0195136047.pdf

4.2 BALANCED THREE-PHASE LOADS 203

A

B

Source Load

VAB VBA A B

IBA

IAB

IA

A A

IB

B B

IC

C C

Figure 4.1.5Notation using subscripts.

Balanced Wye-Connected Load

Let us consider a three-phase, four-wire 208-V supply system connected to a balanced wye-
connected load with an impedance of 10 20°, as shown in Figure 4.2.2(a). We shall solve for
the line currents and draw the corresponding phasor diagram.
Conventionally it is assumed that 208 V is the rms value of the line-to-line voltage of the
supply system, and the phase sequence is positive, orA–B–C, unless mentioned otherwise. The
magnitude of the line-to-neutral (or phase) voltages is given by 208/

√
3, or 120 V. Selecting the
line currents returning through the neutral conductor, as shown in Figure 4.2.2, we have

I ̄A=

V ̄AN

Z ̄

=

( 208 /

√

3 )90°

10  20°

= 12 70° (4.2.1)

I ̄B=

V ̄BN

Z ̄

=

( 208 /

√

3 ) −30°

10 20°

= 12  −50° (4.2.2)

I ̄C=

V ̄CN

Z ̄

=

( 208 /

√

3 ) −150°

10  20°

= 12  −170° (4.2.3)

Note thatV ̄BChas been chosen arbitrarily as the reference phasor, as in Figure 4.1.4(b). Assuming
the direction of the neutral current toward the load as positive, we obtain

I ̄N=−

(

I ̄A+I ̄B+I ̄C

)

=−( 12 70°+ 12  −50°+ 12  170°)= 0 (4.2.4)

That is to say that the system neutral and the star point of the wye-connected load are at the same
potential, even if they are not connected together electrically. It makes no difference whether they
are interconnected or not.
Thus, for abalancedwye-connected load, the neutral current is always zero. The line currents
and phase currents are equal in magnitude, and the line currents are in phase with the corresponding
phase currents. The line-to-line voltages, in magnitude, are

√
3 times the phase voltages, and the
phase voltages lag the corresponding line voltages by 30°.
The phasor diagram is drawn in Figure 4.2.2(b), from which it can be observed that the
balanced line (or phase) currents lag the corresponding line-to-neutral voltages by the impedance