AC Circuit analysis using complex
notation
Solutions Using Transforms
Problem
Transform
Solution
Real, or time
domain
Complicated and difficult
solution process
Inverse
Transform
Transformed
Transformed
Problem
Problem
Relatively simple
solution process, but
using complex numbers
Transformed
Transformed
Solution
Solution
Dr B Rajkumarsingh
Complex or
transform domain
Recap on complex numbers
• Rectangular form: a+jb
a  jb  r(cos  j sin  )
r
• Polar form:
r  a b
2
2
  tan (b / a)
1
Dr B Rajkumarsingh
1. Convert the following from rectangular to polar
form:
a. 1+j15 b. 60+j5 c. 0.01+j0.3 d. 100-j2000
e. -5.6+j86 f. -2.7-j38.6
2. Convert the following from polar to rectangular
form:
a.135
0
b.16087
0
c.76  4
0
d.7.52  125
0
Dr B Rajkumarsingh
1. Perform the following operations. Express your
answer in rectangular and polar form:
a. (1+j15)x( 60+j5)
b. (0.01+j0.3)-(1-j2)
c. (-5.6+j86)/ (-2.7-j38.6)
0
0
d.(135 ).(10  15 )
e.(16087 )  (1305 )
0
0
f.(76  40 )  (13250 )
g.(7.52  1250 ) /(1350 )
Dr B Rajkumarsingh
Alternating Currents
An alternating current such as that produced by a generator has no
direction in the sense that direct current has. The magnitudes vary
sinusoidally with time as given by:
AC-voltage and
current
Emax
imax
E = Emax sin q
i = imax sin q
time, t
Rotating Vector Description
The coordinate of the emf at any instant is the value of Emax sin q.
Observe for incremental angles in steps of 450. Same is true for i.
E
E = Emax sin q
q
1800
450 900 1350
Radius
R = E=maxEmax
2700
3600
Effective AC Current
The average current in a cycle is
zero—half + and half -.
imax
I = imax
But energy is expended, regardless of
direction. So the ―root-mean-square‖
value is useful.
I rms 
The rms value Irms is sometimes
called the effective current Ieff:
I2
I

2 0.707
The effective ac current:
ieff = 0.707 imax
AC Definitions
One effective ampere is that ac current for which the power is the
same as for one ampere of dc current.
Effective current: ieff = 0.707 imax
One effective volt is that ac voltage that gives an effective ampere
through a resistance of one ohm.
Effective voltage: Veff = 0.707 Vmax
Example 1: For a particular device, the house ac voltage
is 120-V and the ac current is 10 A. What are their
maximum values?
ieff = 0.707 imax
imax
ieff
10 A


0.707 0.707
imax = 14.14 A
Veff = 0.707 Vmax
Vmax
Veff
120V


0.707 0.707
Vmax = 170 V
The ac voltage actually varies from +170 V to
from 14.1 A to –14.1 A.
-170 V and the current
Sinusoidal Sources
Amplitude
Period = 1/f
Phase angle
Angular or radian
frequency = 2pf = 2p/T
Sinusoidal voltage source
vs  Vm sin(t  ).
Sinusoidal current source
is  Im Drsin(t
B Rajkumarsingh ).
Example
v
i
+
v
_
circuit
element
i
Voltage and current of a circuit element.
The current leads the voltage by
 radians
OR

The voltage lags the current
by
radians
Dr B Rajkumarsingh
The Transformation
A phasor is a complex number that represents the
magnitude and phase of a sinusoid.
y (t )  Ym cos(t   )  Re Yme
j (t  )
 Y 2 cos(t   )
Time domain

Transformation
Frequency domain
Y  Y 
For practicability purposes the effective value of y is used as the
magnitude of the phasor
Dr B Rajkumarsingh

The Transformation (cont.)
Time domain
i(t )  5 2 sin(100t  30)

Frequency domain
Transformation
I  530
Dr B Rajkumarsingh
Phasor Relationship for R, L, and C Elements
Time domain
v  Ri
Resistor
Frequency domain
V
V  RI or I 
R
Voltage and current are in phase
Dr B Rajkumarsingh
Example (cont.)
v(t )  220 2 sin(100t )
R  200 ,   100 rad / s
I?
V  RI or I 
V
R
V
I
(200  j 0)
2200

