Parallel operation of Single Phase Transformers:

Reference: 1)B.L.Thereja (Article:32.35)

The Transformer is said to be in Parallel Operation when their primary windings are connected
to a common voltage supply, and the secondary windings are connected to a common load.
The connection diagram of the parallel operation of a transformer is shown in the figure below:

Why Parallel Operation of Transformers is required? /Principal reasons for parallel

 It is impractical and uneconomical to have a single large transformer for heavy and large
loads. So, it is necessary to connect a number of transformers in parallel.
 To maximize electrical power system availability parallel operation is necessary. If
numbers of transformers run in parallel, we can shut down any one of them for
maintenance purpose. Other parallel transformers in system will serve the load without
total interruption of power.
 If any one of the transformers run in parallel, is tripped due to fault of other parallel
transformers is the system will share the load, hence power supply may not be interrupted
if the shared loads do not make other transformers over loaded. It maximizes the power
system reliability.
 If the transformers are connected in parallel, so there will be scope in future, for
expansion of a substation to supply a load beyond the capacity of the transformer already
installed.

Necessary Conditions for satisfactory parallel operation:

In order that the transformers work satisfactorily in parallel, the following conditions should be
satisfied:

(i)Primary windings of the transformers should be suitable for the supply system voltage and
frequency.
(ii) Transformers should be properly connected with regard to their polarities.
(iii) The voltage ratings and voltage ratios of the transformers should be the same/identical.
(iv) The per unit or percentage impedances of the transformers should be equal in magnitude
(v) The reactance/resistance ratios (X/R ratio) of the transformers should be the same.

(vi)With transformers having different KVA ratings, the equivalent impedances should be
inversely proportional to the individual KVA rating.

What will happen if the above conditions are violated?

Condition (ii):
Condition (ii) is absolutely essential because wrong connections may result in dead short-circuit.
Fig.1(i) shows the correct method of connecting two single-phase transformers in parallel. It will
be seen that round the loop formed by the secondaries, the two secondary e.m.f.s E A and EB
oppose and there will be no circulating current.

Fig.1(ii) shows the wrong method of connecting two single-phase transformers is parallel. Here
the two secondaries are so connected that their e.m.f.s EA and EB are additive. This may lead to
short-circuit conditions and a very large circulating current will flow in the loop formed by the
two secondaries. Such a condition may damage the transformers unless they are protected by
fuses and circuit breakers.
Fig 1: Connecting transformers using same polarity

Condition (iii):
This condition is desirable for the satisfactory parallel operation of transformers. If this condition
is not met, the secondary e.m.f.s will not be equal and there will be circulating current in the loop
formed by the secondaries. This will result in the unsatisfactory parallel operation of
transformers. Let us illustrate this point-
Consider two single-phase transformer A and B operating in parallel as shown in Fig.2.

Fig: 2
Let EA and EB be their no-load secondary voltages and ZA and ZB be their impedances referred to
the secondary. Then at no-load, the circulating current in the loop formed by the secondaries is

Even a small difference in the induced secondary voltages can cause a large circulating current in
the secondary loop because impedances of the transformers are small. This secondary circulating
current will cause current to be drawn from the supply by the primary of each transformer. These
currents will cause copper losses in both primary and secondary. This creates heating with no
useful output. When load is connected to the system, this circulating current will tend to produce
unequal loading conditions i.e., the transformers will not share the load according to their kVA
ratings. It is because the circulating current will tend to make the terminal voltages of the same
value for both transformers. Therefore, transformer with smaller voltage ratio will tend to carry
more than its proper share of load. Thus, one transformer would tend to become overloaded than
the other and the system could not be loaded to the summation of transformer ratings without
overloading one transformer.

Condition (iv)
This condition is also desirable for proper parallel operation of transformers. If this condition is
not met, the transformers will not share the load according to their kVA ratings. Sometimes this
condition is not fulfilled by the design of the transformers. In that case, it can be corrected by
inserting proper amount of resistance or reactance or both in series with either primary or
secondary circuits of the transformers where the impedance is below the value required to fulfil
Condition (iv).

Condition (v)
If the reactance/resistance ratios (X/R ratios) of the two transformers are not equal, the power
factor of the load supplied by the transformers will not be equal. In other words, one transformer
will be operating with a higher and the other with a lower power factor than that of the load.
Condition (iv) is much more important than condition (v). Considerable deviation from condition
(v) will result in only a small reduction in the satisfactory degree of operation. When desired,
condition (v) also may be improved by inserting external impedance of proper value.

Condition (vi):
If not maintained, circulating current will be produced and the problem of overloading of
transformer will happen in the system. This is explained below-

If conditions i , ii , iii are met, EA = EB, So node A and node B have same voltage (figure a). We
can consider node A and B as a single node A (figure b).
ZA, ZB= impedances of transformer A and B respectively ( winding resistance and leakage
reactance from equivalent circuit)
IA, IB = current from transformer A and B respectively
I = total current
V2= voltage across load
Z= ZA || ZB
From figure b, IA ZA = IB ZB= IZ…….(a)
Using current divider rule,
IA = I ZB / (ZA+ ZB)
IB = I ZA/ (ZA+ ZB)
S=V2I = Total Apparent power supplied by both transformer to the load
Apparent power supplied by transformer A, SA=
V2 IA = V2 I ZB / (ZA+ ZB) = S*ZB / (ZA+ ZB) = S* (1/ZA)/ (1/ZA + 1/ZB) = SA…………….(b)
So, SA is inversely proportional to ZA
Apparent power supplied by transformer B, SB=
V2 IB = V2 I ZA / (ZA+ ZB) = S*ZA / (ZA+ ZB) = S* (1/ZB)/ (1/ZA + 1/ZB) = SB……………(c)
So, SB is inversely proportional to ZB
Suppose ZA> ZB , So if full load is applied SA<SB. If now rated apparent power of transformer A
> rated apparent power of transformer B, from equation (b) and (c) we can conclude that
transformer A will be under loaded and transformer B will be overloaded.
Maths for practice:

From B.L.Thereja:
Example : 32.96-32.103