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Fig. 10-3
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taneous power may be negative, which indicates that the power ows out of the load. of power during one cycle is, however, nonnegative and is called the average power.
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EXAMPLE 10.3 A voltage v 140 cos !t is connected across an impedance Z 5 608. The voltage v results in a current i 28 cos !t 608 . Then, p t vi 140 28 cos !t cos !t 608 980 1960 cos 2!t 608
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Find p t .
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The instantaneous power has a constant component of 980 W and a sinusoidal component with twice the frequency of the source. The plot of p vs. t is similar to that in Fig. 10-4 with  =3.
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The net or average power Pavg hp t i entering a load during one period is called the real power. Since the average of cos 2!t  over one period is zero, from (2) we get Pavg Veff Ieff cos  3
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[CHAP. 10
Fig. 10-4
If Z R jX jZj , then cos  R=jZj and Pavg may be expressed by Pavg Veff Ieff or or Pavg R jZj 4
2 Veff R jZj2
(5) (6)
2 Pavg RIeff
The average power is nonnegative. It depends on V, I, and the phase angle between them. When Veff and Ieff are given, P is maximum for  0. This occurs when the load is purely resistive. For a purely reactive load, jj 908 and Pavg 0. The ratio of Pavg to Veff Ieff is called the power factor pf. From (3), the ratio is equal to cos  and so pf Pavg Veff Ieff 0 pf 1 7
The subscript avg in the average power Pavg is often omitted and so in the remainder of this chapter P will denote average power.
EXAMPLE 10.4 Find P delivered from a sinusoidal voltage source with Veff 110 V to an impedance of Z 10 j8. Find the power factor. Z 10 j8 12:81 38:78 Ieff V 110 eff 8:59 38:78 A Z 12:81 38:78 W pf cos 38:78 0:78
P Veff Ieff cos  110 8:59 cos 38:78 737:43
CHAP. 10]
AC POWER
Alternative Solution We have jZj2 100 64 164.
Then,
2 P Veff R=jZj2 1102 10 =164 737:8 W
The alternative solution gives a more accurate answer.
REACTIVE POWER
If a passive network contains inductors, capacitors, or both, a portion of energy entering it during one cycle is stored and then returned to the source. During the period of energy return, the power is negative. The power involved in this exchange is called reactive or quadrature power. Although the net e ect of reactive power is zero, it degrades the operation of power systems. Reactive power, indicated by Q, is de ned as Q Veff Ieff sin  If Z R jX jZj , then sin  X=jZj and Q may be expressed by Q Veff Ieff or or Q
2 Veff X jZj2
X jZj
9 (10) (11)
2 Q XIeff
The unit of reactive power is the volt-amperes reactive (var). The reactive power Q depends on V, I, and the phase angle between them. It is the product of the voltage and that component of the current which is 908 out of phase with the voltage. Q is zero for  08. This occurs for a purely resistive load, when V and I are in phase. When the load is purely reactive, jj 908 and Q attains its maximum magnitude for given V and I. Note that, while P is always nonnegative, Q can assume positive values (for an inductive load where the current lags the voltage) or negative values (for a capacitive load where the current leads the voltage). It is also customary to specify Q by it magnitude and load type. For example, 100-kvar inductive means Q 100 kvar and 100-kvar capacitive indicates Q 100 kvar.
EXAMPLE 10.5 and Q. The voltage and current across a load are given by Veff 110 V and Ieff 20 508 A. P 110 20 cos 508 1414 W Q 110 20 sin 508 1685 var Find P
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