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Lecture 16
Phasor Circuits, AC Power,
Thevenin
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Kirchhoff’s Laws in Phasor Form
We can apply KVL directly to phasors.
The sum of the phasor voltages equals
zero for any closed path.
V1  V2  V3  0
The sum of the phasor currents entering a
node must equal the sum of the phasor
currents leaving.
I in  I out
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Circuit Analysis Using Phasors
and Impedances
1. Replace the time descriptions of the voltage
and current sources with the corresponding
phasors. (All of the sources must have the
same frequency.)
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
2. Replace inductances by their complex
impedances ZL = jωL. Replace capacitances by
their complex impedances ZC = 1/(jωC).
Resistances have impedances equal to their
resistances.
3. Analyze the circuit using any of the techniques
studied earlier in Chapter 2, performing the
calculations with complex arithmetic.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.3
Find i(t):
VS  100 cos(500 t  30  )  VS  100 30 
Z L  jL  j (500 )( 0.3)  j150 
ZC   j
1
1
j
  j 50 
6
C
(500 )( 40 )(10 )
Z eq  R  Z L  Z C  100  j150  j 50  100  j100
 141 .445 
VS
100 30 


I


0
.
707


15

i
(
t
)

0
.
707
cos(
500
t

15
)

Z
141 .445
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.3
Find the phasor
voltage across each
element:
VS  100 30 
I S  0.707   15 
VR  RI  (100 )( 0.707   15  )
VL  ( jwL )I  (L90  )I  (L90  )( 0.707   15  )  106 .175 
1   1


VC    j
  90  I  (50   90  )( 0.707   15  )  35 .4  105 
I  
C   C


ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.4
Find the voltage vc(t) in steady state:
v s (t )  10 sin( 1000t )  10 cos(1000t  90)  Vs  10  90
Z L  jwL  j (1000)(0.1)  j100
ZC   j
Z RC 

1
1
j
  j100
6
C
(1000)(10 x10 )
1
1
1

R ZC

1
1
1

100  j100

1
0.01  j 0.01
10
 70.71  45  50  j 50
0.0141445
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.4
 Z RC 
 70.71  45  
 70.71  45 

VS  
(10  90 )  
VC  
Z

Z
(
50

j
50
)

j
100


 50  j 50
L 
 RC
 70.71  45  
(10  90  )  (1  90  )(10  90  )  10  180 
 
 
 70.7145 

(10  90  )

vC (t )  10 cos(1000t  180  )  10 cos(1000t )
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.4
Find the phasor current through each element:
VS
10  90 
10  90  10  90 
I



 0.1414  135

Z L  Z RC
j100  (50  j 50) 50  j 50 70.7145
VC 10  180 
IR 

 0.1  180 
R
100
VC 10  180  10  180 
IC 


 0.1  90 
ZC
 j100
100  90
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.5
Use the node voltage technique to find v1(t):
KCL at node 1 :
V1 V1  V2

 2  90   (0.1  j 0.2)V1  j 0.2V2   j 2
10
 j5
KCL at node 2 :
V2 V2  V1

 1.50  
 j 0.2V1  j 0.1V2  1.5
j10
 j5
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.5
(0.1  j 0.2)V1  j 0.2V2   j 2
 j 0.2V1  j 0.1V2  1.5
(0.1  j 0.2)V1  j 0.2V2   j 2
 j 0.4V1  j 0.2V2  3
Adding :
32 j
(0.1  0.2 j )V1  3  2 j  V1 
 14  8 j  16.1229.74 
0.1  0.2 j
v1 (t )  16.1cos(100t  29.7  )
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 5.9
Find i(t):
vS (t )  10 sin(500 t )  10 cos(500 t  90  )  VS  10   90
Z R  250 
Z L  jL  j (500 )( 0.5)  250 j
KVL :
VS
VS
10   90 
3

IZ R  IZ L  VS  I 



28
.
4
x
10


135
Z R  Z L 250  j 250 353 .645 
i (t )  28 .4 x10 3 cos(500 t  135  )
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 5.9
Construct the a phasor diagram showing all three voltages and
the current:
VR  IR  (28 .3 x10 3 )( 250 )  135   7.07   135 

VL  jLI  (500 )( 0.5)90  (28 .3 x10 3 )( 250 )  135 

 7.07   45 
VS  10   90
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 5.10
Find the phasor voltage and phasor current through
each element:
1
1
1
1



Z eff
ZC Z L Z R
ZC   j
1
1
j
 50 j
6
C
(200 )(100 x10 )
Z L  jL  j (200 )(1)  j 200
1
1
1
1
j
j
1
2  j3







Z eff
 50 j j 200 100 50 200 100
200
Z eff 
200
 30 .77  j 46 .15  55 .47   56 .31
2  j3
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 5.10
Find the phasor voltage and current through each element:
VC  VL  VR  IZ eff  (50  )(55.47  56.31 )  277.3  56.31
VC 277.3  56.31

