TECHNIQUES OF
CIRCUIT ANALYSIS
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4.1 Introduction
4.2 The Node-Voltage Method ( Nodal Analysis )
4.3 The Mesh-Current Method ( Mesh Analysis )
4.4 Fundamental Loop Analysis
4.5 Fundamental Cutset Analysis
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4.1 Introduction
A circuit consists of b branches and n nodes.
A direct algebraic approach：2b method
Example.
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4.1 Introduction
b=6，n=4
3 independent equations only (n-1)
KCL：
node A：
I1+I4+I6=0
node B：－I4+I2+I5=0
node C： I3-I5-I6=0
node D：
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I1+I2+I3=0
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4.1 Introduction
KVL：
loop ABDA：V4+V2－V1=0
loop BCDB：V5+V3－V2=0
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loop ACBA：V6－V5－V4=0
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4.1 Introduction
Component model (b branches)
V1 = E1 , V4 = R4I4 ,
V2 = R2I2 , V5 = R5I5 ,
V3 = E3 , V6 = R6I6 ,
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4.1 Introduction
Want to find Vk , Ik , K=1,2,3, … ,6
There are 2b unknowns.
number of KCL eqs：n-1
number of KVL eqs：b-(n-1)
number of component model：b
The total number of equations.
n-1+[b-(n-1)]+b=2b
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4.1 Introduction
Just enough information to find the solution.
Problem：Too many variables !!
It is not efficient !
Question：Is it possible to find an optimal
method with minimum unknowns？
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4.1 Introduction
n
Nodal analysis is based on a systematic application
of KCL and is a general method.
n
Mesh Analysis is based on a systematic application
of KVL and can be used for planar circuits only.
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4.1 Introduction
n
Fundamental loop analysis is based on a
systematic application of KVL to the
fundamental loops. It requires the definition of
tree.
n
Fundamental cutset analysis is based on a
systematic application of KCL to the
fundamental cutsets. It also requires the
definition of tree.
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4.1 Introduction
Node: A point where two or more elements join
n Path: A trace of adjoining basic elements with
no elements included more than once
n Branch: A path that connects two nodes
n Essential node: A node where three or more
elements join
n Essential branch: A path that connects two
essential nodes without
passing through an essential
node
n
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4.1 Introduction
Loop: A path whose last node is the same as the
starting node
n Mesh: A loop that does not enclose any other
loops
n Planar circuit: A circuit that can be drawn on a
plane with no crossing branches
n
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4.1 Introduction
n
Example 4.1.1
The nodes are { a , b , c , d , e , f }
The essential nodes are { b , d , e , f }
The branches are { R1 , R2 , R3 , R4 , R5 , R6 , V7 , R8 }
The essential branches are { R4 , R6 , V7 , R8 ,
R2 − R3 , R1 − R5 }
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4.1 Introduction
n
Example 4.1.1
The meshes are
m1 : b – c – d – e – b
m2 : a – b – e – f – a
m3 : f – e – d – f
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4.1 Introduction
A graph G is defined as a set: G(B , N , F)
B: set of branches
N: set of nodes
F: a function specifying how a branch is connected.
^
Oriented (directed) graph G ,
if each branch of G is assigned a direction.
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4.1 Introduction
Example 4.1.2
n
Gˆ ( B, N , F )
B = { b1 , b2 , b3 , b4 , b5 , b6 , b7 , b8 }
N={a,b,c,d,e,f}
F = { f1 , f2 , f3 , f4 , f5 , f6 , f7 , f8 }
f1 : a → b
f2 : b → c
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4.1 Introduction
^
For each oriented graph G one can construct a tree T.
^
A tree T of G is a connected graph which
(a) contains all the nodes.
(b) without forming any loop.
The branches which do not belong to tree T are called links.
Each tree branch defines a unique cutset (or super node) ,
called fundamental cutset.
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4.1 Introduction
Each link together with its unique path connecting the two
nodes of the link defines a unique loop , called
fundamental loop.
A network with b branches , n nodes , and l links will
satisfy the fundamental theorem of network topology :
b=l+n-1
