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EQUIVALENT Y- AND -CONNECTIONS
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Figure 11-12 shows three impedances connected in a (delta) con guration, and three impedances connected in a Y (wye) con guration. Let the terminals of the two connections be identi ed in pairs as indicated by the labels , ,
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. Then Z1 is the impedance adjoining terminal in the Y-connection, and ZC is the impedance opposite terminal in the -connection, and so on. Looking into any two terminals, the two connections will be equivalent if corresponding input, output, and transfer impedances are equal. The criteria for equivalence are as follows:
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Fig. 11-12
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Z1 Z2 Z1 Z3 Z2 Z3 Z3 Z1 Z2 Z1 Z3 Z2 Z3 ZB Z2 Z1 Z2 Z1 Z3 Z2 Z3 ZC Z1 ZA
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ZA ZB ZA ZB ZC ZA ZC Z2 ZA ZB ZC ZB ZC Z3 ZA ZB ZC Z1
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It should be noted that if the three impedances of one connection are equal, so are those of the equivalent connection, with Z =ZY 3.
SINGLE-LINE EQUIVALENT CIRCUIT FOR BALANCED THREE-PHASE LOADS
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 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 cos . The method may be applied to a balanced -connected load if the load is replaced by its Yequivalent, where ZY 1 Z (Section 11.8). 3
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 VLN 98:0 08 4:90 308 A Z 20 308 Hence,
From Fig. 11-7(b), the phase angles of VAN , VBN , and VCN are 908, 308, and 1508. IA 4:90 608 A IB 4:90 608 A
IC 4:90 1808 A
UNBALANCED DELTA-CONNECTED LOAD
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 10 08  ZBC 10 308  ZCA 15 308 
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[CHAP. 11
Fig. 11-14 Obtain phase and line currents and draw the phasor diagram. IAB VAB 339:4 1208 33:94 1208 A ZAB 10 08
Similarly, IBC 33:94 308 A and ICA 22:63 2708 A. Then, IA IAB IAC 33:94 1208 22:63 2708 54:72 108:18 A Also, IB 65:56 458 A and IC 29:93 169:18 A. The voltage-current phasor diagram is shown in Fig. 11-15(b), with magnitudes and angles to scale.
Fig. 11-15
11.11 UNBALANCED WYE-CONNECTED LOAD Four-Wire The neutral conductor carries the unbalanced current of a wye-connected load and maintains the line-to-neutral voltage magnitude across each phase of the load. The line currents are unequal and the currents on the phasor diagram have no symmetry.
EXAMPLE 11.6 A three-phase, four-wire, 150-V, CBA system has a Y-connected load, with ZA 6 08  ZB 6 308  ZC 5 458 
Obtain all line currents and draw the phasor diagram. IA
See Figure 11-16(a).
VAN 86:6 908 14:43 908 A ZA 6 08 V 86:6 308 14:43 08 A IB BN ZB 6 308 V 86:6 1508 17:32 1058 A IC CN ZC 5 458 IN 14:43 908 14:43 08 17:32 1058 10:21 167:08 A Figure 11-16(b) gives the phasor diagram.
CHAP. 11]
POLYPHASE CIRCUITS
Fig. 11-16
Three-Wire Without the neutral conductor, the Y-connected impedances will have voltages which vary considerably from the line-to-neutral magnitude.
EXAMPLE 11.7 Figure 11-17(a) shows the same system as treated in Example 11.6 except that the neutral wire is no longer present. Obtain the line currents and nd the displacement neutral voltage, VON .
Fig. 11-17 The circuit is redrawn in Fig. 11-17(b) so as to suggest a single node-voltage equation with VOB as the unknown. VOB VAB VOB VOB VBC 0 ZA ZB ZC   1 1 1 150 2408 150 08 VOB 6 08 6 308 5 458 6 08 5 458 from which VOB 66:76 152:858 V. Then, IB VOB 11:13 2:858 A ZB
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