2000
220

0
200
i(t )  1.1 2 sin(100t ) A
Dr B Rajkumarsingh
Inductor
Frequency domain
Time domain
di
vL
dt
V
V
V
V  j LI or I 


0
j L jX l X l 90
Voltage leads current by
Dr B Rajkumarsingh
90
Example (cont.)
I?
v(t )  220 2 sin(100t )
L  2H,   100 rad / s
V
V
V
V  j LI or I 


0
j

L
jX
X

90
l
l
V
I
( j 200)
2200

20090
220

  90
200
i(t )  1.1 2 sin(100t  90) A
Dr B Rajkumarsingh
Capacitor
Frequency domain
Time domain
dv
iC
dt
I
I
I
I  jCV or V 


0
jC jX c X c 90
Voltage lags current by
Dr B Rajkumarsingh
90
Example (cont.)
I?
v(t )  220 2 sin(100t )
C  1 F ,   100 rad / s
I  jCV
4
I  ( j1.10 )V
 (1.104 )900.22000
 220.104 90 A
i(t )  0.022 2 sin(100t  90) A
Dr B Rajkumarsingh
In the circuit below, v = 30 sin(377t - 20°).
a. What is the sinusoidal expression for the
current?
b. Find the power loss in the circuit.
Dr B Rajkumarsingh
v(t )  30sin(377t  20o )  v(t ) 
V
30
2
2 sin(377t  20o )
30
  20o V
2
R  3 ,   377 rad / s
I?
10
 i (t ) 
2
V
V  RI or I 
V R
I
(3  j 0)
30
  20
2

30
10

  20
2
I
V
2 sin(377t  20o )=10 sin(377t  20o ) A
( V, I )    0o
o
P  VI cos   VI cosDr0
 VI 
B Rajkumarsingh
30 10
 150W
2
2
In the circuit below, v =100 sin(157t +30°).
a. Find the sinusoidal expression for i.
b. Find the value of the inductance L.
c. Find the average power loss by the inductor.
Dr B Rajkumarsingh
 100 
 30)  

2 

 100 
 100 
o
E 
E
 30
2 
2 


e  100 sin(157t
XL  50,
I?
E  j LI
2 sin(157t
 30)
  2p f  157 rad / s  f  25Hz
or
E
E
E
I


j L
jX l
X l 900
100
30
2
I 
5090
2

  60
2
 2 
 i (t )  

 2
2 sin(157t  60)  2 sin(157t  60) A
( V, I )    (30  60) o  90o
P  VI cos   VI cos 90o  0W
Dr B Rajkumarsingh
E
I
Impedance
Impedance is defined as the ratio of the phasor voltage
to the phasor current.
Ohm’s law in phasor notation
V
Z
I
magnitude
or
Z  Z q
polar
Z
 Ze
phase
jq
exponential
Dr B Rajkumarsingh
q
 R  jX
rectangular
Graphical representation of impedance
Z  Z q
Z  R X
1 X
q  tan
R
2
Resistor
Inductor
Capacitor
ZR
Z  j L
1
Z
j C
Dr B Rajkumarsingh
R
L
1/C
2
Express the impedances in both polar
and rectangular form
Dr B Rajkumarsingh
Work out
Problems 1,2&3
Dr B Rajkumarsingh
Kirchhoff’s Law using Phasors
KVL
V1  V2  V3    Vn  0
KCL
I1  I2  I3    In  0
Both Kirchhoff’s Laws hold in the frequency domain.
and so all the techniques developed for resistive circuits hold
Superposition
Thevenin &Norton Equivalent Circuits
Source Transformation
Node & Mesh Analysis
etc.
Dr B Rajkumarsingh
Impedances in series
Vn  IZn
Zeq  Z1  Z2  Z3    Zn
Zeq I  V
V
I
Zeq
Dr B Rajkumarsingh
Exercises RLC series
Dr B Rajkumarsingh
Example : R = 9 , L = 10 mH, C = 1 mF i = ?
vs  100 2Cos(100t )
KVL
Zeq  R  Z 2  Z3
Z 2  j L  j1
( R  Z 2  Z3 )I  Vs
1
(9  j1  j10)I  Vs
Z3 
  j10
j C
Vs
1000
I