IC 


5
.
55

33
.
69
ZC
50  90 
VL 277.3  56.31

IL 


1
.
39


146
.
3
ZL
20090 
VR 277.3  56.31

IR 


2
.
77


56
.
31
ZR
1000 
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 5.11
100j
-200j
100j
Solve for the mesh currents:
I1Z L1  ( I1  I 2 ) Z R  VS
I 2 Z C  I 2 Z L2  ( I 2  I1 ) Z R  0
(100  100 j ) I1  100 I 2  VS
 100 I1  (100  j100 ) I 2  0
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 5.11
(100  100 j ) I 1  100 I 2  VS
 100 I 1  (100  j100) I 2  0
(14145) I 1  (1000) I 2  1000
(1000) I 1  (141  45) I 2  0
1000  1000
0
141  45
(1000)(141  45)  (0)( 1000)
I1 

14145  1000
(14145)(141  45)  (1000)( 1000)
 1000 141  45
14100  45
9.97 x10 3  j 9.97 x10 3


 1.009  j1.009  2  45
2
2
9881
(141) 0  (100) 0
i1 (t )  1.41 cos(1000t  45  )
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Power in AC Circuits
V Vm 0
I 
 I m  
Z
Z 

Vm
where I m 
Z
For >0 the load is called “inductive” since Z=jL for an inductor
For <0 the load is “capacitive” since Z=-j/C for a capacitor
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Power in AC Circuits
Current, Voltage and Power for a Resistive Load
Z  R0
v(t )  Vm cos(t )
i (t )  I m cos(t )
p(t )  Vm I m cos 2 (t )
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Power in AC Circuits
Current, Voltage and Power for an Inductive Load
Z  L90 
v(t )  Vm cos(t )  V  Vm 0 I 
Vm 0
L90 
i (t )  I m cos(t  90  )  I m sin(t )

Vm
  90   I m   90 
L
“Reactive” power
Vm I m
sin(2t )
2
using cos(x)sin(x) = (1/2)sin(2x)
p(t )  Vm I m cos(t ) sin(t ) 
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Power in AC Circuits
• For a pure inductance half of the
time the power is positive and
power flows to the inductor
where it is stored as energy in the
magnetic field
• Half of the time the power is
negative and power flows from
the inductor to the source
• Even though the average power is
zero for a pure inductor, current
flows between the source and the
inductor and power is dissipated
in the lines connecting the source
to the inductor
Reactive Power
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Power in AC Circuits
Current, Voltage and Power for a Capacitive Load
Z
1
  90 
C

V

0
m
v(t )  Vm cos(t ) V  Vm 0   I 
 I m 90 
1
  90 
Reactive power for
C
a capacitor
i (t )  I m cos(t  90  )   I m sin(t )
opposite the sign
V I
for an inductor
p(t )  Vm I m cos(t ) sin(t )   m m sin(2t )
2
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Power in AC Circuits
If a load contains
both inductance
and capacitance
with reactive
powers of equal
magnitude, the
reactive powers
cancel.
Capacitor
Inductor
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Power for a General Load
For a general RLC circuit in which the phase can be any
value between -90 to +90
v(t )  Vm cos(t )
i (t )  I m cos(t   i )
p (t )  Vm I m cos(t ) cos(t   i )
Vm I m
Vm I m

cos( i )1  cos(2t ) 
sin( i ) sin(2t )
2
2
 Vm  I m 
Vm I m

 cos( i )
Pav 
cos( i )  
2
 2  2 
 Vrms I rms cos( i ) PF  cos( i )
PF is the “Power Factor”
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Power for a General Load
If the phase angle for the voltage is not zero, we
define the power angle :
Power angle:
   voltage   current
P  Vrms I rms cos( )
PF  cos( )
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
AC Power Calculations
Average Power: P  Vrms I rms cos  W 
Reactive Power: Q  Vrms I rms sin   VAR
Apparent Power: VRMS I RMS
VA
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Power Triangles
Average power
P  Q  Vrms I rms 
2
Average
power
2
2
Reactive
power
Apparent
power
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Load Impedance in the Complex Plane
Z  Z   R  jX
R
cos( ) 
Z
X
sin(  ) 
Z
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Additional Power Relations
PI
2
rms
QI
2
rms
R
X
P
Q
V
2
Rrms
R
V
2
Xrms
X
Average
power
Reactive
power
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
AC Power Calculations
Compute the power and reactive power from the source
in this circuit:
   v   i  90   (135  )  45 
Vsrms 
Vs
I srms 
I

2
2
10
 7.071V
2

0.1414
 0.1A
2
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
AC Power Calculations
P  Vsrms I rms cos( )
 7.071 x0.1cos(45  )  0.5W
Q  Vsrms I rms sin( )
 7.071 x0.1sin(45  )  0.5VAR
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
AC Power Calculations
The reactive power delivered to the inductor and capacitor:
2
QL  I rms
X L  (0.1) 2 (100 )  1.0VAR
2
 0.1 
 (100 )  0.5VAR
QC 
 
 2
Q  QL  QC  1.0VAR  0.5VAR  0.5VAR
I C2rms X C
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
AC Power Calculations
The power delivered to the resistor:
2
PR 
I R2rms R
2
 IR 
 0.1 


