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Known Quantities: Source voltage and resistance; load impedance Find:
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1 Sa = Pava + j Qa for the complex load 2 Sb = Pavb + j Qb for the real load 3 Compare Pav /PS for the two cases
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Schematics, Diagrams, Circuits, and Given Data: VS = 110 (0) V; RS = 4 ; j XL = j 6
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Assumptions: Use rms values for all phasor quantities in the problem Analysis:
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1 The inductor is part of the load De ne the load impedance ZL = RL j L = 10 j 6 = 5145 (103) 10 + j 6
Next, compute the load voltage and current: VL = ZL 5145 (103) 110 = 709 (0444) V VS = RS + Z L 4 + 5145 (103)
VL 709 (0444) IL = = 138 ( 0586) A ZL 5145 (103) Finally, we compute the complex power, as de ned in equation 728: L Sa = VL I = 709 (0444) 138 (0586) = 978 (103) = 503 + j 839 W Therefore Pava = 503 W Qa = +839 VAR
~ IS RS
2 The inductor is removed from the load (Figure 715) De ne the load impedance: ZL = RL = 10 Next, compute the load voltage and current: VL = ZL 10 110 = 786 (0) V VS = RS + Z L 4 + 10
~ + V ~ S
VL 786 (0) = 786 (0) A = IL = ZL 10 Finally, we compute the complex power, as de ned in equation 728: L Sb = VL I = 786 (0) 786 (0) = 617 (0) = 617 W Therefore Pavb = 617 W Qb = 0 VAR
3 Compute the percent power transfer in each case To compute the power transfer we S must rst compute the power delivered by the source in each case, SS = VS I For Case 1: VS VS 110 = 138 ( 0586) A = = IS = Ztotal RS + Z L 4 + 5145 (103) S SSa = VS I = 110 138 ( 0586) = 1,264 + j 838 W = PSa + j QSa and the percent real power transfer is: 100 503 Pa = 398% = PSa 1,264
7
AC Power
For Case 2: VS V 110 = 786 (0) A = = IS = Ztotal RS + RL 4 + 10 S SSb = VS I = 110 786 = 864 + j 0 W = PSb + j QSb and the percent real power transfer is: 100 617 Pb = 714% = PSb 864
Comments: You can see that if it were possible to eliminate the reactive part of the
impedance, the percentage of real power transferred from the source to the load would be signi cantly increased! A procedure that accomplishes this goal, called power factor correction, is discussed next
Focus on Computer-Aided Tools: A le containing the computer-generated solution to
this problem may be found in the CD-ROM that accompanies this book
Power Factor, Revisited The power factor, de ned earlier as the cosine of the angle of the load impedance, plays a very important role in AC power A power factor close to unity signi es an ef cient transfer of energy from the AC source to the load, while a small power factor corresponds to inef cient use of energy, as illustrated in Example 76 It should be apparent that if a load requires a xed amount of real power, P , the source will be providing the smallest amount of current when the power factor is the greatest, that is, when cos = 1 If the power factor is less than unity, some additional current will be drawn from the source, lowering the ef ciency of power transfer from the source to the load However, it will be shown shortly that it is possible to correct the power factor of a load by adding an appropriate reactive component to the load itself Since the reactive power, Q, is related to the reactive part of the load, its sign depends on whether the load reactance is inductive or capacitive This leads to the following important statement:
If the load has an inductive reactance, then is positive and the current lags (or follows) the voltage Thus, when and Q are positive, the corresponding power factor is termed lagging Conversely, a capacitive load will have a negative Q, and hence a negative This corresponds to a leading power factor, meaning that the load current leads the load voltage
Table 72 illustrates the concept and summarizes all of the important points so far In the table, the phasor voltage V has a zero phase angle and the current phasor is referenced to the phase of V The following examples illustrate the computation of complex power for a simple circuit
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