 7.8645
(9  j9) 9 2  45
or i  7.86 2 cos(100t  45) A
Dr B Rajkumarsingh
VOLTAGE DIVIDER RULE
Zn V
Vn 
ZT
The basic format for the voltage divider rule in ac circuits is
exactly the same as that for dc circuits.
Vn is the voltage across one or more elements in series that
have total impedance Zn, V is the total voltage appearing across
the series circuit, and ZT is the total impedance of the series
circuit.
Dr B Rajkumarsingh
Example : Using the voltage divider rule, find the unknown
voltage VR
Dr B Rajkumarsingh
Dr B Rajkumarsingh
Power in an AC Circuit
No power is consumed by inductance or capacitance. Thus power is a
function of the component of the impedance along resistance:
Impedance
XL - XC
Z
P = IV cos 

R
P lost in R only
In terms of ac voltage:
In terms of the resistance R:
P = I2R
The fraction Cos  is known as the power factor.
Example 6: What is the average power loss for the
previous example: V = 120 V,  = -60.50, i = 90.5 mA, and
R = 60 .
P = i2R = (0.0905 A)2(60 )
Resonance XL = XC
0.5 H
Average P = 0.491 W
A
The power factor is: Cos
60.50
Cos  = 0.492 or 49.2%
120 V
? Hz
8 F
60 
The higher the power factor, the more efficient is the circuit in its use
of ac power.
RLC circuit equations
Z L  jX L
ZC   jX C
Zeq  R  jX L  jX C
E
I
Zeq
VR  IR
VL  IZ L
VC  IZ C
power factor=cos (E,I)  cos  (lagging or leading)
1 VL  VC
1 IX L  IX C
1 X L  X C
  tan
 tan
 tan
VR
IR
R
ф
S  EI* = P  jQ VA
P  EI cos  W


Q  EI sin  VAR  E,I : rms values

S  EI VA

Dr B Rajkumarsingh
Problem
Dr B Rajkumarsingh
Dr B Rajkumarsingh
Resonance in RLC circuit
KVL
E  VR  VL  VC
For a given L and C there will be a frequency  that
VL  VC
Resonant frequency
1
L 
C
1
or  
 
LC
1
LC
2
E  VR  IR
E
I
R
Resonance
Dr B Rajkumarsingh
Work out
Odd problems in sections
15.2-15.4
Dr B Rajkumarsingh
AC parallel circuits
Dr B Rajkumarsingh
Admittance is defined as the reciprocal of impedance.
1
1
Y 
 Y  q
Z Z q
conductance
In rectangular form
1
1
R  jX
Y 
 2
 G  jB
2
Z R  jX R  X
Resistor
Inductor
Capacitor
1
YG
R
Y
susceptance
G
1
j L
j
Y  jC 
Xc
Dr B Rajkumarsingh
1/L
C
Admittances in parallel
Yeq  Y1  Y2  Y3    Yn
Dr B Rajkumarsingh
Dr B Rajkumarsingh
Example
Calculate I,IR & IL
Draw the phasor diagram and calculate the
power factor and power drawn by the circuit
Dr B Rajkumarsingh
Dr B Rajkumarsingh
Dr B Rajkumarsingh
Calculate I,IR & IC
Draw the phasor diagram and calculate the
power factor and power drawn by the circuit
Dr B Rajkumarsingh
Dr B Rajkumarsingh
Dr B Rajkumarsingh
Calculate I,IR , IL & IC
Draw the phasor diagram and calculate the
power factor and power drawn by the circuit
Dr B Rajkumarsingh
Dr B Rajkumarsingh
Dr B Rajkumarsingh
Dr B Rajkumarsingh
Using the current divider rule, find the current
through each parallel branch
Dr B Rajkumarsingh
Series parallel ac networks
a. Calculate ZT.
b. Determine Is.
c. Calculate VR and VC.
d. Find IC.
e. Compute the power delivered.
f. Find Fp of the network.
Dr B Rajkumarsingh
Work out
odd problems in sections
15.7-15.9, 16
Dr B Rajkumarsingh
Cramer’s Rule for 2x2 System
•
•
Let A be the coefficient matrix
Linear System
Coeff Matrix
ax+by=e
a bcx+dy=f