 100  0.5W  P

R  
 2
 2


ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Using Power Triangles
Find the power, reactive power and power factor for the
source. Also find the phasor current I:
Load A: 10kVA apparent power, PF=0.5 leading (capacitive)
Load B: 5kW power, PF=0.7 lagging (inductive)
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Using Power Triangles
Load A: 10kVA apparent power, PF=0.5
leading (capacitive)
• Reactive power QA and the power
angle A are negative
Load B: 5kW power, PF=0.7 lagging
(inductive)
• Reactive power QB and the power
angle B are positive
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Using Power Triangles
PA  Vrms I Arms cos( A )  10 4 (0.5)  5kW
QA 
Vrms I A 2  PA2  10 4 2  5000 2
rms
 8.660 kVAR
 B  arccos(0.7)  45 .57 
QB  PB tan( B )  5000 tan(45 .57  )
 5.101 kVAR
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Using Power Triangles
P  PA  PB  5kW  5kW  10 kW
Q  Q A  QB  8.660 kVAR  5.101 kVAR  3.559 kVAR
Q
  3.559 

  arctan   arctan
  19 .59
P
 10 
Vrms I rms  P  Q 
2
2
10    3.559 x10 
3 2
3 2
 10 .61kVA
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Using Power Triangles
Vrms
I rms
V
1414


 1kV
2 1.414
Vrms I rms 10 .61kVA


 10 .61 A
Vrms
1kV
I  2 I rms  (1.414 )(10 .61 A)  15 A
 i   v    30  (19 .59 )  49 .59



I  I  i  15 49 .59 
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Using Power Triangles
V  1414 30

I  1549 .59

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 5.12
Voltage source: V  707.140 
Delivers 5 kW to a load with a power factor of 100 percent.
Find the reactive power and the phasor current:
PF  100 %  cos( )  1   0
Reactive Power Q  Vrms I rns sin( )  0
I rmsVrms  5kW
Vrms
707 .1V

 500V
1.41
I rms
5 x10 3W

 10 A
500V
I  2 I rms  14 .14 A
I  14 .14 40 
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 5.12
If the PF=20% lagging, cos() = 0.20   = arcos(0.20) = 78.46
VrmsIrms
Q = 24.49kVAR

P = 5kW
Q
 tan( )  Q  P tan( )  (5kW ) tan(78 .46  )  24 .49 kVAR
P
P
I rms 
 50 A
Vrms cos( )
I  2 I rms  (1.41)(50 A)  70 .71 A
 i   v    40   78 .46   38 .46 
I  70 .71  38 .46 
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Thevenin Equivalent Circuits
The Thevenin equivalent for an ac circuit consists of a
phasor voltage source Vt in series with a complex
impedance Zt
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Thevenin Equivalent Circuits
The Thévenin voltage is equal to the open-circuit
phasor voltage of the original circuit.
Vt  Voc
We can find the Thévenin impedance by zeroing the
independent sources and determining the impedance
looking into the circuit terminals.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Thevenin Equivalent Circuits
The Thévenin impedance equals the open-circuit
voltage divided by the short-circuit current.
Voc Vt
Z t

I sc I sc
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Norton Equivalent Circuit
The Norton equivalent for an ac circuit consists of a
phasor current source In in parallel with a complex
impedance Zt
I n  I sc
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.9
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.9
Circuit with the voltage and current source zeroed to find
the Thevenin impedance
1
1
10 

Zt 



70
.
71


45
 50  j50

1 / 100  1 /(  j100) 0.01  j 0.01 0.0141445
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.9
Circuit with the output shorted to find the short circuit
current.
I SC  I R  I S
VS 100 0 
IR 

 10 
100
100
I S  190 
I SC  10   190   1  j  1.414   45 
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Example 5.9
Vt  I SC Z t  (1.414  45 )(70.71  45 )  100  90 
I n  I SC  1.414  45
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Maximum Average Power Transfer
The Thevenin equivalent of a two-terminal circuit
delivering power to a load impedance.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Maximum Average Power Transfer
If the load can take on any complex value,
maximum power transfer is attained for a load
impedance equal to the complex conjugate of the
Thévenin impedance.
If the load is required to be a pure resistance,
maximum power transfer is attained for a load
resistance equal to the magnitude of the
Thévenin impedance.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Maximum Average Power Transfer
Find the load for the maximum
power transfer if the load can
have any complex value and if
the load must be a pure
resistance:
Complex load :
Z t  50  j 50
Z t*  50  j 50
Real load
Z t  50  j 50
Z t  50 2  50 2  70 .71
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 5.14
Zero the source
Zeff
Find the Thevenin impedance, the Thevenin voltage and the
Norton current.
1
1
1 j
1
1 j

 


Z eff
jL R 100 100 100
Z eff 
100  100  1 

 
1  j  1  j  1 
j  100  j100
 
 50  j50
j
2
Z t  Z eff  50  j 25  50  j 50  50  j 25  100  j 25
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 5.14
VOC
100
1000 
100 2 0 
1000 


VS 
V



70
.
71


45
S
100  j100
100 245
100 245
245
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Exercise 5.14
I N  I SC
VT 70.71  45 70.71  45





0
.
686


59
.
04
ZT
100  j 25
103.114.04 
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.