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4.1 Introduction
Theorem
Given a connected graph G of n nodes and b
branches, and a tree T of G
n There is unique path along the tree between any
pair of nodes.
n There are n-1 tree branches and b-(n-1) links.
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4.1 Introduction
Every link of T and the unique tree path between
its nodes constitute a unique loop, called the
fundamental loop associated with the link.
n Every tree branch of T together with some links
defines a unique cut set of G, called the
fundamental cut set associated with the tree branch.
n
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4.1 Introduction
Example 4.1.3
^
G (B , N , F) , b=8 , n=5
B={b1 b2 b3 b4 b5 b6 b7 b8}
N={N1 N2 N3 N4 N5}
F=[f1 f2 f3 f4 f5 f6 f7 f8]T
f1: N4 → N1
f2: N1 → N2
f3: N2 → N3
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4.1 Introduction
Example 4.1.4
Fundamental loops :
T={b2 b3 b5 b8}
tree branches b2 , b3 , b5 , b8
no. of tree branches = n-1=4
L={b1 , b4 , b6 , b7}
links b1 , b4 , b6 , b7 no. of
links = b-(n-1)
= 8-(5-1)
=4
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4.1 Introduction
Example 4.1.4
Fundamental loops :
For l1: N1−N5−N4−N1
For l4: N4−N5−N1−N2−N3−N4
For l6: N5−N1−N2−N5
For l7: N5−N1−N2−N3−N5
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4.1 Introduction
Example 4.1.5
Fundamental cutsets :
Cut set of b8 = {b8 , b1 , b4}
Cut set of b8 = {b3 , b4 , b7}
Cut set of b8 = {b2 , b6 , b7 , b4}
Cut set of b8 = {b5 , b1 , b6 , b7 ,
b4}
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4.1 Introduction
Example 4.1.5
n=5 , b=8
T2 = {b5 , b6 , b7 , b8}
Fundamental cut sets :
Cut set of b5 = {b1 , b2 , b5}
Cut set of b6 = {b2 , b3 , b6}
Cut set of b7 = {b3 , b4 , b7}
Cut set of b8 = {b1 , b4 , b8}
Choose tree branch current direction as
reference. Cut only one tree branch.
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4.1 Introduction
Example 4.1.5
Set of links = {b1 , b2 , b3 ,b4}
Fundamental loops :
Loop of
Loop of
Loop of
Loop of
b1 = N1→ N5 → N4 → N1
b2 = N 2 → N 5 → N 1 → N 2
b3 = N 3 → N 5 → N 2 → N 3
b4 = N 4 → N 5 → N 3 → N 4
Choose link current as loop
current. Contain only one link.
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4.2 The Node-Voltage Method
For Linear Resistive Circuits (Example)
n
KCL
n
KVL
n
Component Model (Ohm’s Law)
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4.2 The Node-Voltage Method
n
Instead of using branch voltages and branch currents
as unknowns, one can choose nodal voltages as
unknowns. Only n-1 nodal voltages are required.
n
Each branch voltage can be obtained from nodal voltages
and the corresponding branch currents can be calculated
by its component model.
n
KVL is automatically satisfied.
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4.2 The Node-Voltage Method
One need only find the independent KCL
equations，and the component model is used to
express the branch current in terms of nodal
voltages.
Ge = Is
n
Any unknown can be calculated once the nodal
voltage vector is solved.
n
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4.2 The Node-Voltage Method
Case 1. Containing current source only
(A) independent current source by inspection
(B) with dependent current source express the
controlling parameter in terms of nodal
voltages
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4.2 The Node-Voltage Method
Case 2. Containing voltage source (VS)
(A) VS contains datum node
trivial case
(B) VS does not contain datum node
by supernode concept
(C) for dependent VS
express the controlling parameter in terms
of nodal voltages
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4.2 The Node-Voltage Method
Example 4.2.1. (Case 1A) (Basic Case)
3Ω
1Ω
4Ω
2Ω
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Step 1
Select a node as datum
node and choose nodal
variables e1 , e2 , e3 as
unknowns
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4.2 The Node-Voltage Method
3Ω
1Ω
4Ω
2Ω
Step 2
Apply KCL to n-1 nodes,
use ohm’s law to express
the branch currents in
terms of nodal voltages
∑i
k
=0
leaving
node 1: − i1 − i2 − 3A = 0
node 2 : + i2 + i3 + i4 = 0
node 3 : + i1 − i4 + 4A = 0
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4.2 The Node-Voltage Method
Use ohm’s law
node 1 :
( e1 − e3 ) + ( e1 − e2 ) − 3A = 0
node 2 :
( e2 − e1 ) + ( e2 − 0 ) + ( e2 − e3 ) = 0
node 3 :
3
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1
1
2
4
( e − e1 ) + ( e3 − e2 ) + 4A = 0
+ 3
3
4
1

