Transcript No Slide Title

Chapter 11
AC Power Analysis
Chapter Objectives:
 Know the difference between instantaneous power and average
power
 Learn the AC version of maximum power transfer theorem
 Learn about the concepts of effective or rms value
 Learn about the complex power, apparent power and power factor
 Understand the principle of conservation of AC power
 Learn about power factor correction
Huseyin Bilgekul
Eeng224 Circuit Theory II
Department of Electrical and Electronic Engineering
Eastern Mediterranean University
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Instantenous AC Power
 Instantenous Power p(t) is the power at any instant of time.
v(t )  Vm cos(t  v ) i(t )  I m cos(t  i )
1
1
p(t )  v(t )i (t )  Vm I m cos(v  i )  Vm I m cos(2t  v  i )
2
2
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Instantenous AC Power
 Instantenous Power p(t) is the power at any instant of time.
p(t )  v(t )i(t )
Assume a sinusoidal voltage with phase v , v(t )  Vm cos(t  v )
Assume a sinusoidal current with phase i , i(t )  I m cos(t  i )
1
1
p(t )  v(t )i(t )  Vm I m cos(v  i )  Vm I m cos(2t  v  i )
2
2
p(t )  CONSTANT POWER+SINUSOIDAL POWER (frequency 2 )
1
1
p(t )  v(t )i (t )  Vm I m cos( v  i )  Vm I m cos(2t   v  i )
2
2
 The instantaneous power is composed of two parts.
• A constant part.
• The part which is a function of time.
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Instantenous and Average Power
 The instantaneous power p(t) is composed of a constant part (DC) and a time
dependent part having frequency 2ω.
p(t )  v(t )i (t )
v(t )  Vm cos(t  v )
i (t )  I m cos(t  i )
1
1
p(t )  Vm I m cos(v  i )  Vm I m cos(2t  v  i )
2
2
Instantenous Power p(t)
Average Power
P  12 Vm I m cos(v  i )
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Instantenous and Average Power
p(t )  12 Vm I m cos(v i )  12 Vm I m cos(2t v i )  p1 (t )  p2 (t )
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Average Power
The average power P is the average of the instantaneous power over one period .
p (t )  v(t )i (t ) Instantaneous Power
1 T
P   p (t )dt Average Power
T 0
v(t )  Vm cos(t   v ) i (t )  I m cos(t  i )
1
P
T

T
0
1
p (t )dt 
T

T
1
0 2
1
Vm I m cos( v   i )dt 
T

T
1
0 2
Vm I m cos(2 t   v   i )dt
1 T
1 T
1
P  Vm I m cos(v  i )  dt  2 Vm I m  cos(2t  v  i )dt
T 0
T 0
= 12 Vm I m cos(v  i )  0
(Integral of a Sinusoidal=0)
1
2
P  12 Vm I m cos(v  i )
1
P  Re  VI   Vm I m cos(v  i )
2
1
2

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Average Power
The average power P, is the average of the instantaneous power over one period .
P  12 Vm I m cos(v  i )
1
P  Re  VI   Vm I m cos(v  i )
2

1
2
 A resistor has (θv-θi)=0º so the average power becomes:
PR  Vm I m  I m R 
1
2
1.
2.
3.
4.
1
2
2
1
2
2
I R
P is not time dependent.
When θv = θi , it is a purely resistive load case.
When θv– θi = ±90o, it is a purely reactive load case.
P = 0 means that the circuit absorbs no average power.
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Instantenous and Average Power
 Example 1 Calculate the instantaneous power and
average power absorbed by a passive linear network if:
v(t )  80cos (10 t  20)
i(t )  15 sin (10 t  60)
1
1
p(t )  Vm I m cos( v  i )  Vm I m cos(2t   v  i )
2
2
=385.7  600cos(20t  10) W
P= 385.7 W is the average power flow
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Average Power Problem
 Practice Problem 11.4: Calculate the average power absorbed by each of the five
elements in the circuit given.
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Average Power Problem
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Maximum Average Power Transfer
 Finding the maximum average power which can be transferred from
a linear circuit to a Load connected.
a) Circuit with a load
b) Thevenin Equivalent circuit
• Represent the circuit to the left of the load by its Thevenin equiv.
• Load ZL represents any element that is absorbing the power generated
by the circuit.
• Find the load ZL that will absorb the Maximum Average Power from
the circuit to which it is connected.
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Maximum Average Power Transfer Condition
• Write the expression for average power associated with ZL: P(ZL).
ZTh = RTh + jXTh
ZL = RL + jXL
I
VTh
VTh

