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Chapter 11
AC Circuit Power Analysis
11.1 Instantaneous Power
11.2 Average Power
11.3 Effective Values of Current and Voltage
11.4 Apparent Power and Power Factor
11.5 Complex Power
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11.1 Instantaneous Power
Instantaneous Power Product of the time-domain voltage and time-domain
current associated with the element or network of interest.
∶
1
∶
1
assume
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0
∞
0
1
∶
∞
1
assume
2
11.1 Instantaneous Power
1
1
1
1
Total power
delivered by the
source
Power delivered to the resistor
1
1
Power absorbed by the inductor
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At all times t,
power supplied = power absorbed
3
11.1 Instantaneous Power
Power due to Sinusoidal Excitation
cos
tan
cos
cos
2
cos
cos
2
cos 2
Example 11.1 Find the power being absorbed by the capacitor and the resistor at t=1.2ms.
⇒
200
5
300
/
10
1
,
300
⇒
1.2
100
1.2
300
60
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,
0
40 ,
= 100
60
0
200
0.3
40  60u(t )
5
200
.
300
1.2
200
1.633
1.2
90.36 81.93
7.403
4
11.2 Average Power
1
Average power selected on a general interval
1
Average power for periodic waveforms
1
1
,
1, 2, 3, ⋯
/
lim
1
→
/
/
/
Average power in the sinusoidal steady state
cos
cos
cos
1
2
cos
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cos
1
2
cos 2
Average power
1
2
cos
5
11.2 Average Power
Example 11.2 Find both the average power and an expression for the instantaneous power.
4 cos
⇒
6
4∠0°
4∠0°
2∠60°
Phasor current
Average power
1
cos 0°
2
60°
4 cos
•
6
2∠
60°
2
2 cos
60°
6
2
4 cos
3
60° Average power absorbed by an ideal resistor
1
2
•
2 ∠60°
cos 0°
1
2
1
2
1
2
cos
cos 0°
1
Instantaneous
power
Phase-angle
difference is zero
across a pure
resistor
Average power absorbed by purely reactive elements
Average power for (L , C )= 0
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cos
cos 90°
0
0
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11.2 Average Power
Example 11.3 Find the average power being delivered to an impedance ZL by I .
  520 A
 L  8  j11 
 P
1 2
1
I m R  528  100 W
2
2
Example 11.4 Find the average power absorbed by each of the three passive elements.
5
10
5
5
7.071∠
5
1
2
11.18∠
5∠
1
5
2
63.53°
45°
90°
2
25
Absorb
1
2
cos
1
20 11.18 cos 0
2
1
cos
2
1
10 7.071 cos 0
2
50
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25
63.53
45
50 25 Deliver
Absorb
25
7
11.2 Average Power
Maximum Power Transfer
An independent voltage source in series
with an impedance Zth delivers a
maximum average power to that load
impedance ZL which is the conjugate of
Zth:
=
=
1
2
cos
ZL = Zth*
1
2
cos tan
Example 11.5 Find Z for maximum power deliver. Source voltage is 3cos(100t-3).
*
?  ZTh
  500  j3  500  j3 
*
This impedance can be constructed by series
connection of 500 resistor and 3.333mF capacitor.
1
1
j
  j3
jC
100  3.333 103
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11.2 Average Power
Average Power for Non-periodic Functions
Usually, the sum of periodic functions is periodic except the ratio of periods is irrational
number.
sin
sin
2
sin
sin
sin
100
2
sin
sin
2
lim
→
Generally
Non-periodic
sin 3.14
sin
100
sin
sin 3.14
sin 3.14
314
with
1
2
/
lim
/
cos
→
cos
100
Periodic with period 100
1
1
/
sin
sin
2 sin sin
/
⋯
cos
1
2
Example 11.6 Find the average power delivered to a 4 resistor.
1
2 cos10
3 20 2
3 4
2
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1
2
1
2
0
1
⋯
26
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11.3 Effective Values of Current and Voltage
Power outlet of 220 V do Not mean the instantaneous voltage of 220 V, but
effective value of the sinusoidal voltage out.
The same power is delivered to the resistor in the circuits shown.
1
1
1
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The effective value is often
referred to as the root-meansquare or RMS value.
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11.3 Effective Values of Current and Voltage
Effective (RMS) value of a sinusoidal waveform
cos
2
with
/
1
cos
2
1
2
1
2
2
2
2
Use of RMS values to compute Average Power
1
2
1
2
1
2
cos
2
1
2
2
2
cos
∴ cos
Effective Value with Multiple-Frequency Circuits
cos
cos
⋯
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⋯
cos
⇒ ⇒ 1
2
⋯
⋯
11
11.4 Apparent Power and Power Factor
Let
cos
,
cos
phaseangle:
Apparent power
Average power
1
cos
2
unit: VA
cos
Power factor
Sinusoidal case
pure resistive load:
pure reactive load:
cos
1
90° →
0
Example 11.8 Calculate the average power delivered to each of the two impedances.
60∠0°
3
4
12∠
12
12
60 12
Apparentpower:
Load: inductive
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12
53.13° 2
288
1
144
720
288 144
720
432
720
0.6
12
11.5 Complex Power
Define the complex power S as
•
•
S  Veff I*eff  Veff I eff e j(  )  P  jQ
the real part of S is P, the average power
the imaginary part of S is Q, the reactive power, which represents the flow of energy
back and forth from the source (utility company) to the inductors and capacitors of
the load (customer).
Average power
∗
cos
sin
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reactive power
unit: VAR
∗
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11.5 Complex Power
The Power Triangle
0: powerfactorislagging
inductiveload
0: powerfactorisleading
capacitiveload
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•
Real power (average power) : the magnitude of
voltage phasor  the magnitude of current phasor
which is in phase with the voltage.
•
Reactive power (quadrature power) : the
magnitude of voltage phasor  the magnitude of
current phasor which is 90 out of phase with the
voltage.
14
11.5 Complex Power
Power Measurement
S  VI*  V(I1  I 2 )*  V(I1*  I*2 )  S1  S2
Example 11.9 An industrial consumer is operating a 50 kW induction motor at a lagging
PF of 0.8. The source voltage is 230 V rms. In order to obtain lower electrical rates, the
customer wishes to raise the PF to 0.95 lagging. Specify the suitable solution.
S1 : 50 kW induction motor at a lagging PF of 0.8
∗
230 V rms
50
∠ cos
0.8
In order to achieve PF of 0.95
50
∠ cos
0.95
⇒
50
0.95
16.43
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50
0.8
∠
50 ∠36.9°
0.8
⇒
50
16.43
37.5
21.07
⇒
∠
50
∗
⇒
37.5
∗
21.07 10
230
230
91.6
91.6
2.51Ω
15
11.5 Complex Power
Homework : 11장 Exercises 5의 배수 문제. 기말고사 칠 때 제출.
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이 자료에 나오는 모든 그림은 McGraw·hill 출판사로부터 제공받은 그림을 사
용하였음.
All figures at this slide file are provided from The McGraw-hill company.
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