Engineering 43
Thevenin/Norton
AC Power
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering-43: Engineering Circuit Analysis
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Thevenin’s Equivalence Theorem
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A
RTH


vTH

i
a
vO
b
_

i
LINEAR CIRCUIT
PART A
b
 Resistance to
Impedance
Analogy
vO  VO
i I
a
vO
_
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART B
PART B
VTH  VTH
RTH  ZTH
Thevenin Equivalent Circuit
for PART A
Engineering-43: Engineering Circuit Analysis
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
AC Thevenin Analysis
 Same Circuit,
Find IO by Thevenin
 Take as “Part B” Load
the 1Ω Resistor Thru
Which IO Flows
 The Thevenin Ckt
 Now Use Src-Xform
Vx  2 A0  (1  j )
Vx  (2  j 2)V
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
AC Thevenin Analysis cont.
 The Xformed
Circuit
 Another
Source
Xform
 Now the Combined 8+j2
Source and all The
Impedance are
IN SERIES
8 2j
 Find ZTH by Zeroing the
8+j2V-Source
• Thus VOC Determined by
V-Divider
VOC
1 j
10  j 6

(8  2 j ) 
(1  j )  (1  j )
2
Engineering-43: Engineering Circuit Analysis
4
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
ReDraw the Voltage Divider
Z2
Z1

VO

 The V-Divider Formula: VO  VS Z 2 Z1  Z 2 
VOC
1 j
1 j 8  8 j  2 j  2 j2
 8  2 j 
 8  2 j 

(1  j )  (1  j )
2
2
8  8 j  2 j  2 j 2 8  2   8 j  2 j  10  6 j
VOC 


 53j
2
2
2 Bruce Mayer, PE
Engineering-43: Engineering Circuit Analysis
5
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
AC Thevenin Analysis
 Find ZTH by Zeroing the
8+j2 V-Source
ZTH  (1  j ) || (1  j )
ZTH

1  j 1  j 

1  j   1  j 
ZTH
1 j2 2

  1
11 2
Engineering-43: Engineering Circuit Analysis
6
 So The Thevenin
Equivalent Circuit
 IO is Now Simple
VOC
VTH
IO 

ZTH  1 1  1
53j
IO 
( A)
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Norton’s Equivalence Theorem
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A

RN
a
vO
b
_

iN
i
i
LINEAR CIRCUIT
PART A
b
 Resistance to
Impedance
Analogy
vO  VO
i I
a
vO
_
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART B
PART B
iN  I N
RN  Z N
Norton Equivalent Circuit
for PART A
Engineering-43: Engineering Circuit Analysis
7
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
AC Norton Analysis
 Same Circuit,
Find IO by Norton
 Take as the “Part B”
Load the 1Ω Resistor
Thru Which IO Flows
 Shorting the Load
Yields Isc= IN
 Possible techniques to
Find ISC:
• Loops or Nodes
• Source Transformation
• SuperPosition
 Choose Nodes
Engineering-43: Engineering Circuit Analysis
8
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
AC Norton Analysis cont
 Short-Ckt Node On The
Norton Circuit
I SC
60 8  2 j
 20 

( A)
1 j
1 j
 As Before Deactivate
Srcs to Find ZN=ZTH
 Use ISC=IN, and
ZN to Find IO
• See Next Slide
ZTH  (1  j) || (1  j)  1
Engineering-43: Engineering Circuit Analysis
9
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
AC Norton Analysis cont.2
 The Norton
Equivalent Circuit
with Load
Reattached
 Find IO By I-Divider
I O  I SC
4  j 1 j
IO 
1 j 1 j
IO


4  j    4 j  j 2  5  j 3


12  j 2
Engineering-43: Engineering Circuit Analysis
10
 Same as all the Others
I SC 4  j
1