−1
1 + 3

 −1 1 + 1 + 1

2 4

1
1
 −
−
 3
4

3Ω
1Ω
4Ω
2Ω
1 
3  e   3A 
 1
1   

−
e2  =  0 


4   
− 4 A 
 e
1 1  3  
+
3 4 
−
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4.2 The Node-Voltage Method
 I s1 
 e1 
 
( G jk )  e2  = G e =  I s 2  = I s
e 
I 
 3
 s3 
G
: conduction matrix
Gkk
: sum of the conductances connected to node k
Gkj=Gjk : negative of the sum of conductances directly
connecting nodes k and j , k≠
k≠ j
ek
: nodal voltage of the kth node
Isk
: sum of independent current sources entering to node k
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4.2 The Node-Voltage Method
Example 4.2.2. (Case 1B)
1
Ω
4
1Ω
e1
5A
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e2
ix
1
Ω
2
e3
1
Ω
3
2ix
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4.2 The Node-Voltage Method
Step 1 Same as Example 1
−1
−4   e1   5 A 
1 + 4

  

 −1 1 + 2 + 3 −3   e2  =  0 
 −4
−3
3 + 4   e3   −2ix 

Step 2 express -2ix in terms of nodal voltages
−2ix = −2
( e1 − e2 )
1
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4.2 The Node-Voltage Method
Step 3 Move the unknowns to the left hand side of the equation
−1
−4   e1   5 
 5

   
6
−3   e2  =  0 
 −1
 −4 + 2 −3 − 2 3 + 4   e   0 

 3   
1
Ω
4
1Ω
Ge = I s
1
Ω
2
1
Ω
3
Note : G matrix is no longer symmetric.
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4.2 The Node-Voltage Method
Example 4.2.3. (Case 2A) (trivial case)
Step 1
Select datum node .
choose nodal voltages
e1 , e2 , e3
3Ω
4Ω
1Ω
2Ω
5Ω
Step 2
KCL eqs.
e1=10V , already known
Need 2 KCL eqs. only!
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4.2 The Node-Voltage Method
node 1 : e1 = 10V
e2 − 10 e2 − 0 e2 − e3
+
+
= −4 A
1
2
4
e − 10 e3 − e2 e3
node 3 : 3
+
+ = 4A
3
4
5
node 2 :
 1 1
1 + 2 + 4

 −1

4

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3Ω
4Ω
1Ω
5Ω
2Ω
1 
10 − 4 
4   e2  

   = 10
1 1 1   e3   + 4 
+ + 
 3

3 4 5
−
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4.2 The Node-Voltage Method
Example 4.2.4. (Case 2B) (supernode)
n－1＝4－1＝3
independent nodal voltages
1 Ω
4
1Ω
1
Ω
2
1
Ω
3
1 voltage source
Only 2 unknowns are needed.
Say , choose ea & eb
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4.2 The Node-Voltage Method
Node a : ( ea − eb ) + ( ea − ec ) 4 = 1A − 5 A LL (1)
Let the current through the voltage source be ix .
Node b : ( eb − ea ) + eb ( 2 ) + ix = 0
Node c : − ix + ec ( 3 ) + ( ec − ea ) 4 = 5 A
( eb − ea ) + eb ( 2 ) + ec ( 3) + ( ec − ea ) 4 = 5 A LL ( 2 )
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4.2 The Node-Voltage Method
Which is the KCL equation written for the
supernode including nodes b and c .
From the known voltage source , one can get
ec = eb − 5V
LLLLLLLLLLL ( 3)
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4.2 The Node-Voltage Method
Substituting (3) into (1) & (2)
LLLL ( 4 )
( ea − eb ) + ( ea − eb + 5) 4 = −4
( eb − ea ) + 2eb + 3 ( eb − 5) + ( eb − 5 − ea ) 4 = 5 ( 5)
From eqs.(4) and (5)
 5 −5   ea   −20 − 4 

  = 

 −5 10   eb  15 + 20 + 5 
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4.2 The Node-Voltage Method
Example 4.2.5 (Case 2C)
1
Ω
3
2vx
ea
1A
eb
1Ω
1
Ω
2
vx
2A
ec
1
Ω
4
n-1=4-1=3, nodal voltages
There is one (dependent)
voltage source.
Only two unknown are needed.
Say ,choose eb and ec.
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4.2 The Node-Voltage Method
For the supernode containing nodes a and b:
eb(1) + (eb-ec)2 + (ea-ec)3 = 1A + 2A
……(1)
For node c:
(ec-eb)2 + ec(4) + (ec-ea)3 = 0 ….……...…(2)
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4.2 The Node-Voltage Method
Express the controlling parameters in terms of nodal
voltages.
1
3
2vx = 2(eb-ec)
2vx
ea
Also, for the known voltage source
1A
Ω
eb
1
Ω
2
vx
1Ω
ec
1
Ω
4
2A
ea = 2vx + eb = 2(eb-ec) + eb
= 3eb-2ec ………………………………….(3)
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4.2 The Node-Voltage Method
Substituting (3) into (1) and (2) yields
eb + 2(eb-ec) + (3eb-2ec-ec)3 = 3A ……….…(4)
(ec-eb)2 + 4ec + (ec-3eb+2ec)3 = 0 ………….(5)
1
Ω
3
 12 -11 eb   3A 
-11 15   e  =  0 

