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電路學(二)
Chapter 11
AC Power Analysis
1
AC Power Analysis
Chapter 11
11.1
11.2
11.3
11.4
11.5
11.6
11.7
11.8
Instantaneous and Average Power
Maximum Average Power Transfer
Effective or RMS Value
Apparent Power and Power Factor
Complex Power
Conservation of AC Power
Power Factor Correction
Power Measurement
2
11.1 Instantaneous and
Average Power (1)
• The instantaneously power, p(t)
p(t )  v(t ) i(t )  Vm I m cos (w t   v ) cos (w t  i )
1
cos A cos B  [cos( A  B)  cos( A  B)]
2
1
1
 Vm I m cos ( v  i )  Vm I m cos (2w t   v  i )
2
2
Constant power
Sinusoidal power at 2wt
p(t) > 0: power is absorbed by the circuit; p(t) < 0: power is absorbed by the source.
3
11.1 Instantaneous and
Average Power (2)
• The average power, P, is the average of the instantaneous
power over one period.
1
P
T

T
0
1
p(t ) dt  Vm I m cos ( v   i )
2
1. P is not time dependent.
2. When θv = θi , it is a purely
resistive load case.
3. When θv– θi = ±90o, it is a
purely reactive load case.
4. P = 0 means that the circuit
absorbs no average power.
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11.1 Instantaneous and
Average Power (3)
1 * 1
VI  Vm I m  v  i
2
2
1
 Vm I m [cos( v  i )  j sin( v  i )]
2
1
1
*
P  Re[VI ]  Vm I m cos( v  i )
2
2
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11.1 Instantaneous and
Average Power (4)
Example 1
Given that
v(t )  120 cos ( 377 t  45)
i(t )  10 sin ( 377 t  10)
find the instantaneous power and average power
absorbed by a passive linear network .
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11.1 Instantaneous and
Average Power (5)
Example 2
Calculate the instantaneous power and average
power absorbed by a passive linear network if:
v(t )  80 cos (10 t  20)
i(t )  15 sin (10 t  60)
Answer: 385.7  600cos(20t  10)W, 387.5W
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11.1 Instantaneous and
Average Power (6)
Example 3
Calculate the average power absorbed by an
impedance Z = 30  j70  when a voltage
V = 1200 is applied across it.
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11.1 Instantaneous and
Average Power (6)
Example 4
A current I  10  30 flows through an
impedance Z  20  22Ω . Find the average
power delivered to the impedance.
Answer: 927.2W
9
11.2 Maximum Average Power
Transfer (1)
ZTH  R TH  j XTH
ZL  R L  j XL
The maximum average power
can be transferred to the load if
XL = –XTH and RL = RTH
Pmax 
If the load is purely real, then R L 
VTH
2
8 R TH
2
2
R TH
 XTH
 ZTH
10
11.2 Maximum Average Power
Transfer (2)
Example 5
For the circuit shown below, find the load impedance ZL that
absorbs the maximum average power. Calculate that maximum
average power.
Answer: 3.415 – j0.7317, 1.429W
11
11.3 Effective or RMS Value (1)
The total power dissipated by R is given by:
1 T 2
R T 2
2
P   i Rdt   i dt  I rms
R
0
0
T
T
Hence, Ieff is equal to: I eff 
1
T

T
i 2 dt  I rms
0
The effect value of a periodic signal is its
root-mean-square (rms) value.
The effective of a periodic current is the dc current that delivers the
same average power to a resistor as the periodic current.
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11.3 Effective or RMS Value (2)
The rms value of a sinusoid i(t) = Imcos(wt)
is given by:
I rms
Im

2
The rms is a constant itself which depending on
the shape of the function i(t).
The average power can be written in terms of
the rms values:
1
Vm I m cos (θv  θi )  Vrms I rms cos (θv  θi )
2
R
2
2
 I rms
R  Vrms
R2  X 2
Peff 
Note: If you express amplitude of a phasor source(s) in rms, then all the
answer as a result of this phasor source(s) must also be in rms value.
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11.3 Effective or RMS Value (3)
Example 6
Determine the rms value of the current waveform. If the current
is passed through a 2- resistor, find the average power
absorbed by the resistor.
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11.3 Effective or RMS Value (4)
Example 7
The waveform is a half-wave rectified sine wave. Find the rms
value and the average power dissipated in a 10- resistor.
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11.4 Apparent Power and
Power Factor (1)
• Apparent Power 視功率, S, is the product of the r.m.s.
values of voltage and current.
• It is measured in volt-amperes or VA to distinguish it from
the average or real power which is measured in watts.
P  Vrms I rms cos (θ v  θi )  S cos (θ v  θi )
Apparent Power, S
Power Factor, pf
• Power factor is the cosine of the phase difference between
the voltage and current. It is also the cosine of the angle
of the load impedance.
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11.4 Apparent Power and
Power Factor (2)
Purely resistive
θv– θi = 0, pf = 1
load (R)
Purely reactive θv– θi = ±90o, pf
load (L or C)
=0
Resistive and
reactive load
(R and L/C)
θv– θi > 0
θv– θi < 0
P/S = 1, all power are
consumed
P = 0, no real power
consumption
• Lagging - inductive
load
• Leading - capacitive
load
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11.4 Apparent Power and
Power Factor (3)
Example 8
A series-connected load draws a current i(t)=4cos(100t + 10) A
when the applied voltage is v(t)=120cos(100t  20) V. Find the
apparent power and the power factor of the load. Determine the
element values that from the series-connected load.
18
11.4 Apparent Power and
Power Factor (4)
Example 9
Determine the power factor of the entire circuit of the figure as
seen by the source. Calculate the average power delivered by the
source.
19
11.5 Complex Power (1)
Complex power S is the product of the voltage and the
complex conjugate of the current:
V  Vmθ v
I  I mθi
1

