ssrs barcodelib Figure 10-5(d) shows pT t for a load with a leading power in Software

Generator Denso QR Bar Code in Software Figure 10-5(d) shows pT t for a load with a leading power

Figure 10-5(d) shows pT t for a load with a leading power
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COMPLEX POWER, APPARENT POWER, AND POWER TRIANGLE
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The two components P and Q of power play di erent roles and may not be added together. However, they may conveniently be brought together in the a vector quantity called complex power S p form of and de ned by S P jQ. The magnitude jSj P2 Q2 Veff Ieff is called the apparent power S and is expressed in units of volt-amperes (VA). The three scalar quantities S, P, and Q may be represented geometrically as the hypotenuse, horizontal and vertical legs, respectively, of a right triangle
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AC POWER
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(called a power triangle) as shown in Fig. 10-6(a). The power triangle is simply the triangle of the 2 impedance Z scaled by the factor Ieff as shown in Fig. 10-6(b). Power triangles for an inductive load and a capacitive load are shown in Figs. 10-6(c) and (d), respectively.
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Fig. 10-6
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I eff
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It can be easily proved that S Veff I , where Veff is the complex amplitude of e ective voltage and eff is the complex conjugate of the amplitude of e ective current. An equivalent formula is S I2 Z. eff In summary, Complex Power: Real Power: Reactive Power: Apparent Power:
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2 S Veff I P jQ Ieff Z eff
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(13) (14) (15) (16)
P Re S Veff Ieff cos  Q Im S Veff Ieff sin  S Veff Ieff
EXAMPLE 10.10 (a) A sinusoidal voltage with Veff 10 V is connected across Z1 1 j as shown in Fig. 10-7(a). Find i1 , I1;eff , p1 t , P1 , Q1 , power factor pf1 , and S1 . (b) Repeat part (a) replacing the load Z1 in (a) by Z2 1 j, as shown in Fig. 10-7(b). (c) Repeat part (a) after connecting in parallel Z1 in (a) and Z2 in (b) as shown in Fig. 10-7(c). p Let v 10 2 cos !t.
AC POWER
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(a) See Fig. 10-7(a). p Z1 2 458 i1 10 cos !t 458 p I1;eff 5 2 458 p p1 t 100 2 cos !t cos !t 458 p 50 50 2 cos 2!t 458 W P1 Veff I1;eff cos 458 50 W Q1 Veff I1;eff sin 458 50 var S1 P1 jQ1 50 j50 p S1 jS1 j 50 2 70:7 VA pf1 0:707 lagging (c) See Fig. 10-7(c).
See Fig. 10-7(b) p Z2 2 458 i2 10 cos !t 458 p I2;eff 5 2 458 p p2 t 100 2 cos !t cos !t 458 p 50 50 2 cos 2!t 458 W P2 Veff I2;eff cos 458 50 W Q2 Veff I2;eff sin 458 50 var S2 P2 jQ2 50 j50 p S2 jS2 j 50 2 70:7 VA pf2 0:707 leading
1 j 1 j 1 Z Z1 kZ2 1 j 1 j p i 10 2 cos !t Ieff 10 p t 200 cos2 !t 100 100 cos 2!t W P Veff Ieff 100 W Q 0 S P 100 S jSj 100 VA pf 1
Fig. 10-7
CHAP. 10]
AC POWER
Fig. 10-7 (Cont.)
AC POWER
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The results of part (c) may also be derived from the observation that for the Z1 kZ2 combination, i i1 i2 and, consequently, p t p1 t p2 t p p 50 50 2 cos 2!t 458 50 50 2 cos 2!t 458 100 100 cos 2!t W P P1 P2 50 50 100 W Q Q1 Q2 50 50 0 S 100 < S1 S2 The power triangles are shown in Figs. 10-7(a), (b), and (c). three loads. Figure 10-7(d) shows the plots of v, i, and p for the
EXAMPLE 10.11 A certain passive network has equivalent impedance Z 3 j4
and an applied voltage v 42:5 cos 1000t 308 Give complete power information. 42:5 Veff p 308 V 2 p V 42:5= 2 308 8:5 p 23:138 A eff Z 5 53:138 2 V
Ieff
S Veff I 180:6 53:138 108:4 j144:5 eff Hence, P 108:4 W, Q 144:5 var (inductive), S 180:6 VA, and pf cos 53:138 0:6 lagging.
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