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Lesson 12_et332b.pptx

3. Parallel Operation of Transformers
Lesson 12_et332b.pptx
3
When voltage ratios are not equal,
currents circulate between the
windings of each transformer without a
load connected. Circulating currents
reduce the load capacity of transformer
EA≠EB
Currents circulate
between A and B based
on the voltage
difference and
transformer
impedance even with
no load
Where: EA = operating voltage of
transformer A
EB = operating voltage of
transformer B
ZA = series impedance of A
ZB = series impedance of B
B
A
B
A
c
Z
Z
E
E
I




4. Capacity Loss Due to Circulating
Currents
Lesson 12_et332b.pptx
4
Find effects using superposition Transformer A current
IA+Ic
c
A
TA I
I
I 

IB-Ic
c
B
TB I
I
I 

Transformer B current
Ic driven by EA – EB
Adding circulating
current to transformer
A increases total
current in winding. Not
seen in load current.
Can cause overload
ILoad

5. Circulating Current Example
Lesson 12_et332b.pptx
5
Example 12-1: Two 100 kVA single phase transformer operated in
parallel.
Nameplate data:
Transformer V-ratio %R %X
A 2300-460 1.36 3.50
B 2300-450 1.40 3.32
Find Ic magnitude and Ic as percent of transformer secondary ratings

6. Example 12-1 Solution (1)
Lesson 12_et332b.pptx
6
Use per unit method – Vbase = secondary voltage

7. Example 12-1 Solution (2)
Lesson 12_et332b.pptx
7
Use formula
Convert per unit to percent
29.55% of Transformer A’s
capacity is consumed by Ic.
Ans
Now convert this to amps using a base
current

8. Load Division Between Parallel Transformers
Lesson 12_et332b.pptx
8
When turns ratios are equal, the load current divides following the
winding impedance of the transformers. More current flows through
the lowest impedance.
VT
ZA
ZB
Zk
Zn
Load
Iin
IA
IB
Ik
In
All Transformer Z's and Load Z referred
to the same side of transformer or all per
unit (%) quantities
Circuit model
n
n
k
k
B
B
A
A
Z
1
Y
...
Z
1
Y
...
Z
1
Y
,
Z
1
Y 



Use current divider rule
n
k
B
A
p Y
...
Y
...
Y
Y
Y 













p
k
in
k
Y
Y
I
I Finds the current in
the kth transformer

9. Parallel Transformer Example
Lesson 12_et332b.pptx
9
Example 12-2: A 100 kVA transformer is to be paralleled with a
200 kVA transformer. Each transformer has rated voltages of 4160 -
240 V. Their percent impedances based on the ratings
of each are:
Z% = 1.64+3.16j % 100 kVA
Z% = 1.10+4.03j % 200 kVA
Find: a) rated high side current of each transformer
b) % of total bank current drawn by each transformer
c) maximum bank load that can be handled without
overloading either transformer

10. Example 12-2 Solution (1)
Lesson 12_et332b.pptx
10
a) Rated current of both transformers
Transformer A: 100 kVA
Transformer B: 200 kVA
b) Percent current drawn by each transformer
Convert %Z to actual ohms. Need base impedances

11. Example 12-2 Solution (2)
Lesson 12_et332b.pptx
11
Convert p.u. to ohms

12. Example 12-2 Solution (3)
Lesson 12_et332b.pptx
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Find the admittance
Total admittance.
Now use current divide rule to find flows through
each transformer.

13. Example 12-2 Solution (4)
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Find IA and IB in terms of Iin

14. Example 12-2 Solution (5)
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14
Transformer A carries 37.17% of the
total load
Transformer B carries 63.35% of the
total load
c) Find the maximum load of the parallel transformers without an overload
Let IA=IratedA=24.04 A and compute Iin using relationships above. Then
find flow through other transformer

15. Example 12-2 Solution (6)
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15
TX B not
overloaded
IratedB=48.08
Let IB=IratedB=48.08 A. Find Iin and
then compute the flow in
transformer A

16. Example 12-2 Solution (7)
Lesson 12_et332b.pptx
16
Max load, Iin=64.68 A
Find bank power

17. Autotransformers
Lesson 12_et332b.pptx
17
Autotransformers use a single taped coil to change voltage levels
and current levels – They provide no electrical isolation
NLS = number of turns
"embraced" by low side
NHS = number of turns on high
side
Polarity of induced voltages
determined by direction of
current and winding wraps.
If NLS = 20 and NHS = 80
LS
HS
LS
HS
V
V
N
N
a 
 4
20
80
a 
 V
30
4
120
a
V
V
so
V
120
V HS
LS
HS 



Step-down action

18. Autotransformers: Step-Down Operation
Lesson 12_et332b.pptx
18
Some load is transferred via conduction from one side to the other
and some is transferred by transformer action
Autotransformer connected in step-
down mode. Note direction of Itr
Like two winding transformers
LS
LS
HS
HS
LS
HS
I
V
I
V
S
S




Itr = the current from
transformer action
Low side current must
increase to maintain
power balance so:
Transformed
current
tr
HS
LS I
I
I 


19. Autotransformer Current Ratio and
Step-Up Operation
Lesson 12_et332b.pptx
19
Current ratio of autotransformer
LS
HS
LS
HS
N
N
a
Where
a
1
I
I


Autotransformer In Step-up Mode
Note: direction of Itr reversed to maintain power balance
Coils in these diagrams are
series aiding
(induced voltages add)

20. Autotransformers from Two-Winding
Transformers
Lesson 12_et332b.pptx
20
Autotransformer action can be obtained by proper connection of
two winding transformer coils
N2
N1
VHS
Load
For step-down mode
2
LS
2
1
HS
N
N
N
N
N



LS
HS
2
2
1
LS
HS
V
V
N
N
N
N
N
a 



Where: N1 = number of turns in primary
(HV)
N2 = number of turns in
secondary (LV)

21. Autotransformers from Two-Winding
Transformers
Lesson 12_et332b.pptx
21
Step-Down Connections
Find VLS with VHS=120 V, N1=500 and N2 =100
LS
HS
2
2
1
LS
HS
V
V
N
N
N
N
N
a 


 6
100
100
500
a 


V
120
V
Where
V
V
a HS
LS
HS


V
20
6
V
120
a
V
V HS
LS 



22. Autotransformer Example
Lesson 12_et332b.pptx
22
Example 12-3: 400 turn autotransformer operating at a
25% tap supplies a 4.8 kVA load at 0.85 lagging P.F. VHS =
2400 V
Find:
a) load current b) incoming line current c) Itr d) apparent
power transformed and conducted

23. Example 12-3 Solution (1)
Lesson 12_et332b.pptx
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Find turns ratio
Find secondary voltage
a) Find ILS
Ans

24. Example 12-3 Solution (2)
Lesson 12_et332b.pptx
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b) Find high-side current
Current must decrease to
maintain power balance
c) Find transformed current
Ans
d) Find transformed and conducted apparent powers
Ans
Ans

25. ET 332b Ac Motors, Generators and Power Systems
Lesson 12_et332b.pptx
25