0195136047.pdf

116 TIME-DEPENDENT CIRCUIT ANALYSIS

φ = 24.23°

S = VRMSIRMS = 456 VA

PR = 415.9 W = PS

QS = 187.1 VAR

(d)

QC = −20.8 VAR

P^2 + Q^2

QL = 207.9 VAR

Figure E3.1.5Continued

Note that the power factor of the circuit is cos(24.23°)=0.912 lagging, since the current is

lagging the voltage in this inductive load.

PR=VR^2 rms/R=Irms^2 R=( 4. 56 )^220 = 415 .9W

A resistor absorbs real power.

QL=VL^2 rms/XL=Irms^2 XL=( 4. 56 )^210 = 207 .9VAR

An inductor absorbs positive reactive power.

QC=VC^2 rms/XC=Irms^2 XC=( 4. 56 )^2 (− 1 )=− 20 .8VAR

A capacitor absorbs negative reactive power (or delivers positive reactive power).

The power triangle is shown in Figure E3.1.5(d). Note thatPS(=PR) is delivered by the

source; andQS(=QL+QC) is delivered by the source. Note thatQCis negative.

EXAMPLE 3.1.6

Draw the phasor diagrams for anRLCseries circuit supplied by a sinusoidal voltage source with

a lagging power factor and aGLCparallel circuit supplied by a sinusoidal current source with a

leading power factor.

Solution

For theRLCseries circuit in the frequency domain in Figure E3.1.6(a), the phasor diagram for

the case of a lagging power factor is shown in Figure E3.1.6(b).

For theGLCparallel circuit in the frequency domain in Figure E3.1.6(c), the phasor diagram

for the case of a leading power factor is shown in Figure E3.1.6(d).

−

+

+−+

+

−

−

V

VR VL

I VC

jXL = jωL

jXC === j()−

R

(a)

jωC

1

ωC

1

ωC

−j

Figure E3.1.6(a)RLCseries circuit in

frequency domain.(b)Phasor diagram

for lagging power factor.(c)GLCparal-

lel circuit in frequency domain.(d)Pha-

sor diagram for leading power factor.