0195136047.pdf

3.1 SINUSOIDAL STEADY-STATE PHASOR ANALYSIS 117

φ

ωC

1

VR = R I

VR + VL + VC = V VL + VC = j(ωL − ) I

I = I ∠ 0 ° (Reference)

(b)

Figure E3.1.6Continued

I V

−

+

(c)

R

GR =^1

IR IL IC

ωL

jB^1

L^ =^ j(− ) jBC^ =^ jωC

φ

ωL

1

IR = V/R

IR + IL + IC = I IC + IL = j(ωC − ) V

V = V ∠ 0 ° (Reference)

(d)

EXAMPLE 3.1.7

Two single-phase 60-Hz sinusoidal-source generators (with negligible internal impedances) are
supplying to a common load of 10 kW at 0.8 power factor lagging. The impedance of the feeder
connecting the generatorG 1 to the load is 1. 4 +j 1. 6 , whereas that of the feeder connecting
the generatorG 2 to the load is 0. 8 +j 1. 0 . If the generatorG 1 , operating at a terminal voltage
of 462 V (rms), supplies 5 kW at 0.8 power factor lagging, determine:

(a) The voltage at the load terminals;

(b) The terminal voltage of generatorG 2 ; and

(c) The real power and the reactive power output of the generatorG 2.

Solution

(a) From Equation (3.1.37) applied toG 1 , 462I 1 ( 0. 8 )= 5 × 103 ,orI 1 = 13 .53 A. WithV ̄ 1

taken as reference, the phasor expression forI ̄ 1 is given byI ̄ 1 = 13. 53  −cos−^10. 8 =

13. 53  − 36 .9° A, sinceG 1 supplies at 0.8 lagging power factor.

The KVL equation yields

V ̄L= 462  0°−( 13. 53  − 36 .9°)( 1. 4 +j 1. 6 )= 433. 8  − 0 .78° V

The magnitude of the voltage at the load terminals is 433.8 V (rms).