Introduction to Electric Circuits

5.5 Power in balanced three-phase circuits 115

Also there is a phase difference of 30 ~ between the line and phase currents.

Example 5.3

Three impedances, each of impedance Z = (5 - j12) 1) are connected in delta
across a three-phase supply whose line voltage is 110 V. Determine (1) the
voltage across each impedance, (2) the line current drawn from the supply, (3)
the power factor.

Solution

Figure 5.14

I

VL = ,(

IL

.r 5~//'~ IpH

12f2 ..._ 12f2

~~~ 5~

The arrangement is shown in Fig. 5.14.

1 The voltage across each impedance is the phase voltage which, for a delta

connection, is the same as the line voltage: VpH = VL = 110 V.

2 The phase current is

IpH-- VpH/ZPH-- 110/~v/( 52+ 122)-- 110/13 = 8.46 A

3 The power factor is cos ch - R/Z = 5/13 = 0.385 leading.

5.5 POWER IN BALANCED THREE-PHASE CIRCUITS

The total power (P) in any three-phase system whose phases are A, B and C is

given by the sum of the powers in each of the three phases. If these are

respectively PA, PB, and Pc then

P = PA + PB + Pc (5.5)

For a balanced system the power in each phase is the same, so that the total

power is simply three times the power in one phase.

We saw in Chapter 4 that the power in a single-phase circuit is given by

P = VI cos 4~ where V and I are the rms voltage and current, respectively, and

cos 4~ is the power factor. In a balanced three-phase circuit therefore the total

power is given by:

P = 3 VpHIpH COS ~) (5.6)