ZN 1 11 1 j
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Outline – AC Steady State Power
 Instantaneous Power Concept
• For The Special Case Of Steady State
Sinusoidal Signals
 Average Power Concept
• Power Absorbed Or Supplied During an
Integer Number of Complete Cycles
 Maximum Average Power Transfer
• When The Circuit is in
Sinusoidal Steady State
Engineering-43: Engineering Circuit Analysis
11
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Outline – AC SS Power cont.
 Power Factor
• A Measure Of The Angle Between the
CURRENT and VOLTAGE Phasors
 Power Factor Correction
• Improve Power Transfer To a Load By
“Aligning” the I & V Phasors
 Single Phase Three-Wire Circuits
• Typical HouseHold Power Distribution
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Instantaneous Power
 Consider in the Time
Domain a Voltage
Source Supplying
Current to an
Impedance Load
 Now Recall From Chp1
The Eqn for Power
p  vi
 Then at any Instant for
Time-Varying Sinusoidal
Signals
pt   vt   it 
 VM I M cos t  v  cos t   i 
 In the General Case
v(t )  VM cos t  v 
i (t )  I M cos t   i 
Engineering-43: Engineering Circuit Analysis
13
 Now Use Trig ID
cos x cos y 
1
cos( x  y)  cos( x  y)
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Instantaneous Power cont.
 Rewriting the Power Relation for Sinusoids
VM I M
cos( v  i )  cos(2 t   v  i )
p (t ) 
2
VM I M
VM I M
p (t ) 
cos( v   i ) 
cos( 2 t   v   i )
2
2
• The First Term is a
CONSTANT, or DC
value; i.e. There is No
Time Dependence
• The Second Term is a
Sinusoid of TWICE the
Frequency of the Driving
Source
 Examine the TWO Terms of the Power Equation
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Example
 For the Single Loop Ckt
 Use Phasors To find I
V 4V60
I 
 2 A30
Z 230
 Assume
 To Obtain the Time
Domain current Take
the Real Part of the
Phasor Current
v(t )  4V cos t  60
i(t )  Re230  Re2e30e jt 
or V  4V60
it   2 Amp  cos t  30
and Z  230
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Example cont
 Thus for This case
 In the Power Equation
p (t )  vt i t 
 4W cos60  30  cos2 t  60  30
 Or
4V cos t  60
1.732  j
 The Amplitudes and
Phase Angles
VM  4V
I M 2A
 v  60
 i  30
Engineering-43: Engineering Circuit Analysis
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p(t )  3.46W  4W cos2 t  90
 See Next Slide for Plots
• v(t)
• i(t)
• p(t) = v(t)•i(t)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Sinusoidal Power Example 9.1
8
p(t) Calculated by p(t)=v(t)•i(t)
Max-p = 7.46W
6
v(t) or i(t) or p(t)
4
Avg-p = 3.46W
2
0
NEGATIVE POWER – Inductive Load can
Release stored Energy to the Circuit
-2
v(t) (V)
-4
0.000
0.003
0.006
0.009
0.012
i(t) (V)
0.015
P(t) (V)
0.018
0.021
0.024
0.027
Time (S)
file =Sinusoid_Lead-Lag_Plot_0311.xls
p = 0 if either i or v are zero
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
0.030
Average Power
 For ANY Periodic Function, It’s Average Value Can be
calculated by Integrating Over at Least ONE COMPLETE
PERIOD, T, and Then Dividing the Integrated Value by
t T
the Period
1 0
X
x(t )dt
t0  an arbitary BaseLine time

T t0
 Then the Average POWER for Electrical Circuits With
Sinusoidal Excitation
1 T  t0
1 T  t0
P 
pt dt   vt   i t dt
t
T 0
T t0
1 T  t0
  VM I M cos t  v  cos t   i dt
T t0
 And Recall
T  2 
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Average Power cont
 Also For ANY Periodic Function, the Average value
May be Calculated over any INTEGER number of
periods. This is, in Fact, How most Electrical Power
Values are MEASURED. Mathematically:
1
P
nT

nT  t 0
t0
VM I M cos t  v  cos t   i dt
 Now Sub Into the Average Power Integral The
Simplified (n =1) Expression for Instantaneous Power
1 T t0 VM I M
cos( v   i )  cos( 2 t   v   i )dt
P 
t
T 0
2
1 VM I M T t0
1 VM I M T t0
P
cos( v   i )dt 
cos( 2 t   v   i )dt


