Advanced Circuit Theory

1. REVISION
Contingency Plan

2. Motivation Word

3. What now?
3
More advanced circuit analysis!
Nodal Analysis
 Extension of KCL concepts
Loop Analysis
 Extension of KVL concepts
Thevenin / Norton equivalent circuits
 Ways to make complex circuits simple

4. NODAL ANALYSIS

10. Ref.
Node/point,
Datum Node,
ground node
NON-Ref.
Node/point

11. Apply kcl at
NON-Ref.
Node/point

29. 1
2

41. SUMMARY

43. EXAMPLE

48. What Do You Think?

49. Test Yourself

51. `

52. Nodal Analysis
52
 How do we tackle a circuit like this?
 Not just one loop or one node 
 Simultaneous equations!
 Use node voltages as variables :
Nodal Analysis.
 If we can find the node voltages,
everything else is easy
 Ohm’s law etc

53. Nodal Analysis
53
 Choose one node as the reference.
 Usually the node with the most
connections.
 Commonly called ground.
 At ground-zero potential
 Might (should!) represent the chassis or
ground line in an actual circuit.
 Assume all voltages are positive with
respect to ground.
 For N nodes, we need N – 1 linearly
independent KCL equations to solve
(with respect to ground).

54. Nodal Analysis
Circuits with independent current sources
54
Consider the network shown.
There are three nodes, so we need two equations.
We can use the bottom node as the reference (ground), so we still need to
find v1 and v2.
Use KCL and Ohm’s law to write two
nodal equations.
N – 1 linearly independent KCL

55. Nodal Analysis
Circuits with independent current sources
55
Solution 1) Use Gaussian Elimination to find the unknowns, if
IA = 1 mA, R1 = 12 kΩ, R2 = 6 kΩ, IB = 4 mA and R3 = 6 kΩ.

56. Nodal Analysis
Circuits with independent current sources
56
Solution 2) Use Matrix Analysis to find the unknowns, if
IA = 1 mA, R1 = 12 kΩ, R2 = 6 kΩ, IB = 4 mA and R3 = 6 kΩ.

57. Example
57
Oh oh! We can’t write KCL for V1 or V2 (we don’t know the
current through the voltage source…)

58. Example
58
But we can write V1 – V2 = 6 V. There are 3 nodes, so we need 2
equations: now we have one equation, so we only need one
more.

59. Summary:
Problem Solving Strategy
59
Step 1)
Determine the number of nodes, select one as the
reference (ground). Assign voltage names to the
other nodes.
Step 2)
Write an equation for each voltage source (i.e.: V2 =
12 V, or V1 – V4 = -6 V). Each of these equations is
one of the linearly independent equations we will
use. Identify supernodes when a voltage source
connects between two non-reference nodes.
Step 3)
Use KCL to write the remaining equations. For N
nodes, you will need a total of N – 1 equations.
Solve to find the unknown voltages.

60. Chap 1 – Steady State Analysis

61. Sinusoidal Steady State Analysis
• Any steady state voltage or current in a linear circuit with a
sinusoidal source is a sinusoid
– All steady state voltages and currents have the same frequency
as the source
• In order to find a steady state voltage or current, all we need to
know is its magnitude and its phase relative to the source (we
already know its frequency)
• We do not have to find this differential equation from the circuit, nor
do we have to solve it
• Instead, we use the concepts of phasors and complex impedances
• Phasors and complex impedances convert problems involving
differential equations into circuit analysis problems
 Focus on steady state; Focus on sinusoids.

62. Characteristics of Sinusoidal
Key Words:
• Period: T ,
• Frequency: f , Radian frequency 
• Phase angle
• Amplitude: Vm Im
Sinusoidal Steady State Analysis

63. Characteristics of Sinusoidal
  tVv mt sin iI1
I1
I1 I1 I1 I1
R1
R1
R
5
5
+
_
IS 
E
I1
U1
+
-
U
I

iI1 I1 I1 I1
R1
R1
R
5
5
-
+
IS 
E
I1
U1
+
-
U
I

v、i
tt1 t20
Both the polarity and magnitude of voltage are changing.
Sinusoidal Steady State Analysis

64. Characteristics of Sinusoidal
Radian frequency(Angular frequency):  = 2f = 2/T (rad/s）
Period: T — Time necessary to go through one cycle. (s)
Frequency: f — Cycles per second. (Hz)
f = 1/T
Amplitude: Vm Im
i = Imsint， v =Vmsint
v、i
t 20
Vm、Im
Sinusoidal Steady State Analysis

65. Phasors
E.g. voltage response
A sinusoidal v/i
Complex transform
Phasor transform
By knowing angular
frequency ω rads/s.
Time domain
Frequency domain
  eR v t
Complex form:
   cosmv t V t  
Phasor form:
   j t
mv t V e  

Angular frequency ω is
known in the circuit.
Sinusoidal Steady State Analysis
 || mVV
 || mVV