V I  Vrms I rms  θ v  θ i
2
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11.5 Complex Power (2)
1
S  V I  Vrms I rms  θ v  θ i
2
 S  Vrms I rms cos (θ v  θi )  j Vrms I rms sin (θ v  θi )
S =
P
+ j
Q
P: is the average power in watts delivered to a load and it is
the only useful power.
Q: is the reactive power exchange between the source and
the reactive part of the load. It is measured in VAR.
• Q = 0 for resistive loads (unity pf).
• Q < 0 for capacitive loads (leading pf).
• Q > 0 for inductive loads (lagging pf).
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11.5 Complex Power (3)
 S  Vrms I rms cos (θ v  θi )  j Vrms I rms sin (θ v  θi )
S =
P
+ j
Q
Apparent Power, S = |S| = Vrms*Irms = P 2  Q 2
Real power 實功率, P = Re(S) = S cos(θv – θi)
Reactive Power 虛功率, Q = Im(S) = S sin(θv – θi)
Power factor 功率因子,
pf = P/S = cos(θv – θi)
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11.5 Complex Power (4)
 S  Vrms I rms cos (θ v  θi )  j Vrms I rms sin (θ v  θi )
S =
Power Triangle Impedance Triangle
P
+ j
Power Factor
Q
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11.5 Complex Power (5)
Example 10
The voltage across a load is v(t)=60cos(wt  10) V and the
current through the element in the direction of the voltage drop is
i(t)=1.5cos(wt + 50) A. Find: (a) the complex and apparent
powers, (b) the real and reactive powers, and (c) the power
factor and the load impedance.
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11.5 Complex Power (6)
Example 11
A load Z draws 12 kVA at a power factor of 0.856 lagging from a
120-V rms sinusoidal source. Calculate: (a) the average and
reactive powers delivered to the load, (b) the peak current, and
(c) the load impedance.
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11.6 Conservation of AC Power (1)
The complex real, and reactive powers of the sources
equal the respective sums of the complex, real, and
reactive powers of the individual loads.
For parallel connection:
S
1 * 1
1
1
VI  V(I1*  I*2 )  VI1*  VI *2  S1  S2
2
2
2
2
The same results can be obtained for a series connection. 26
11.6 Conservation of AC Power (2)
Example 12
The figure shows a load being fed by a voltage source through a
transmission line. The impedance of the line is represented by the
(4 + j2)  impedance and a return path. Find the real power and
reactive power absorbed by: (a) the source, (b) the line, and (c)
the load.
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11.6 Conservation of AC Power (3)
Example 13
In the circuit, Z1 = 6030  and Z2 = 4045 . Calculate the
total: (a) apparent power, (b) real power, (c) reactive power, and
(d) pf, supplied by the source and seen by the source.
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11.7 Power Factor Correction (1)
Power factor correction is the process of increasing the
power factor without altering the voltage or current to
the original load.
Power factor correction is necessary for economic reason.
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11.7 Power Factor Correction (2)
Qc = Q1 – Q2
= P (tan θ1 - tan θ2)
= ωCV2rms
Q1 = S1 sin θ1
= P tan θ1
P = S1 cos θ1
Q2 = P tan θ2
C 
Qc
P (tan θ1  tan θ 2 )

2
2
ωVrms
ω Vrms
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11.7 Power Factor Correction (3)
Example 14
When connected to a 120-V (rms), 60-Hz power line, a load
absorbs 4 kW at a lagging power factor of 0.8. Find the value of
capacitance necessary to raise the pf to 0.95.
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11.8 Power Measurement (1)
The wattmeter is the instrument for measuring the average
power.
The basic structure
If
Equivalent Circuit with load
v(t )  Vm cos(wt   v ) and i(t )  I m cos(wt   i )
P  Vrms Irms cos (θ v  θi ) 
1
2
Vm Im cos (θ v  θi )
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