c d 


•
If detA

0, then the system has exactly one solution:
e b
f d
x
det A
and
a e
c f
y
det A
Example 1- Cramer’s Rule 2x2
• Solve the system:
• 8x+5y=2
• 2x-4y=-10
The coefficient matrix is:
So:
2
5
 10  4
x
 42
8 5 
 2  4


and
and
8 5
 (32)  (10)  42
2 4
8
2
2  10
y
 42
2
5
 10  4  8  (50)
42
x


 1
 42
 42
 42
8 2
2  10  80  4  84
y


2
 42
 42
 42
Solution: (-1,2)
MESH ANALYSIS
Using the format approach to mesh analysis, find the
current I1
Example 10.9-2
v=?
Nodal analysis
10 2Cos1000t
KCL
V
V
V


 100
10 10  j10  j10
0.1V  (0.05  j 0.05)V  j 0.1V  10
10
V
 63.3  18.4
0.15818.4
or
v  63.3 2 cos(1000t  18.4) V
Dr B Rajkumarsingh
Node Voltage & Mesh Current using Phasors
is  I m 2 cos t
C  100 μF, L  5 mH
  1000 rad/s
va = ? vb = ?
1
Z1 
  j10
jC
Y2 
1
1
1

 (1  j )
5 j L 5
Z3  10
Dr B Rajkumarsingh
KCL at node a
Va Va  Vb

 Is
Z1
Z3
KCL at node b
Vb  Va Vb

0
Z3
Z2
Rearranging
( Y1  Y3 ) Va  (  Y3 ) Vb  I s
(  Y3 ) Va  ( Y2  Y3 ) Vb  0
 Y3   Va   I s 
( Y1  Y3 )
 
 Y



Y2  Y3   Vb   0 
3


Admittance matrix
Y matrix
Dr B Rajkumarsingh
If Im = 10 √2 A and
Is  Is 0  100
Using Cramer’s rule to solve for Va
100(3  2 j )
Va 
4 j
100(3  2 j )(4  j )

17
100

(10  11 j )
17
 87.5  47.7
Therefore the steady state voltage va is
va  87.5 2 cos(1000t  47.7 ) V
Dr B Rajkumarsingh
Determine the voltage across the inductor for the
Network using nodal analysis
Dr B Rajkumarsingh
INDEPENDENT VERSUS DEPENDENT
(CONTROLLED) SOURCES
Dependent sources
Example 10.10-2
i1 = ?
Dependent
source
vs  10 2 cos( t  45)
C  5 mF, L  30 mH
  100 rad/s
Z L  j L  j 3
1
ZC 
  j2
jC
Vs  1045
(3  j 3)I1  j 3I2  Vs
(3  j 3)I1  ( j 3  j 2)I 2  0
Dr B Rajkumarsingh
Work out
odd problems in sections
17.2-17.6
Dr B Rajkumarsingh
Superposition, Thevenin & Norton Equivalents
and Source Transformations
Example 10.11-1
i=?
vs  10 2 cos10t V
is  3 A
C  10 mF, L  1.5 H
Vs  100
I s  30
Consider the response to the voltage source acting alone = i1
Dr B Rajkumarsingh
Example 10.11-2 (cont.)
Substitute
Vs
I1 
5  j L  Z p
ZC R
Zp 
 5(1  j ) and  L  15
R  ZC
100
I1 
5  j15  (5  j5)
10
10


  45
10 Dr B Rajkumarsingh
j10
200
Example 10.11-2 (cont.)
Consider the response to the current source acting alone = i2
 0
10
I2   (3)  2 A
15
Using the principle of superposition
i  0.71 2 cos(10t  45)  2 A
Dr B Rajkumarsingh
Source Transformations
VI
IV
Dr B Rajkumarsingh
Example 10.11-2
IS = ?
vs  10 2 cos(t  45) V
  100 rad/s
Z s  10  j10
Vs  1045
1045
Is 
20045
10