66. Phasors
Complex Numbers
jbaA  — Rectangular Coordinates
  sincos jAA 
j
eAA  — Polar Coordinates
j
eAAjbaA 
conversion： 22
baA 
a
b
arctg
jbaeA j

cosAa 
sinAb 

a
b
real axis
imaginary axis
jjje j

090sin90cos90 
Sinusoidal Steady State Analysis

67. Phasors
Phasors
A phasor is a complex number that represents the magnitude and
phase of a sinusoid:
  tim cos  mII
Phasor Diagrams
• A phasor diagram is just a graph of several phasors on the complex
plane (using real and imaginary axes).
• A phasor diagram helps to visualize the relationships between
currents and voltages.
Sinusoidal Steady State Analysis

68. 68
.Steady-State Sinusoidal Analysis – Complex Impedances

69. Phasor Relationships for R, L and C
Resistor:
Q ， , R=10，Find i and P。tv 314sin311
 V
V
V m
220
2
311
2

 A
R
V
I 22
10
220

ti 314sin222  WIVP 484022220 
Sinusoidal Steady State Analysis

70. Phasor Relationships for R, L and C
Inductor:
Q，L = 10mH，v = 100sint，Find iL when f = 50Hz and 50kHz.
  
14.310105022 3
fLX L
 
   Atti
A
X
V
I
L
L

90sin25.22
5.22
14.3
2/100
50



  
31401010105022 33
fLX L
 
   mAtti
mA
X
V
I
L
L
k

90sin25.22
5.22
14.3
2/100
50



Sinusoidal Steady State Analysis

71. Phasor Relationships for R, L and C
Capacitor:
Q，Suppose C=20F，AC source v=100sint，Find XC and I forf = 50Hz, 50kHz。
 159
2
11
Hz50
fCC
Xf c

A38.1
2

c
m
c X
V
X
V
I
  159.0
2
11
KHz50
fCC
Xf c

 A1380
2

c
m
c X
V
X
V
I
Sinusoidal Steady State Analysis

72. Phasor Relationships for R, L and C
Review (v-I relationship)
Time domain Frequency domain
iRv  IRV  
I
Cj
V  

1
ILjV   
dt
di
LvL 
dt
dv
CiC 
C
XC

1

LXL ,
,
, v and i are in phase.
, v leads i by 90°.
, v lags i by 90°.
R
C
L
Sinusoidal Steady State Analysis

73. Phasor Relationships for R, L and C
Summary
 R： RX R  0
L： ffLLXL   2
2

  iv
C：
ffcc
XC
1
2
11

 2

  iv
 IXV 
 Frequency characteristics of an Ideal Inductor and Capacitor:
A capacitor is an open circuit to DC currents;
A Inducter is a short circuit to DC currents.
Sinusoidal Steady State Analysis

74. Impedance
Complex Impedance
Phasors and complex impedance allow us to use Ohm’s law with
complex numbers to compute current from voltage and voltage
from current
20k
+
-
1F10V  0 VC
+
-
 = 377
Find VC
Q.
• How do we find VC?
• First compute impedances for resistor and capacitor:
ZR = 20k = 20k  0
ZC = 1/j (377 *1F) = 2.65k  -90
Sinusoidal Steady State Analysis

75. Impedance
Complex Impedance
20k
+
-
1F10V  0 VC
+
-
 = 377
Find VC
Q.
20k  0
+
-
2.65k  -9010V  0 VC
+
-
Now use the voltage divider to find VC:




46.82V31.1
54.717.20
9065.2
010VCV
)
0209065.2
9065.2
(010 





kk
k
VVC
Sinusoidal Steady State Analysis

76. Impedance
Impedance allows us to use the same solution techniques
for AC steady state as we use for DC steady state.
• All the analysis techniques we have learned for the linear circuits are
applicable to compute phasors
– KCL & KVL
– node analysis / loop analysis
– superposition
– Thevenin equivalents / Norton equivalents
– source exchange
• The only difference is that now complex numbers are used.
Complex Impedance
Sinusoidal Steady State Analysis

77. Impedance
Phasor Diagrams
• A phasor diagram is just a graph of several phasors on the complex
plane (using real and imaginary axes).
• A phasor diagram helps to visualize the relationships between
currents and voltages.
2mA  40
–
1F VC
+
–
1k VR
+
+
–
V
I = 2mA  40, VR = 2V  40
VC = 5.31V  -50, V = 5.67V  -29.37
Real Axis
Imaginary Axis
VR
VC
V
Sinusoidal Steady State Analysis

78. Q1
• Find the time-domain voltage over the
capacitor using phasor domain techniques.
The voltage and current source are both AC
(sine) sources with frequency of 1MHz.