t
t
0
0
T 2
T 2
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Average Power cont.2
 Examine the Two Terms from the Average Power Eqn
T t0
1 VM I M T t0
1 VM I M
P1 
cos( v   i )dt 
cos( v   i )  1dt

t
t0
0
T 2
T 2
1 VM I M
VM I M
T t0
 P1 
cos v   i t t0  P1 
cos( v   i )
T 2
2
 Thus the First Term is a CONSTANT that depends on
the Relative Phase Angle
 Also by Trig: cos(−) = cos()
• Thus for the First Term It Does NOT Matter if the Current
LEADS or LAGs the Voltage
cos v  i   cos v  i   cos  v  i   cosi   v 
Engineering-43: Engineering Circuit Analysis
20
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Average Power cont.3
 The Second Term from the Average Power Eqn
1 VM I M
P2 
T 2

T t 0
t0
cos( 2 t   v   i )dt  0
• As the The Integral of a sin or cos over an
INTEGER number of periods is ZERO
 Thus the Average Power is Described by the
FIRST TERM ONLY
VM I M
VM I M
P
cos( v   i )  P 
cos( i   v )
2
2
VM I M
P
cos  
2
Engineering-43: Engineering Circuit Analysis
21
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Resistive & Reactive Power
 For a Purely
RESISTIVE Circuit
There is NO Imaginary
Component of the
Impedance and thus no
Phase Shift Between
i & v. So for sinusoids
VM I M
Pres 
2
 Thus the Resistors
Absorb, or Dissipate,
Power (as Heat) only
Engineering-43: Engineering Circuit Analysis
22
 For A Purely REACTIVE
Circuit i&v are ±90° out
of phase
• So then Avg Power eqn
Preact
VM I M

cos 90  0!
2
 Thus Purely reactive
Impedances absorb NO
Power on Average
• They STORE Energy
over one Half-Cycle, and
then RELEASE it over
the NEXT
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Example  Avg Power
 For This Circuit
 The Power Parameters
VM  10V
 v  60
I M  3.54 A
 i  15
 The Power Calculation
VM I M
P
cos( v   i )
2
10V  3.54 A
P
cos60  15
2
P  17.7W cos 45  12.5W
 FIND: The
Dissipated Power
 Use Phasor Algebra to
Find the Current
I
10V60
1060

 3.5415( A)
2  j 2 2.828445
Engineering-43: Engineering Circuit Analysis
23
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Example  Avg Power cont
 Since Only the Resistor
Dissipates Power,
Check the Previous
Calc by using Pres=vres•i

VR

 Alternatively by OHM
VR  RI  23.54 A15  7.07V15
 For a Resistor the
Current & Voltage
WITHIN the Resistor
are IN-PHASE; thus
P
 Use Phasors for VR
2
VR 
10V60  7.07V15(V )
2  j2
Engineering-43: Engineering Circuit Analysis
24
VM , R I M
cos( v   i )
2
7.07V  3.54 A
P
cos15  15
2
P  25.00W 2  12.5W
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx

Example  Cap Circuit
 For This Circuit
I
 Find the Total
Impedance Across the
V-Source
4( j 4) 8  j8  j16

4  j4
4  j4
25.3  71.6

 4.4721  26.565
4 2  45
ZT  2 
I2
 Find The Dissipated
Power in Each Resistor
 Start by Finding The
Current In the Ckt
Branches that Contain
the Resistors
Engineering-43: Engineering Circuit Analysis
25
 Then The total Current I
I
V
12V60

ZT 4.47  26.6
 2.68 A86.6
• i(t) LEADS vS(t)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Example  Cap Circuit cont
I
P2  
I2
 So the 2Ω Resistor
Power Dissipation
PR