0
200
Dr B Rajkumarsingh
Dr B Rajkumarsingh
Procedure
1. Remove that portion of the network across which the
Thévenin equivalent circuit is to be found.
2. Mark (, ●, and so on) the terminals of the remaining
two-terminal network.
3. Calculate ZTh by first setting all voltage and current
sources to zero (short circuit and open circuit,
respectively) and then finding the resulting impedance
between the two marked terminals.
4. Calculate ETh by first replacing the voltage and
current sources and then finding the open-circuit voltage
between the marked terminals.
5. Draw the Thévenin equivalent circuit with the portion
of the circuit previously removed replaced between the
terminals of the Thévenin
equivalent circuit.
Dr B Rajkumarsingh
Example
Dr B Rajkumarsingh
Dr B Rajkumarsingh
NORTON’S THEOREM
Norton’s theorem allows us to replace any two-terminal linear
bilateral ac network with an equivalent circuit consisting of a
current source and an impedance
Dr B Rajkumarsingh
Procedure
1. Remove that portion of the network across which the Norton
equivalent circuit is to be found.
2. Mark (, ●, and so on) the terminals of the remaining twoterminal network.
3. Calculate ZN by first setting all voltage and current sources to
zero (short circuit and open circuit, respectively) and then
finding the resulting impedance between the two marked
terminals.
4. Calculate IN by first replacing the voltage and current sources
and then finding the short-circuit current between the marked
terminals.
5. Draw the Norton equivalent circuit with the portion of the
circuit previously removed replaced
Dr B Rajkumarsingh
Example: Determine the Norton equivalent circuit for the
network external to the 6Ω resistor. Determine current in the 6Ω
resistor.
Dr B Rajkumarsingh
IN
ZN
IR 
IN
6  ZN
 2.31  61.07o
Dr B Rajkumarsingh
Maximum Power Transfer in ac circuits
 Finding the maximum power which can be transferred from a linear circuit to a Load
connected.
• 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 Power from the circuit to which it is
connected.
Dr B Rajkumarsingh
Maximum Power Transfer Condition
• Write the expression for power associated with ZL: P(ZL).
ZTh = RTh + jXTh
ZL = RL + jXL
I
VTh
VTh

ZTh  Z L ( RTh  jX Th )  ( RL  jX L )
2
VTh
2
P  I RL 
RL
2
2
( RTh  RL )  ( X Th  X L )
Ajust R L and X L to get maximum P
2 VTh RL ( X Th  X L )
dP

dX L ( R  R ) 2  ( X  X ) 2  2
L
Th
L
 Th

2
2
2
dP VTh ( RTh  RL )  ( X Th  X L )  2 RL ( RTh  RL ) 

2
2 2
dRL
( RTh  RL )  ( X Th  X L ) 
2
dP
 0  X L   X Th
dX L
dP
0
dRL
 RL  RTh 2  ( X Th  X L ) 2  RTh
Z L  RL  jX L  RTh  jX Th  ZTh
Pmax  I RL
2
Dr B Rajkumarsingh
Example
ZL
(a) Find the values of RL and XL for maximum power transfer to
the load and the value of this maximum power
(b) Calculate the resistive load (XL=0) needed for maximum power
transfer and the maximum power.
Dr B Rajkumarsingh
(a)
For maximum power transfer Z L  Z*th  17.81  j 24.57
 RL  17.81 and X L  j 24.57
I
Vth
Vth
Vth


Zth  Z L Zth  Z*th
2 Rth
I 
Vth
2 Rth
2
 V 
 35.98 
P  I RL   th  RL  
 x17.81  18.2W
 2 x17.81 
 2 Rth 
2
Dr B Rajkumarsingh
2
(b)
For maximum power transfer R L  Z*th  17.81  j 24.57

RL 
)
17.812  24.57 2  30
Vth
35.98  31.91o
35.98  31.91o
I


Zth  RL 17.81  j 24.57  30
53.7527.2o
 0.669 A  4.7o
P  I RL  0.6692 x30  13.4W
2
Dr B Rajkumarsingh
Phasor Diagrams (cont.)
KVL
Vs  VR  VL  VC
For a given L and C there will be a frequency  that
VL  VC
1
L 
C
Resonant frequency
1
or  
 
LC
2
Resonance
Vs  VR
Dr B Rajkumarsingh
1
LC
Summary
Sinusoidal Sources
Steady-State Response of an RL Circuit for Sinusoidal Forcing Function
Complex Exponential Forcing Function
The Phasor Concept
Impedance and Admittance
Electrical Circuit Laws using Phasors
Dr B Rajkumarsingh
Work out
odd problems in sections 18.2-18.5