ZTh  Z L ( RTh  jX Th )  ( RL  jX L )
P
VTh
2
RL
1 2
2
I RL 
2
( RTh  RL ) 2  ( X Th  X L ) 2
Ajust R L and X L to get maximum P
VTh RL ( X Th  X L )
2
P

X L ( R  R ) 2  ( X  X ) 2  2
L
Th
L
 Th

2
2
P VTh ( RTh  RL )  ( X Th  X L )  2 RL ( RTh  RL ) 

2
2 2
RL
2 ( RTh  RL )  ( X Th  X L ) 
2
P
 0  X L   X Th
X L
P
0
RL
 RL  RTh 2  ( X Th  X L ) 2  RTh
Z L  RL  jX L  RTh  jX Th  ZTh
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Maximum Average Power Transfer Condition
• Therefore: ZL = RTh - XTh = ZTh will generate the maximum power
2
transfer.
2
I L RL VTh

• Maximum power Pmax Pmax 
2
8 RTh
 For Maximum average power transfer to a load impedance ZL we must
choose ZL as the complex conjugate of the Thevenin impedance ZTh.
Z L  RL  jX L  RTh  jX Th  Z Th
Pmax 
VTh
2
8RTh
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Maximum Average Power Transfer
 Practice Problem 11.5: Calculate the load impedance for maximum power
transfer and the maximum average power.
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Maximum Average Power Transfer
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Maximum Average Power for Resistive Load
 When the load is PURELY RESISTIVE, the condition for maximum power
transfer is:
XL  0
 RL  RTh 2  ( X Th  X L ) 2  RTh 2  X Th 2  ZTh
 Now the maximum power can not be obtained from the Pmax formula given before.
 Maximum power can be calculated by finding the power of RL when XL=0.
●
RESISTIVE
LOAD
●
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Maximum Average Power for Resistive Load
 Practice Problem 11.6: Calculate the resistive load needed for maximum power
transfer and the maximum average power.
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Maximum Average Power for Resistive Load
RL
 Notice the way that the maximum power is calculated using the Thevenin
Equivalent circuit.
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Effective or RMS Value
 The EFFECTIVE Value or the Root Mean Square (RMS) value of a periodic
current is the DC value that delivers the same average power to a resistor as the
periodic current.
a) AC circuit
b) DC circuit
1 T
R T
2
P   i (t ) Rdt   i (t ) 2 dt  I eff 2 R  I Rms 2 R
T 0
T 0
I eff  I Rms
1 T
2

i
(
t
)
dt

0
T
Veff  VRms
1 T
2

v
(
t
)
dt

0
T
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Effective or RMS Value of a Sinusoidal
 The Root Mean Square (RMS) value of a sinusoidal voltage or current is equal
to the maximum value divided by square root of 2.
I Rms
1 T 2
2

I
cos
tdt 
m

0
T
Im2
T

T
0
I
1
(1  cos 2t )dt  m
2
2
P  12 Vm I m cos(v i )  VRms I Rms cos(v i )
 The average power for resistive loads using the (RMS) value is:
2
V
PR  I Rms 2 R  Rms
R
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Effective or RMS Value
 Practice Problem 11.7: Find the RMS value of the current waveform. Calculate
the average power if the current is applied to a 9  resistor.
4t
0  t 1
 4t
i(t )  
8  4t 1  t  2
I
2
rms
1 T 2
1
  i dt  
T 0
2 
16 
2
I rms  
2
I rms


T 2
2

(4
t
)
dt

(8

4
t
)
dt
0
1

1
2
2

2
2
t dt  (4 4t t ) dt 
0
1

1
8-4t
2
16

 2.309A
3
2
I rms
3
1 
 2  16
t
2
 8    4t  2t   1  
3  3
3 
PI
2
rms
 16 
R   (9)  48W
3
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