V I

M ,R
2
M
cos v   i 
1
RI M I M cos0
2
1
PR  RI M2
2
PR 
Engineering-43: Engineering Circuit Analysis
26
1 2 1
RI M   2  2.682  7.20W
2
2
 Now Find I2 by Current
Divider
 j4
4  90
I
 2.6886.6
4  j4
4 2  45
 1.9041.6
I2 
 Then the Power
Absorbed by the 4Ω R
1
P4   4 1.902 (W )
2
 P4  7.20W
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Maximum Avg Power Transfer
 Recall From the Study
of Resistive Ckts The
Criteria for Max Power
Xfer to a Load Resistor
 Consider This General
Thevenin Equivalent Ckt
RL ,max pwr  RTH
 Where RTH is the
Thevenin Equivalent
Resistance for the
Driving Ckt
 Now Try to Develop a
Similar Relationship for
Impedances
Engineering-43: Engineering Circuit Analysis
27
 For This Ckt the Avg Pwr
Delivered to the Load
1
PL  VL I L cos( vL   iL )
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Max Avg Power Transfer cont
 Now by Phasors
VOC
IL 
ZTH  Z L
VL  VOC
ZL
ZTH  Z L
 Where
V  Z  VM Z M V   z 
Z L  RL  jX L
 VZ  VM Z M
ZTH  RTH  jX TH
and VZ  V  Z
Engineering-43: Engineering Circuit Analysis
28
 Now By Euler Rln Recall
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Max Avg Power Transfer cont.2
 Then the Load Voltage
& Current Magnitudes
ZL
| VL |
VOC
Z L  ZTH
ZL
or V L
VOC
Z L  Z TH
 Similarly
 And
VL
IL 
ZL
 Also by Euler
ZL  RL  jX L  tan Z L X L RL
 Now a Useful Trig ID
I L  VL Z L  I L   I L  VL  Z L cos   1
But
VL  VL   I L  VL  Z L
 VL   I L  Z L
Engineering-43: Engineering Circuit Analysis
29
1  tan 2 
  VL   I L  tan   tan VL   I L
But




tan VL   I L  tan Z L  X L RL
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Max Avg Power Transfer cont.3
 ReArrange Trig ID to
Find
cos(VL   I L ) 
1
1   X L RL 
 cos(VL   I L ) 
2
RL
RL2  X L2
 Again the Power Eqn
1
PL  VL I L cos( vL   iL )
2
 Substitute to Find
1  ZL
PL  
2  Z L  ZTH
 ZL

V
  Z L  ZTH OC 
RL

VOC  
ZL
 RL2  X L2



Engineering-43: Engineering Circuit Analysis
30
2
Z L VOC
2 Z L  ZTH 2
 Or P  1
L
RL
RL2  X L2
 And
Z L  ZTH  ( RL  RTH )  j ( X L  X TH )
Z L  ZTH 2  ( RL  RTH ) 2  ( X L  X TH )2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Max Avg Power Transfer cont.4
 Finally The Power Eqn
Restated
2
VOC
RL
1
PL 
2 ( RL  RTH ) 2  ( X L  X TH ) 2
 Now To Maximize the
Power Transfer, Set the
Partial Derivatives to 0
PL

 0
X L
  X L   X TH


PL
R

R
L
TH


0

RL
Engineering-43: Engineering Circuit Analysis
31
 at LAST The Optimized
opt
*
Load
Z L  ZTH
• The Complex Conjugate
 And the Power
Transferred at Optimum
2

1  VOC
max

PL  
2  4 RTH 
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Max Avg Power Transfer cont.5
 Check
2
VOC
RL
1
PL 
2 ( RL  RTH ) 2  ( X L  X TH ) 2
PL

 0
X L
  X L   X TH


PL
 0   RL  RTH

RL
 At the Max Condition
2
V
1
max
OC RTH
PL 
2 ( RTH  RTH ) 2  ( X TH  X TH ) 2
PLmax
2
2
2
RTH
RTH
VOC
1 VOC
1 VOC



2
2
2
2 (2 RTH )  0
2 4 RTH
8 RTH
PLmax
2
VOC

8 RTH
Engineering-43: Engineering Circuit Analysis
32

Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Example  Maximum Power Xfer
 Find ZL for the
Maximum Power Xfer
 And The Max Pwr Xfer
2
PLmax  VOC
8RTH
 Need to find
• VOC = VTH
• RTH
 Recall The Max Power
Criteria
opt
*
Z L  ZTH
Engineering-43: Engineering Circuit Analysis
33
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Example  Max Pwr Xfer cont.
 Remove Load and Find
VOC by Loop Current
 And by Ohm’s Law in
the Frequency Domain
VOC  j 2I  120
  j 2  9(1  j )  12
 j18  6  18.974V71.56