 c  
ea
1A
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2vx
eb
1Ω
1
Ω
2
vx
2A
ec
1
Ω
4
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4.2 The Node-Voltage Method
node-to-branch incidence matrix Aa
B
A
2
C
3
5
4
1
D
b1 b2
b3 b4 b5
A
1
1
0
0
0
B
0
-1
1
0
1
C
0
0
-1
1
0
D
-1
0
0
-1
-1
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4.2 The Node-Voltage Method
1
Aa={aik}, aik=
if
branch i-k leaves node i
-1 if
branch i-k enters node k
0
branch i-k is not
if
incident with node i and k
∧
An oriented graph G can be represented by its
incident matrix Aa in computer
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4.2 The Node-Voltage Method
Consider the general case
Example 4.2.5
b=5 , n=4
node to branch incidence matrix
b1 b2 b3 b4
b5
b6
Aa = a
1
1
1
0
0
0
b
0
-1
0
1
0
1
c
0
0
-1
0
1
-1
d -1
0
0
-1
-1
0
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4.2 The Node-Voltage Method
Reduced node to branch incidence matrix
If node d is chosen as datum node
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4.2 The Node-Voltage Method
Since each branch voltage can be expressed in terms of
nodal voltages：
V1 = ea – 0
,
Let
V2 = ea – eb
VbT =[V1 V2 V3 V4 V5 V6 ]
V3 = ea – ec
eT =[ea eb ec ]
V4 = eb – 0
Then
V b = N T e ………(1)
V5 = ec – 0
V3 = eb – ec
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4.2 The Node-Voltage Method
• KCL, N Ib=0……….(2)
I bT = [ I 1
I2
I3
I4
I 5 I6 ]
‚ KVL, automatically satisfied
Example, loop abd
v2 + v4 − v1 = 0
⇒ (ea − eb ) + (eb − 0) − (ea − 0) = 0
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4.2 The Node-Voltage Method
ƒ
General computer model
I k = I sk + (Vk − Vsk ) Gk
Gk =
1
Rk
K = 1, 2,.........., b
using matrix notation
(
)
I b = I s + [G ] Vb − Vs .......... ( 3)
where I = [ I s1 I s 2 LL I sb ]
T
s
VsT = [Vs1 Vs 2 LLVsb ]
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4.2 The Node-Voltage Method
Multiply both sides of (3) by N
0 = N I s + N [G ] (Vb -Vs ) .................( 4 )
substitute (1) into
0 = N I s + N [G ]N T e - N [G ]Vs
∴ N [G ] N T e = N [G ]Vs - N I s ,
AX = b
A = N[G]N T
b = N[G]Vs - NI s
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4.3 The Mesh Current Method
(a) Another general method for analyzing planar
circuits.
(b) A planar circuit is one that can be drawn in a
plane with no branches crossing one
another；otherwise , it is nonplanar.
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4.3 The Mesh Current Method
(c) A mesh in a planar circuit is a loop that does not
contain any other loop within it.
(d) Instead of choosing branch currents as unknown
variables , one can choose mesh currents as
unknown variables to reduce the number of
equations which must be solved simultaneously.
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4.3 The Mesh Current Method
Example 4.3.1 :
mesh currents
i1 : mesh A B E F A
i2 : mesh B C D E B
i3 : mesh E D G H E
i4 : mesh F E H K F
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4.3 The Mesh Current Method
(e) If the mesh currents are known , then any branch
current can be obtained from mesh currents and
the corresponding branch voltage can be obtained
from the component model directly.
(f) For a planar circuit consists of b branches and n
nodes, one can have b-(n-1) l independent
meshes , where l is the number of links.
(g) KCL is automatically satisfied.
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4.3 The Mesh Current Method
Mesh Analysis Method
Step 1：Assign mesh currents, i1, i2, …il
as unknowns.
Step 2：Apply KVL to each mesh and use Ohm’s
law to express the voltages in terms of the
mesh currents.
Step 3：Solve the algebraic equation
A x= b
and find the desired answer.
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4.3 The Mesh Current Method
Examples:
Case 1：Planar circuits with voltage sources
(A) Only independent voltage sources.
： by inspection.
(B) With dependent voltages sources.
： Need one more step, namely
express the controlling parameter in
terms of the mesh currents.
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4.3 The Mesh Current Method
Case 2：Planar circuits with current sources
(A) Independent current source exist only
in one mesh.
：Trivial case.
(B) Independent current source exists
between two meshes.
：Use supermesh concept
(C) With dependent current source
：Need one more step, namely express
the controlling parameter in terms of
the mesh currents.
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4.3 The Mesh Current Method
Example 4.3.2：Case 1A(Basic case)
2Ω
4Ω
10V
5V
1Ω
5Ω
3Ω
b=7 , n =6
l = 7-(6-1) = 2
Choose i1 & i2 mesh currents.
mesh 1：KVL , i1 × 1 + i1 × 2 + 5V + (i1 - i2 ) × 3 = 0
mesh 2：KVL , (i2 -i1 ) × 3 + ( - 5) + i2 × 4 + 10V + i2 × 5 = 0
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4.3 The Mesh Current Method
