Download AC Circuit Power Analysis

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
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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
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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
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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
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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
⋯
⋯
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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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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.
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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
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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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