VOC
I
 Find ZTH by Source
Deactivation (Zeroing)

 Using KVL on the Loop
 240  120  j 2I  2I  0
36(2  j 2)
 I
 9(1  j )  12.73  45
8
Engineering-43: Engineering Circuit Analysis
34
ZTH
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Example  Max Pwr Xfer cont.2
 Then ZTH
4j
ZTH   j 2  (2 || j 2)   j 2 

2  j2
 or ZTH  4  8  j8  1  j ()
2  j2
8
ZTH
 Taking The Conjugate
Z
opt
L
 1  j ()
 Then The Power
Transferred to this Load
Engineering-43: Engineering Circuit Analysis
35
max
L
P
V
2
OC
8RTH

18.974V 

2
8 1
 45W
360

W
8
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Power Factor
 Consider a Complex
Current Thru a Complex
Impedance Load
I M  i

Z L V 
M
v

 The Current and
Load-Voltage Phasors
(Vectors) Can Be
Plotted on the
Complex Plane
Engineering-43: Engineering Circuit Analysis
36
V
v
z
I
i
 By Ohm & Euler
V  ZI  V  Z  I
v   z  i
or  z   v   i
 in the Electrical Power
Industry θZ is the Power
Factor Angle, or Simply
the Phase Angle
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Power Factor cont
 The Phase Angle Can
Be Positive or Negative
Depending on the
Nature of the Load
(capacitiv e)
V  VM 0
(inductive )
V is the
BaseLine
I  V Z
 0 Z  Z
Engineering-43: Engineering Circuit Analysis
37
• Large Electric Motors are
Essentially Inductors
 Now Recall The
General Power Eqn
 90   z  0
current leads
0   z  90
current lags
 Typical Industrial Case
is the INDUCTIVE Load
1
V I
P  VM I M cos( v   i )  M M cos  Z
2
2 2
P  Vrms I rms cos( v   i )  Vrms I rms cos  Z
 Measuring the Load
with an AC DMM yields
• Vrms
• Irms
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Power Factor cont.2
 The Product of the
DMM Measurements is
the APPARENT Power
Papparent  Vrms I rms
 The Apparent Power is
NOT the Actual Power,
and is thus NOT stated
in Watts.
• Apparent Power
Units = VA or kVA
Engineering-43: Engineering Circuit Analysis
38
 Now Define the Power
Factor for the Load
Pactual
pf 
 cos( v   i )  cos  z
Papparent
and
Pactual  V rmsI rms pf
 Some Load Types
pf
z
0
 90
pure capacitive
0  pf  1  90   z  0 leading or capacitive
1
0
resistive
0  pf  1 0   z  90 lagging or inductive
0
90
pure inductive
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
pf – Why do We Care?
 Consider this case
• Vrms = 460 V
• Irms = 200A
• pf = 1.5%
 Then
• Papparent = 92kVA
• Pactual =1.4 kW
  This Load
requires The Same
Power as a Hair
Dryer or Toaster
Engineering-43: Engineering Circuit Analysis
39
 However, Despite
the low power
levels, The WIRES
and CIRCUIT
BREAKERS that
feed this small Load
must be Sized for
200A!
• The Wires would be
nearly an INCH in
Diameter
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Example  Power Factor
 The Local Power
Company Services this
Large Industrial Load
0.1
510V0
Power company
 100 kW
I lags V
 Find Irms by Pwr Factor
P  Vrms  I rms  pf
I rms  P  pf  Vrms 
I rms  100kW 0.707  480Vrms 
Engineering-43: Engineering Circuit Analysis
40
 Then the I2R Loses in
the 100 mΩ line
2
P
Rline
1
2
Plosses  I rms Rline 
 2
2
Vrms
pf
1010  0.1
1
Plosses ( pf  0.707) 

(W )
2
2
480
0.707
 4.34kW  2
 Improving the pf to 94%
1010  0.1
1
Plosses ( pf  0.94) 

