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Complex power
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where the asterisk denotes the complex conjugate (see Appendix A) You may easily verify that this de nition leads to the convenient expression S = V I cos + j V I sin = I 2 R + j I 2 X = I 2 Z or S = Pav + j Q The complex power S may be interpreted graphically as a vector in the complex plane, as shown in Figure 711 The magnitude of S, |S|, is measured in units of volt-amperes (VA) and is called apparent power, because this is the quantity one would compute by measuring the rms load voltage and currents without regard for the phase angle of the load Note that the right triangle of Figure 711 is similar to the right triangle of Figure 74, since is the load impedance angle The complex power may also be expressed by the product of the square of the rms current through the load and the complex load impedance: S = I 2Z or I 2R + j I 2X = I 2Z or, equivalently, by the ratio of the square of the rms voltage across the load to the complex conjugate of the load impedance: S= V2 Z (731) (730) (729)
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S Z Pav
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~ ~ |S| = Pav2 + Q2 = V I ~~ Pav =VI cos ~~ Q =VI sin
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Figure 711 The complex power triangle
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The power triangle and complex power greatly simplify load power calculations, as illustrated in the following examples
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EXAMPLE 74 Complex Power Calculations
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Problem
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Use the de nition of complex power to calculate real and reactive power for the load of Figure 712
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Part I
Circuits
Solution
Known Quantities: Source, load voltage and current Find: S = Pav + j Q for the complex load
+ ~
i(t) = 2 cos( t 0262) A
Schematics, Diagrams, Circuits, and Given Data: v(t) = 100 cos( t + 0262) V; Figure 712 Assumptions: Use rms values for all phasor quantities in the problem Analysis: First, we convert the voltage and current into phasor quantities:
100 V = (0262) V 2
2 I = ( 0262) A 2
Next, we compute real and reactive power using the de nitions of equation 713: 200 cos(0524) = 866 W Pav = |V I| cos( ) = 2 200 Q = |V I| sin( ) = sin(0524) = 50 VAR 2 Now we apply the de nition of complex power (equation 728) to repeat the same calculation: 100 2 S = VI = (0262) ( 0262) = 100 (0524) 2 2 = 866 + j 50 W Therefore Pav = 866 W time
Focus on Computer-Aided Tools: A le containing the computer-generated solution to
Q = 50 VAR
Comments: Note how the de nition of complex power yields both quantities at one
this problem may be found in the CD-ROM that accompanies this book
EXAMPLE 75 Real and Reactive Power Calculations
Problem
Use the de nition of complex power to calculate real and reactive power for the load of Figure 713
RS + RL ~ VS + ~ ~ VL C Source
Solution
Known Quantities: Source voltage and resistance; load impedance Find: S = Pav + j Q for the complex load
~ IL
RL = 5
Schematics, Diagrams, Circuits, and Given Data: VS = 110 0 V; RS = 2 ; C = 2,000 F
Assumptions: Use rms values for all phasor quantities in the problem
Load
7
AC Power
Analysis: De ne the load impedance:
Z L = RL +
1 = 5 j 1326 = 5173 ( 0259) j C
Next, compute the load voltage and current: VL = ZL 5 j 1326 110 = 799 ( 0072) V VS = RS + Z L 7 j 1326
VL 799 ( 0072) = = 1544 (0187) A IL = ZL 5 ( 0259) Finally, we compute the complex power, as de ned in equation 728: L S = VL I = 799 ( 0072) 1544 ( 0187) = 1, 233 ( 0259) = 1,192 j 316 W Therefore Pav = 1,192 W Q = 316 VAR
Comments: Is the reactive power capacitive or inductive 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
Although the reactive power does not contribute to any average power dissipation in the load, it may have an adverse effect on power consumption, because it increases the overall rms current owing in the circuit Recall from Example 72 that the presence of any source resistance (typically, the resistance of the line wires in AC power circuits) will cause a loss of power; the power loss due to this line resistance is unrecoverable and constitutes a net loss for the electric company, since the user never receives this power The following example illustrates quantitatively the effect of such line losses in an AC circuit
EXAMPLE 76 Real Power Transfer for Complex Loads
Problem
~ IS
Use the de nition of complex power to calculate real and reactive power for the load of Figure 714 Repeat the calculation when the inductor is removed from the load, and compare the real power transfer between source and load for the two cases
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