The Polarity Of Connection Of Single Phase Transformers And Its Two Cases

The polarity of connection in the case of single phase transformers can be either same or opposite. Inside the loop formed by the two secondaries the resulting voltage must be zero. If wrong polarity is chosen the two voltages get added and short circuit results. In the case of polyphase banks it is possible to have permanent phase error between the phases with substantial circulating current. Such transformer banks must not be connected in parallel. The turns ratios in such groups can be adjusted to give very close voltage ratios but phase errors cannot be compensated. Phase error of 0.6 degree gives rise to one percent difference in voltage. Hence poly phase transformers belonging to the same vector group alone must be taken for paralleling. Transformers having _30 angle can be paralleled to that having +30 angle by reversing the phase sequence of both primary and secondary terminals of one of the transformers. This way one can overcome the problem of the phase angle error. Polarity of connection The polarity of connection in the case of single phase transformers can be either same or opposite. Inside the loop formed by the two secondaries  the resulting voltage must be zero. If wrong polarity is chosen the two voltages get added and short circuit results. In the case of polyphase banks it is possible to have permanent phase error between the phases with substantial circulating current. Such transformer banks must not be connected in parallel. The turns ratios in such groups can be adjusted to give very close voltage ratios but phase errors cannot be compensated. Phase error of 0.6 degree gives rise to one percent difference in voltage. Hence poly phase transformers belonging to the same vector group alone must be taken for paralleling. Transformers having _30angle can be paralleled to that having +30 angle by reversing the phase sequence of both primary and secondary terminals of one of the transformers. This way one can overcome the problem of the phase angle error.


are not the same. These are discussed now in sequence Phase sequence

 The phase sequence of operation becomes relevant only in the case of poly phase systems. The poly phase banks belonging to same vector group can be connected in parallel. A transformer with +30 phase angle however can be paralleled with the one with _30 phase angle, the phase sequence is reversed for one of them both at primary and secondary terminals. If the phase sequences are not the same then the two transformers cannot be connected in parallel even if they belong to same vector group. The phase sequence can be found out by the use of a phase sequence indicator. Performance of two or more single phase transformers working in parallel can be computed using their equivalent circuit. In the case of poly phase banks also the approach is identical and the single phase equivalent circuit of the same can be used. Basically two cases arise in these problems. Case A: when the voltage ratio of the two transformers is the same and Case B: when the voltage ratios.

CASE A: Equal voltage ratios

Always two transformers of equal voltage ratios are selected for working in parallel. This way one can avoid a circulating current between the transformers. Load can be switched on subsequently to these bus bars. Neglecting the parallel branch of the equivalent circuit the above connection can be shown as in Fig. 38(a),(b). The equivalent circuit is drawn in terms of the secondary parameters. This may be further simplified as shown under Fig. 38(c). The voltage drop across the two transformers must be the same by virtue of common connection
at input as well as output ends. By inspection the voltage equation for the drop can be

 From the above it is seen that the transformer with higher impedance supplies lesser load current and vice versa. If transformers of dissimilar ratings are paralleled the transformer with larger rating shall have smaller impedance as it has to produce the same drop as the other transformer, at a larger current. Thus the ohmic values of the impedances must be in the inverse ratio of the ratings of the transformers.

(IAZA = IBZB ) and IA//IB = ZB /ZA
.

Expressing the voltage drops in p.u basis, we aim at the same per unit drops at any load for the transformers. The per unit impedances must therefore be the same on their respective bases. Fig. 39 shows the phasor diagram of operation for these conditions. The drops are magnified and shown to improve clarity. It is seen that the total voltage drop inside the

transformers is v but the currents IA and IB are forced to have a different phase angle due to the difference in the internal power factor angles IA and IB. This forces the active and reactive components of the currents drawn by each transformer to be different ( even in the case when current in each transformer is the same). If we want them to share the load current in proportion to their ratings, their percentage ( or p.u) impedances must be the same. In order to avoid any divergence and to share active and reactive powers also properly, φA = φB. Thus the condition for satisfactory parallel operation is that the p.u resistances and p.u reactance must be the same on their respective bases for the two transformers. To determine the sharing of currents and power either p.u parameters or ohmic values can be used. 

CASE-B UNEQUAL VOLTAGE RATIO

One may not be able to get two transformers of identical voltage ratio in spite of ones best efforts. Due to manufacturing differences, even in transformers built as per the same design, the voltage ratios may not be the same. In such cases the circuit representation for parallel operation will be different as shown in Fig. 40. In this case the two input voltages cannot be merged to one, as they are different. The load brings about a common connection at the output side. EA and EB are the no-load secondary emf. ZL is the load impedance at the secondary terminals. By nspection the voltage equation can be written as below:

EA = IAZA + (IA + IB)ZL = V + IAZA

EB = IBZB + (IA + IB)ZL = V + IBZB                                         (95)

Solving the two equations the expression for IA and IB can be obtained as