Rearrange the above equations
1+2+3 Ω
-3Ω
-3Ω
3+4+5 Ω
i1
i2
＝
[ R ik ] i @ R i = VS
i : mesh current vector
VS : sourse voltage vector
-5V
5-10
2Ω
4Ω
10V
5V
1Ω
5Ω
3Ω
R : resistance matrix
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4.3 The Mesh Current Method
Rkk = sum of all the resistances in mesh k
Rik = sum of the resistances between meshes i and
k and the algebraic sign depends on the
relative direction of meshes i and k , plus
(minus)sign for same(opposite) direction.
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4.3 The Mesh Current Method
Example 4.3.3：Case 1B
IO
24Ω
10Ω
4Ω
24V
12Ω
Step 1. Choose mesh
currents i1, i2, i3
Step 2. Same as example 1
4IO
10+12
-10
-12
i1
-10
-12
10+4+24
-4
-4
4+12
i2
24
＝
i3
0
-4I0
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4.3 The Mesh Current Method
Step 3. Express Io in terms of mesh currents Io=i1-i2
Substitute into the above mesh equations
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10+12
-10
-12
i1
-10
-12+4
10+4+24
-4-4
-4
4+12
i2
i3
24
＝
0
0
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4.3 The Mesh Current Method
∴ i1 = 2.25A
i2 = 0.75A
IO
24Ω
10Ω
i3 = 1.5A
4Ω
24V
IO = i1 - i2 = 1.5A
12Ω
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4IO
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4.3 The Mesh Current Method
Example 4.3.4：Case 2A
3Ω
4Ω
10V
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5Ω
6Ω
5A
b= 6, n=4
l= 6-(4-1)=3
Step 1. Choose mesh
currents i1, i2, i3
Step 2. For case 2A
i3=-5A, answer
70
4.3 The Mesh Current Method
KVL for mesh 1：
3Ω
(i1 - i2 ) × 4 + (i1 - i3 ) × 6 = 10V
KVL for mesh 2：
4Ω
5Ω
i2 × 3 + (i2 - i3 ) × 5 + (i2 - i1 ) × 4 = 0
Substitute i3 = -5A into KVL equations
10V
6Ω
5A
(i1 - i2 ) × 4 + (i1 + 5) × 6 = 10V
i2 × 3 + (i2 + 5) × 5 + (i2 - i1 ) × 4 = 0
4+6
-4
-4
3+5+4 Ω
i1
i2
＝
10-30
-25
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4.3 The Mesh Current Method
Example 4.3.5：Case 2B (Supermesh concept)
6Ω
Step 1. Choose mesh currents
i1 , i2
10Ω
2Ω
20V
4Ω
vx
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6A
Step 2. Algebraic solution
approach：
Assume the voltage across
the current source is v x
72
4.3 The Mesh Current Method
KVL for mesh 1：
6Ω
10Ω
i1 × 6 + (i1 - i2 ) × 2 + (-v x ) = 20V
2Ω
KVL for mesh 2：
20V
4Ω
v x + (i2 - i1 ) × 2 + i2 × 10 + i2 × 4 = 0
vx
6A
Add two KVL equations
i1 × 6 + i2 ×10 + i2 × 4 = 20V.......(A)
This is the KVL equation for the supermesh!
Need one more equation form the current source：
6A = i2 - i1................................(B)
From (A) and (B) , one can get the solution.
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4.3 The Mesh Current Method
Example 4.3.6：Case 2C
Step1: choose i1, i2, i3, i4 mesh currents.
There are two current sources , one is independent
and the other is dependent. And the number of
unknowns can be reduced by 2 , namely only two
2Ω
equations are required.
i1
5A
4Ω
2Ω
I0
+
6Ω
i2
3I0 i3
8Ω
i4
10V
-
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4.3 The Mesh Current Method
Step2: We can write KVL equations for mesh 4
and the supermesh as indicated in Fig.
mesh 4: (i4 - i3)*8 + i4*2 = -10 V
(A)
supermesh: i2*6 + i1*2 + i3*4 + (i3-i4)*8 = 0
(B)
2Ω
i1
5A
4Ω
2Ω
I0
+
6Ω
i2
3I0 i3
8Ω
i4
10V
-
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4.3 The Mesh Current Method
For two current sources, we can get two equations
to eliminate two unknowns
5A = i2-i1 => i2=i1+5
3I0 = i2-i3 = i1+5-i3 => i3 = i1+5-3I0
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(C)
(D)
76
4.3 The Mesh Current Method
For this case with dependent source, we need one
more step, namely to express the controlling
parameter Io in terms of mesh currents.
I0 = -i 4
(E)
Substitute (C), (D), and (E) into (A) and (B) :
[i4 – (i1 + 5 + 3i4)]*8 + i4*2 = -10
(i1 + 5)*6 + i1*2 + (i1 + 5 + 3i4)*4 + [(i1 + 5 + 3i4) - i4]*8=0
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4.3 The Mesh Current Method
Equivalently
 -8 -14   i1  =  30 
 28 36   i4   -90 
∴ i1 = -7.5 A
i2 = -2.5 A
i3 = 3.93 A
i4 = 2.143 A
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4.3 The Mesh Current Method
(i) General Case
b1 b2 b3 b4 b5 b6
mesh1 -1 1 0 1
0
0
mesh2
0 -1 1 0 -1 0
mesh3
0 0 0 -1 1
outer mesh 1 0 -1 0
1
0 -1
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4.3 The Mesh Current Method
Mik = 1 (-1) if branch k is included in mesh i and the
branch current direction is in the same
(opposite) direction as the mesh current.
Mik = 0
if branch k is not included in mesh i.
Reduced mesh to branch matrix M
M=
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-1
1
0
1
0
0
0
-1
1
0
-1
0
0
0
0
-1
1
1
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4.3 The Mesh Current Method
• KCL is automatically satisfied
‚ KVL
MV B = 0
(A)
VB = [ v1 v2 v3 v4 v5 v6 ] T
 -1 1 0
 0 -1 1

 0 0 0
1 0 0
0 -1 0 
-1 1 1 
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 v1 
v 
 2  0 
 v3   
  = 0 
 v4  0 
 v5   
 
 v6 
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4.3 The Mesh Current Method
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-v1 + v2 + v4 = 0
-v2 + v3 – v5 = 0
-v4 + v5 + v6 = 0
:
:
:
KVL for mesh 1
KVL for mesh 2
KVL for mesh 3
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4.3 The Mesh Current Method
The branch currents can be expressed in terms of
mesh currents
IB = MTJ
(B)
IB [ i1 i2 i3 i4 i5 i6 ]T
J = [ j1 j2 j3 ]T
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4.3 The Mesh Current Method
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i1 = -j1
i2 = j1-j2
i3 = j2
i4 = j1-j3
i5 = -j2+j3
i6 = j3
84
4.3 The Mesh Current Method
ik
A
vk
+
-
B
vk = ESK + (ik - Isk) RK
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4.3 The Mesh Current Method
In matrix form
VB = [R]IB + ES - [R]IS
(C)
[R] = diag [R1 R2 …… Rb]
branch resistance matrix
ES = [ES1 ES2 …… ESb]T branch voltage source vector
IS = [IS1 IS2 …… ISb]T
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branch current source vector
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4.3 The Mesh Current Method
Multiply M on both sides of (C) and use relation (A) to
get
0 = MVB = M[R] IB + M ES – M[R]IS
(D)
Substitute (B) into (D)
M[R]MTJ = M[R]IS – MES
AX = b
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4.3 The Mesh Current Method
(j) Nodal Versus Mesh Analysis
Both methods provide a systematic way of analyzing
a complex network.
Which method should be chosen for a particular problem?
The choice is based on two considerations.
(1)First consideration: choose the method which
contains minimum number of unknowns that
requires simultaneous solution.
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4.3 The Mesh Current Method
(1) First consideration: choose the method which
contains minimum number of unknowns that
requires simultaneous solution.
For a circuit contains b branches, n nodes,
among them there are mv voltage sources
and mi current sources, including dependent
source, then the number of unknowns to be
solved simultaneously is
n-l-mv : for nodal analysis
b-(n-1)-mi : for mesh analysis
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4.3 The Mesh Current Method
(2)Second consideration: Depends on the solution
required, if nodal voltages are required, it may be
expedient to use nodal analysis. If branch or mesh
currents are required it may be better to use mesh
analysis.
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4.4 Fundamental Loop Analysis
n
Two other variations besides nodal and mesh
analysis are introduced, namely the fundamental
loop analysis and the fundamental cutset analyses.
n
They are useful for understanding how to write
the state equation of a circuit.
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4.4 Fundamental Loop Analysis
Theorem
Given an oriented graph G of n nodes and b
branches, and a tree T of G, then
1. Between any pair of nodes there is a unique path
along the tree.
2. There are n-1 tree branches and b-n+1 links.
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4.4 Fundamental Loop Analysis
Theorem
3. Every link of T and the corresponding unique tree
path between its two terminals constitute a unique
loop, called the fundamental loop associated with
this link.
4. Every tree branch of T together with some links
defines a unique cutset of G, called the
fundamental cutset associated with this tree
branch.
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4.4 Fundamental Loop Analysis
Given the oriented
graph G of n (= 4) nodes
and b (= 6) branches.
G
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4.4 Fundamental Loop Analysis
I2
Step 1
J2
I6
I4
`
J3
J1
I1
I5
I3
Pick an arbitrary tree T.
Number the links from 1
to l and then tree branches
from l +1 to b.
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4.4 Fundamental Loop Analysis
Step 2
Choose link currents J1, J2, J3 as unknowns
and adopt each link current direction as the
reference of the corresponding fundamental
loop current.
Write down KVL equations for the l
fundamental loops.
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4.4 Fundamental Loop Analysis
Loop1  1 0 0 -1 -1 0 


Loop2  0 1 0 -1 0 -1
Loop3  0 0 1 0 1 -1
V1 
V 
 2  0
V3   
  = 0
V4  0 
V5   
 
V6 
branch voltage vector VBT @ [V1 V2 V3 V4 V5 V6 ]
T
branch current vector I B @ [ I1
I2
I3
I4
I5
I6 ]
In matrix form L VB = 0 .................... ( A)
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4.4 Fundamental Loop Analysis
Step 3 KCL
If every branch current is formed by linear combination of
loop currents, then KCL will be satisfied automatically.
I1 = J 1
I2 = J2
In matrix form
I3 = J 3
I4 = - J1 - J 2
I B = LT J .............. ( B )
I5 = - J1 + J 3
I6 = - J 2 - J 3
J T = [ J1
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J2
J 3 ] Link current vector
98
4.4 Fundamental Loop Analysis
Step 4 Component model
branch k
k
k
sk
sk
K
∴ Vk = ( I k - I sk ) Rk + Vsk , k = 1, 2,LL , b
or VB = [ R ] ( I B - I s ) + Vs .................... ( C )
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4.4 Fundamental Loop Analysis
Multiply (C) by L matrix and substitute (A) and (B)
L VB = L [ R ] I B - L [ R ] I s + L Vs
0 = L [ R ] LT J - L [ R ] I s + L Vs
∴ L [ R ] LT J = L [ R ] I s - L Vs
⇒ Z J @ es ............... ( D )
[R] ：diagonal matrix
Z ：l × l matrix, called loop impedance matrix
es ：equivalent loop voltage source vector
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4.4 Fundamental Loop Analysis
Step 5
Solve equation (D) and calculate IB and VB.
Note that
1. Link currents are used as the unknowns.
2. The number of links is the same as that of the
mesh currents. Because they are all optimal
methods in terms of using minimum number of
unknown current variables.
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4.4 Fundamental Loop Analysis
3. The resulting circuit loop equations depend on
the choice of a tree.
4. Compared with mesh analysis, the fundament
loop analysis is more general, i.e. not restricted
to planar circuits, but requires defining a tree.
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4.5 Fundamental Cutset Analysis
Use the same example as an demonstration.
2
+ e1 - b + e3 -
a
4
1
+
e2
-
5
6
3
Given the oriented graph G.
Step 1
c
Pick a tree T. Number the
links from 1 to l and then
tree branches from l +1 to b.
d
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4.5 Fundamental Cutset Analysis
Step 2
Choose the tree branch voltages e1, e2, e3 as
unknowns and adopt the tree branch current as the
reference direction of the associated fundamental
cutset. Write down KCL equations for the n-1
fundamental cutsets.
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Cutset 1 for branch 4
i4 + i1 + i2 = 0
Cutset 2 for branch 5
i5 + i1 - i3 = 0
Cutset 3 for branch 6
i6 + i2 + i3 = 0
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4.5 Fundamental Cutset Analysis
In matrix form
1 1 0 1 0 0 
1 0 -1 0 1 0 


0 1 1 0 0 1 
 i1 
i 
 2  0 
i3   
  = 0 
i4  0 
i5   
 
i6 
C I B = 0 .................... ( A)
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4.5 Fundamental Cutset Analysis
Step 3 KVL
Every branch voltage can be expressed in terms of
the tree branch voltages.
In matrix form
VB = C T e ....................... ( B )
eT = e1 e2 e3 
Tree branch voltage vector
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4.5 Fundamental Cutset Analysis
Step 4 Component model
branch k
I k = Gk (Vk - Vsk ) + I sk , k = 1, 2, 3,L , b
In matrix form
I B = [G ] (VB - Vs ) + I s ................... ( C )
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4.5 Fundamental Cutset Analysis
From (A), (B) and (C)
C I B = C [G ]VB - C [G ]Vs + C I s
0 = C [G ] C T e - C [G ]Vs + C I s
C [G ] C T e = C [G ]Vs - C I s
⇒ Y e @ J s ....................... ( D )
Y ：cutset admittance matrix
[G] ：diagonal conductance matrix
Js ：cutset current source vector
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4.5 Fundamental Cutset Analysis
Step 5
Solve equation (D) and calculate IB and VB.
Note that
1. Tree branch voltages are used as unknowns.
2. The number of tree branches is the same as that
of the nodal analysis. Because they are all
optimal methods in terms of using minimum
number of unknown voltages.
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4.5 Fundamental Cutset Analysis
3. The resulting cutset equations depend on the
choice of a tree..
4. Compared with nodal analysis, the fundament
loop method requires defining a tree.
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Summary
Table 1 Comparison of Nodal and Fundamental Cutset Analyses
Methods
Nodal Analysis
KCL
NI B = 0
KVL
Component Model
CI B = 0
VB = C T et
VB = N T e
I B = I S + [G ](VB − VS ) I B = I S + [G ](VB − VS )
Ax = b
Equation to be Solved
Ax = b
A = N [G ]N
b = N [G ]VS − NI S
T
x=e
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Fundamental
Cutset Analysis
A = C[G ]C T
b = C[G ]VS − CI S
x = et
e : node voltage vector
et : tree branch voltage vector
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Summary
Table 2 Comparison of Mesh and Fundamental Loop Analyses
Methods
Nodal Analysis
Fundamental
Cutset Analysis
KCL
IB = M T Jm
I B = LT J
KVL
MVB = 0
LVB = 0
Component Model
VB = ES + [ R]( I B − I S )
Ax = b
Equation to be Solved
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A = M [ R]M
VB = VS + [ R ]( I B − I S )
Ax = b
b = M [ R]I S − MES
A = L[ R ]LT
b = L[ R ]I S − LVS
x = Jm
x=J
Jm : mesh current vector
J : link current vector
T
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Summary
An interesting and practical circuit
[Problem 4.16]
R
R
R
R
vo
v1
v2
v3
vn
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Summary
By using node-voltage method:
Choose node b as the datum node .
Then
vo − v1 vo − v2 vo − v3
v −v
+
+
+ ... + o n = 0
R
R
R
R
∴ n vo = (v1 + v2 + v3 + ... + vn )
1 n
∴ vo = ∑ vk
n k =1
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Summary
It is an averaging circuit used to find the average
value of n voltages.
Note that there is a common node and all resisters
are equal , called sharing bus method.
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Summary
nObjective 1 : Understand and be able to use the
node-voltage method to solve a
circuit.
nObjective 2 : Understand and be able to use the
mesh-current method to solve a
circuit.
nObjective 3 : Be able to decide whether the nodevoltage method or the mesh –current
method is preferred approach to
solving a particular circuit.
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Summary
nObjective 4 : Understand how a circuit is stored in
a computer and some basic graph
theory.
nObjective 5 : Understand the basic principle of
fundamental loop analysis.
nObjective 6 : Understand the basic principle of
fundamental cutset analysis.
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Summary
n Problem : 4.9
4.17
4.24
4.31
4.39
4.43
n Due within one week.
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