(W )
4802
0.942
 4.34kW 1.13
Psaved  0.87  4.34kW  3.77kW
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Example - Power Factor cont
 For This Ckt The
Effect of the Power
Factor on Line
Losses
0.1
510V0
 100 kW
Engineering-43: Engineering Circuit Analysis
41
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Complex Power
 Consider a general Ckt
with an Impedance Ld
 Mathematically
S  Vrms v   I rms i 
*
S  Vrms v   I rms   i 
S  Vrms I rms v   i 
recall :  v   i    Z
 For this Situation
Define the Complex
Power for the Load:
SV I
*
rms rms
Engineering-43: Engineering Circuit Analysis
42
 Converting to
Rectangular Notation
S  Vrms I rms cos( v   i )  j Vrms I rms sin( v   i )
P
Active Power
Q
Reactive Power
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Complex Power cont
 Thus S in Shorthand
S  P  jQ
 Alternatively,
Reconsider the General
Sinusoidal Circuit
 S & Q are NOT Actual
Power, and Thus all
Terms are given
Non-Watt Units
• S→ Volt-Amps (VA)
• Q → Volt-Amps, Reactive
(VAR)
 P is Actual Power and
hence has Units of W
Engineering-43: Engineering Circuit Analysis
43
 First: U vs. Urms
U  U M 
and
U rms  U M
 U rms  U rms
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
2
Complex Power cont.2
 Now in the General Ckt
By Ohm’s Law
Z  Vrms I rms
Vrms v Vrms
Z

 v   i 
I rms i I rms
and Vrms I rms  Z  Z
so
ReZ j ImZ 
Z cos v   i   jZ sin  v   i 
 In the Last Expression
Equate the REAL and
Imaginary Parts
Engineering-43: Engineering Circuit Analysis
44
ReZ
cos v   i  
Z
ImZ
sin  v   i  
Z
 And Again by Ohm
I rms  Vrms Z
I rms 
Vrms v
V
 rms  i
Z v   i 
Z
So I rms  Vrms Z
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Complex Power cont.3
 And by Complex Power
Definition
S  P  jQ then
P  ReS  Vrms I rms cos v   i 
Q  ImS  Vrms I rms sin  v   i 
 Using the Previous
Results for P
 ReZ
P  ReS  Vrms I rms 
 Z 
Vrms

I rms ReZ
Z
2
2
P  I rms
ReZ  I rms
R
Engineering-43: Engineering Circuit Analysis
45
 Similarly for Q
QI
2
rms
2


Im Z  I rms X
 So Finally the
Alternative Expression
for S
SI
2
rms
Z
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Complex Power Triangle
 The Expressions for S
S  P  jQ
SI
2
rms
Z
 Plotting S in the
Complex Plane
 From The Complex
Power “Triangle”
Observe
Q
tan  v   i  
P
 Note also That Complex
Power is CONSERVED
2
Stot   Sk   I rms
,k Z k
Engineering-43: Engineering Circuit Analysis
46
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Example - Complex Power
 For the Circuit At Right
•
•
•
•
•
Zline =0.09 Ω + j0.3 Ω
Pload = 20 kW
Vload = 2200°
pf = 80%, lagging
f = 60 Hz → ω = 377s-1
 Lagging pf → Inductive
inductive
 From the Actual Power
P  ReS | S | cos(v  i )  S  pf
 Thus
P 20kW
 SL 

 25kVA
pf
0.8
 And Q from Pwr Triangle
Q 2  S L  P 2  Q  15 kVAR
2
capacitive
Engineering-43: Engineering Circuit Analysis
47
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Example - Complex Power cont
 Then SL
S L  20  j15kVA  25kVA36.87
 Recall the S
Mathematical Definition
S L  VL I*L
 Note also that [U*]* = U
 In the S Definition,
Isolating the Load
Current and then
Conjugating Both Sides
Engineering-43: Engineering Circuit Analysis
48
*
 S   25kVA36.87 
 IL   L   