79. ANS:

80. Q2
• 2. Write the time-domain (differential) node
voltage equations for the circuit shown. Do
not solve.

81. ANS:

82. Q3
• Set-up the phasor-domain node-voltage equations
for the following circuit. Do not solve.

83. ANS:

84. Q4
• In the circuit shown below, find the current flowing down
through the inductor. The frequency of the source is 2
Mrads/sec and the phase is 0 degrees. You must use phasor
analysis, but your final answer must be converted back to
time-domain.

85. ANS:

86. Q5
• Write the time-domain (integral-differential) node-
voltage equations for the circuit shown, using the
ground (0) node as your reference. Do not solve.

87. Ans:

88. Examples for Sinusoidal Circuits Analysis
v1=120sint v2

i3
 i1  i2
Q., Find 1I
2I 3I
2V in the circuit of the following fig.
Complex Numbers Analysis
Sinusoidal Steady State Analysis

89. 89
Additional Example:
represent the circuit shown in the
frequency domain using impedances
and phasors.

90. 90
Additional Example: represent the
circuit shown in the frequency domain
using impedances and phasors.

91. 91
Example – Steady-State AC Analysis of a Series Circuit
* Find the steady-state current, the phasor voltage across each element, and
construct a phasor diagram.
   






15t500cos707.0ti
15707.0
454.141
30100
Z
454.141100j10050j150j100ZZRZ
50j
C
1
jZ,150jLjZ,30100
s
CLeq
CLs
V
I
V 





1054.3515707.09050
C
1
j
751.10615707.090150Lj
157.7015707.0100R
C
L
R
IV
IV
IV



92. 92
Example – Series/Parallel Combination of Complex Impedances
* Find the voltage across the capacitor, the phasor current through each
element, and construct a phasor diagram
     




































901.0
90100
18010
100j
18010
Z
V
1801.0
100
18010
R
V
1351414.0
50j50
9010
50j50100j
9010
ZZ
t1000cos10180t1000cos10tv
18010
50j50100j
4571.70
9010
ZZ
Z
50j50Z
4571.70
4501414.0
01
)100j(11001
1
Z1R1
1
Z
100j
C
1
jZ,100jLjZ,90-10
C
C
C
C
R
RCL
s
C
RCL
RC
sC
RC
C
RC
CLs
I
I
V
I
VV
V 



93. 93
Circuit Analysis
Example – Steady-State AC Node-Voltage Analysis
* Find the voltage at node 1 using nodal analysis
 
   







7.29t100cos1.16tvor7.291.16
:forSolve
5.11.0j2.0j
2j2.0j2.0j1.0
05.1
5j10j
90-2
j5-
-
10
2nodeandnode1atKCLWrite
11
1
21
21
122
211
V
V
VV
VV
VVV
VVV

94. 94
Circuit Analysis
Exercise – Steady-State AC Mesh-Current Analysis
* Solve for the mesh currents

95. Chap 2 – Mutual Inductance
• Dot Convention

97. In Practical
• Mutual inductance can exist even in places where we would
rather it not. Take for instance the situation of a "heavy"
(high-current) AC electric load, where each conductor is
routed through its own metal conduit.
• The oscillating magnetic field around each conductor induces
currents in the metal conduits, causing them to resistively
heat (Joule's Law, P = I2 R):

98. • It is standard industry practice to avoid running the
conductors of a large AC load in separate metal
conduits. Rather, the conductors should be run in the
same conduit to avoid inductive heating:

99. Explain why this wiring technique
eliminates inductive heating of the
conduit.
• Now, suppose two empty metal conduits stretch between the location of a
large electric motor, and the motor control center (MCC) where the circuit
breaker and on/off "contactor" equipment is located.
• Each conduit is too small to hold both motor conductors, but we know
we're not supposed to run each conductor in its own conduit, let the
conduits heat up from induction.
• What do we do, then?

100. Answer
• Use terminal blocks to split up" the
conductors from one pair into two pairs:

101. Example 1
• Calculate the phasor currents I1 and I2.

102. Solution

103. Example 2
• Determine V0 in the circuit below.
I2= 0.06 -90 degree x 10ohm

104. Example 3
• Determine the phasor currents I1 and I2 in the
circuit below.
I1=0.132 – j2.136 , I2=0.193+j3.214

105. Ex 4
• Determine the phasor currents I1 and I2 in the
circuit below.
-j8 ohm

106. Ex 5
• Calculate Vo.

107. Ex 5 : Ans

108. Ex 6
• Calculate I1 and I2.
• (Note that j1 and j2 are reactances and not
inductances.

109. Ex 6:Ans

110. Chap 3 - Laplace
• Q1. Evaluate the Vo(t) by using Nodal Analysis.
Show all the steps to solve this problem.

111. Ans:

112. • Q2. Solve for i(t) for the circuit, given that
V(t) = 10 sin5t V, R = 4 Ω and L = 2 H.
Assume i0=i(0)=0.
i(t)

113. Ans

114. • Q3. Evaluate the Vth(s) and Zth(s) by using
Tevenin Theorem. Show all the steps to solve
this problem.
Vth

115. Ans: