0195136047.pdf

204 THREE-PHASE CIRCUITS AND RESIDENTIAL WIRING

ZB

Y–∆ transformation

For the balanced case,

A ZAB

C

B

ZA

ZC

ZAB =

ZAZB + ZBZC + ZCZA

ZC

ZBC =

ZAZB + ZBZC + ZCZA

ZA

ZCA =ZAZB^ +^ ZBZC^ +^ ZCZA

ZB

ZAB = ZBC = ZCA = Z∆ = 3 ZY

∆–Y transformation

For the balanced case,

1

3

ZA =

ZABZCA

ZAB + ZBC + ZCA

ZB =

ZABZBC

ZAB + ZBC + ZCA

ZC =

ZCAZBC

ZAB + ZBC + ZCA

ZA = ZB = ZC = ZY = Z∆

ZCA

ZBC

Figure 4.2.1Wye–delta and delta–wye transformations.

angle (20° in our example). The load power factor is given by cos 20° for our problem, and it is

said to belaggingin this case, as the impedance angle is positive and the phase current lags the

corresponding phase voltage by that angle.

The problem can also be solved in a simpler way by making use of a single-line equivalent

circuit, as shown in Figure 4.2.2.(c),

I ̄L=

V ̄L−N

Z ̄ =

( 208 /

√

3 )0°

10 20°

= 12  −20° (4.2.5)

in whichV ̄L−Nis chosen as the reference for convenience. The magnitude of the line current

and the power factor angle are known; the negative sign associated with the angle indicates that

the power factor is lagging. By knowing that the line (or phase) currentsI ̄A, I ̄B, I ̄Clag their

respective voltagesV ̄AN,V ̄BN, andV ̄CNby 20°, the phase angles of various voltages and currents,

if desired, can be obtained with respect to any chosen reference, such asV ̄BC.

Balanced Delta-Connected Load

Next, let us consider the case of a balanced delta-connected load with impedance of 5 45°

supplied by a three-phase, three-wire 100-V system, as shown in Figure 4.2.3(a). We shall

determine the line currents and draw the corresponding phasor diagram.

With the assumed positive phase sequence (A–B–C) and withV ̄BCas the reference phasor,

the line-to-line voltagesV ̄AB, V ̄BC, andV ̄CAare shown in Figure 4.2.3. The rms value of the

line-to-line voltages is 100 V for our example. Choosing the positive directions of the line and

phase currents as in Figure 4.2.3(a), we have

I ̄AB=

V ̄AB

Z ̄ =

100  120°

5 45°

= 20  75° (4.2.6)

I ̄BC=

V ̄BC

Z ̄

=

100  0°

5  45°

= 20  −45° (4.2.7)

I ̄CA=

V ̄CA

Z ̄ =

100  240°

5 45°

= 20  195° (4.2.8)

By the application of Kirchhoff’s current law at each of the vertices of the delta-connected

load, we obtain

I ̄A=I ̄AB+I ̄AC= 20  75°− 20 195°= 34. 64  45° (4.2.9)

I ̄B=I ̄BA+I ̄BC=− 20  75°+ 20  −45°= 34. 64  −75° (4.2.10)