 VL   220V0 
I L  113.64  36.86( A)
 Alternatively
 20,000  j15,000 
IL  

220


I L  90.91  j 68.18( A)
*
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
*
Example - Complex Pwr cont.2
 Now Determine VS
4.86
VS  Vline  VL
VS  (0.09  j 0.3)I L  2200
VS  (0.09  j 0.3)(90.91  j 68.18)  220(V )
 Then VS
VS  28.63  j 21.14  220
VS  248.63  j 21.14
VS  249.53Vrms4.86
 To find the Src Power
Factor, Draw the I & V
Phasor Diagram
Engineering-43: Engineering Circuit Analysis
49
IL
VS
 36.86
 Then The Phase Angle
 v   i  4.86   36.86  41.72
 I Lags V  Inductive Load
and also
pf  cos v   i   cos 41.72
 pf  0.7464
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Example - Complex Power kVAR
0.1
 For the Circuit At
Right, Determine
j 0.25
40kW
pf  0.84 lagging
• Real & Reactive Power
losses in the Line
• Real & Reactive Power
at the Source
 From the Actual Power
 Lagging pf → Inductive P  ReS | S | cos(v  i )  S  pf
inductive
 Thus
P 40kW
 SL 

 47.62kVA
pf
0.84
 And by S Definition
capacitive
Engineering-43: Engineering Circuit Analysis
50
S  VI *  I L 
SL
 216.45( A) rms
VL
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Example - Complex kVAR cont.
0.1
 Also from the S
Relation & Pythagorus
j 0.25
40kW
pf  0.84 lagging
| Q L | | S L |2  P 2  25,839(VAR)
 Now the Power
Factor Angle
 Then for Line Loses
Sline  VlineI*line  (ZlineI L )I*L
I Lagging V
 pf = cos(θv − θi); hence
v  i  acos0.84  32.86
Engineering-43: Engineering Circuit Analysis
51
 ZlineI L2
 Quantitatively
Sline  (0.1  j 0.25)( 216.45) 2
 Sline  4685  j11713 VA
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Example - Complex kVAR cont.2
0.1
 Find Power Supplied
by Conservation of
Complex Power
j 0.25
40kW
pf  0.84 lagging
S Supplied  S line  S Load
 In this Case
 Then to Summarize the
S Sup  4.685  j11.713  40  j 25.839
Answer
 4.685  40  j 11.713  25.839
 44.685  j 37.552 kVA
 58.3740.04 kVA
Engineering-43: Engineering Circuit Analysis
52
•
•
•
•
Pline = 4.685 kW
Qline = 11.713 kVAR
PS = 44.685 kW
QS = 37.552 kVAR
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Power Factor Correction
 As Noted Earlier, Most
Industrial Electrical
Power Loads are
Inductive
• The Inductive
Component is Typically
Associated with Motors
 To Arrive at the Power
Factor Correction
Strategy Consider A
Schematic of a typical
Industrial Load
 The Motor-Related
Lagging Power Factor
Can Result in Large
Line Losses
 The Line-Losses can
Be Reduced by Power
Factor Correction
Engineering-43: Engineering Circuit Analysis
53
Intentionally Added
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Power Factor Correction cont.
 Prior to The Addition
of the Capacitor
S old  Pold  jQold | Sold |  old
pf old  cos( old )
 For The Capacitive
Load
V
I C  L  VL  jC  VL  L  C90
ZC
 I C  CVL
2



CV
L
QC  I C2 ImZ C  
 CVL2
 C
 After Addition of the
Capacitor
Engineering-43: Engineering Circuit Analysis
54
S new  S old  S C
 Pold  jQold  jQC
| S new |  new
pf new  cos( new )
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Power Factor Correction cont.2
 Find θnew
• Cap is a Purely
REACTIVE Load
tan  new
Qold  QC Qnew


Pold
Pold
 The Vector Plot Below
Shows Power Factor
Correction Strategy
 Use Trig ID to find QC to
give desired θnew
cos  
QL
Qnew
L-QC
P
Engineering-43: Engineering Circuit Analysis
55
QC
1
1  tan 
2

Qold  QC
1

1
2
Pold
cos  new
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Trig ID Digression
 Start with the ID
cos  
1
1  tan 2 
 Solve for tan
1
tan  
1
2
cos 
2
 Recall tanθnew
Qold  QC
tan  new 
Pold
 Substituting
Engineering-43: Engineering Circuit Analysis
56
Qold  QC
1

