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CHAP. 11]



Y-to- Transformation

-to-Y Transformation

Z1 Z2 þ Z1 Z3 þ Z2 Z3 Z3 Z1 Z2 þ Z1 Z3 þ Z2 Z3 ZB ¼ Z2 Z1 Z2 þ Z1 Z3 þ Z2 Z3 ZC ¼ Z1 ZA ¼

ZA ZB ZA þ ZB þ ZC ZA ZC Z2 ¼ ZA þ ZB þ ZC ZB ZC Z3 ¼ ZA þ ZB þ ZC Z1 ¼

It should be noted that if the three impedances of one connection are equal, so are those of the equivalent connection, with Z =ZY ¼ 3.



Figure 11-13(a) shows a balanced Y-connected load. In many cases, for instance, in power calculations, only the common magnitude, IL , of the three line currents is needed. This may be obtained from the single-line equivalent, Fig. 11-13(b), which represents one phase of the original system, with the line-to-neutral voltage arbitrarily given a zero phase angle. This makes IL ¼ IL , where  is the impedance angle. If the actual line currents IA , IB , and IC are desired, their phase angles may be found by adding  to the phase angles of VAN , VBN , and VCN as given in Fig. 11-7. Observe that the angle on IL gives the power factor for each phase, pf ¼ cos . The method may be applied to a balanced -connected load if the load is replaced by its Yequivalent, where ZY ¼ 13 Z (Section 11.8).

Fig. 11-13 EXAMPLE 11.4 Rework Example 11.3 by the single-line equivalent method. Referring to Fig. 11-14 (in which the symbol Y indicates the type of connection of the original load), IL ¼

VLN 98:0 08 ¼ ¼ 4:90 308 A Z 20 308

From Fig. 11-7(b), the phase angles of VAN , VBN , and VCN are 908, 308, and 1508. IA ¼ 4:90 608 A


IB ¼ 4:90 608 A


IC ¼ 4:90 1808 A


The solution of the unbalanced delta-connected load consists in computing the phase currents and then applying KCL to obtain the line currents. The currents will be unequal and will not have the symmetry of the balanced case. EXAMPLE 11.5

A three-phase, 339.4-V, ABC system [Fig. 11-15(a)] has a -connected load, with ZAB ¼ 10 08 

ZBC ¼ 10 308 

ZCA ¼ 15 308