vb.net barcode reader code Computing Average and Instantaneous AC Power in Software

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EXAMPLE 71 Computing Average and Instantaneous AC Power
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Compute the average and instantaneous power dissipated by the load of Figure 75
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R v(t) = 1414 sin ( t) + ~ ( = 377 rad/s) _
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Known Quantities: Source voltage and frequency, load resistance and inductance values Find: Pav and p(t) for the RL load Figure 75
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L = 8 mH
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Schematics, Diagrams, Circuits, and Given Data: v(t) = 1414 sin(377t) V; R = 4 Assumptions: Use rms values for all phasor quantities in the problem
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AC Power
Analysis: First, we de ne the phasors and impedances at the frequency of interest in the problem, = 377 rad/s: Z = R + j L = 4 + j 3 = 5 (0644) V = 10 2 10 V 2 = 2 ( 2215) = I= Z 5 (0644)
The average power can be computed from the phasor quantities: Pav = VI cos( ) = 10 2 cos(0644) = 16 W The instantaneous power is given by the expression: p(t) = v(t) i(t) = 2 10 sin(377t) 2 2 cos(377t 2215) W The instantaneous voltage and current waveforms and the instantaneous and average power are plotted in Figure 76
Voltage and current waveforms for Example 71 20 10 0 10 20 0 0005 001 0015 t (s) 002 0025 003 0035
v(t) i(t) ) Volts ( ), amps (
Instantaneous and average power for Example 71 40 30 p(t)
Watts
20 10 0 10 0 0005 001 0015 t (s) 002 0025 003 0035 Pav
Figure 76 Comments: Please pay attention to the use of rms values in this example: It is very
important to remember that we have de ned phasors to have rms amplitude in power calculation This is a standard procedure in electrical engineering practice Note that the instantaneous power can be negative for brief periods of time, even though the average power is positive
EXAMPLE 72 Computing Average AC Power
Problem
Compute the average power dissipated by the load of Figure 77
Part I
Circuits
Solution
+ ~ + V ~ S ~ VL
Known Quantities: Source voltage, internal resistance and frequency, load resistance and inductance values Find: Pav for the RC load
C RL
Schematics, Diagrams, Circuits, and Given Data: Vs = 110 0; RS = 2 16 ; C = 100 F
Assumptions: Use rms values for all phasor quantities in the problem
; RL =
= 377 rad/s
Analysis: First, we compute the load impedance at the frequency of interest in the
problem, = 377 rad/s: ZL = R RL 1 16 = = 137 ( 0543) = j C 1 + j CRL 1 + j 06032
Next, we compute the load voltage, using the voltage divider rule: VL = ZL 137 ( 0543) 110 (0) = 976 ( 0067) V VS = RS + ZL 2 + 137 ( 0543) 9762 |VL |2 cos( ) = cos( 0543) = 595 W |ZL | 137
Knowing the load voltage, we can compute the average power according to: Pav =
or, alternatively, we can compute the load current and calculate average power according to the equation below: VL = 71 (0476) A IL = ZL Pav = |IL |2 |ZL | cos( ) = 712 137 cos( 0543) = 595 W
Comments: Please observe that it is very important to determine load current and/or
voltage before proceeding to the computation of power; the internal source resistance in this problem causes the source and load voltages to be different
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
EXAMPLE 73 Computing Average AC Power
Problem
R v(t) + ~ L C
Compute the average power dissipated by the load of Figure 78
Solution
Known Quantities: Source voltage, internal resistance and frequency, load resistance, capacitance and inductance values Find: Pav for the complex load
~ V + ~
An AC circuit
R 1 jvC jvL
005 H; C = 470 F
Schematics, Diagrams, Circuits, and Given Data: Vs = 110 0 V; R = 10
Assumptions: Use rms values for all phasor quantities in the problem
Its complex form
7
AC Power
Analysis: First, we compute the load impedance at the frequency of interest in the
problem, = 377 rad/s: 1 = ZL = (R + j L) j C (R + j L) j C 1 R + j L + j C 2 LC R + j L = 116 j 718 + j CR
116 ~ V + ~ j718
= 727 ( 141) Note that the equivalent load impedance consists of a capacitive load at this frequency, as shown in Figure 79 Knowing that the load voltage is equal to the source voltage, we can compute the average power according to: Pav = 1102 |VL |2 cos( ) = cos( 141) = 266 W |ZL | 727
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 The phase angle of the load impedance plays a very important role in the absorption of power by a load impedance As illustrated in equation 713 and in the preceding examples, the average power dissipated by an AC load is dependent on the cosine of the angle of the impedance To recognize the importance of this factor in AC power computations, the term cos( ) is referred to as the power factor (pf) Note that the power factor is equal to 0 for a purely inductive or capacitive load and equal to 1 for a purely resistive load; in every other case, 0 < pf < 1 (720)
Two equivalent expressions for the power factor are given in the following: pf = cos( ) = Pav VI Power factor (721)
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