1
2
Pold
cos  new
 Or
Qold  QC
cos 2  new
1


2
Pold
cos  new cos 2  new
1  cos 2  new
1  cos 2  new


2
cos  new
cos  new
 But: cosθnew = pfnew
tan  new
1  pf

pf new
2
new
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Example  pf Correction
 Kayak Centrifugal
Injection-Molding
Power Analysis
Roto-molding
process
• Improve Power Factor
to 95%
 Find Sold
P  ReS  S cos( v  i )  S  pf
 Sold
pf  0.8 lagging
 Adding A Cap Does
NOT Change P
• Use Trig ID to Find Tan(new)
P 50kW


 62.5kVA
2
1  pf new
Qnew
pf
0.80
cos  new  0.95  tan  new 

 0.329
P
 Now Qold
pf new
 And by S Relation
| Qold | | Sold |  P  37.5(kVAR)
2
2
Engineering-43: Engineering Circuit Analysis
57
50kW ,VL  2200rms

Qnew
 Qnew  0.329  P  16.43kVAR
P
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Example  pf Correction cont
 Then the Needed QC
 QC  Qold  Qnew  37.5  16.43
QC  21.07kVA
 Recall The Expression
for QC
Roto-molding
process
50kW ,VL  2200rms
pf  0.8 lagging
QC | VL || I C | VL2C
 Then C from QC
QC
21.07 103
C

2
VL (2  60)  (220) 2
C  0.001155( F )  1155 F
Engineering-43: Engineering Circuit Analysis
58
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
All Done for Today
Power
Factor
Correction
Engineering-43: Engineering Circuit Analysis
59
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
WhiteBoard
Work
 Let’s Work Problem similar
to P5.78
• Determine at the input
SOURCE
– Voltage & Current
– Complex Power
– Δθ & pf
60 kVA
Engineering-43: Engineering Circuit Analysis
60
220Vrms 19
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Engineering-43: Engineering Circuit Analysis
64
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Engineering-43: Engineering Circuit Analysis
65
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
WhiteBoard Work
 Let’s Work This Nice
Problem
Ix
j1 
2
-j2 
ZL
120 V
2Ix
Figure P 9.32
 Find ZL for Pav,max, &
the value of Pav,max
Engineering-43: Engineering Circuit Analysis
74
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Example
V1
 Use Nodal Analysis
to Find VO
 The KCL at Node-1
V1  1230 V1 V1 V1  VO
 

0
2
1 j2
j
 Now KCL At VO
VO  V1 VO

0
j
1
 V1  1  j VO
 Multiply Node-1 KCL
by 2j to Obtain
Engineering-43: Engineering Circuit Analysis
75
j (V1  1230)  j 2V1  V1  2(V1  VO )  0
 Subbing for V1 Yields
2VO  (1  2  2 j  j )(1  j )VO  12 j30
 Factoring out VO Yields
(2  (1  3 j )(1  j ))VO  1901230
 Isolating VO
12120
12120
VO 

44j
5.65745
VO  2.121V75
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Example Alternative
 This Time Use
Thevenin to Find VO
 Take Cap & Res in
Series as the Load
 Deactivate V-Src
to Find ZTH
ZTH
ZTH
ZTH

4 j 3
 2 || 1 || j 2 
2 3  j 2
j4
j 42  j 6


2  j6
22  62
 0.6  j 0.2
Engineering-43: Engineering Circuit Analysis
76
 Find VOC by V-Divider
1 || j 2
1230
2  (1 || j 2)
j2

1230
2(1  2 j )  2 j
VOC 
VOC
24120 12120

26j
1 3 j
 3.795V48.435
VOC 
VOC
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
Example Alternative cont.
 Now Apply the
Thevenin Equivalent
Circuit to the Load
 Calculate VO By
V-Divider
VO 
Z TH
 j1
ZTH
0 .6  j 0 . 2
VOC
+
-

1

1
VOC
1 j
1
 Same as Before
VOC
0.6  j 0.2  1  j
1
VO 
VOC
1.6  j 0.8
VO  0.55926.565  3.795V48.435
VO 

VO  2.121V75.00
Engineering-43: Engineering Circuit Analysis
77
V0
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
P5.57 Graphics
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx
P5.81 Graphics
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx