9

Network Topology

9.1 INTRODUCTION
The purpose of network analysis is to find voltage across and current through all the elements. When the
network is complicated and has a large number of nodes and closed paths, network analysis can be done
conveniently by using ‘Network Topology’. This theory does not make any distinction between different
types of physical elements of the network but makes the study based on a geometric pattern of the network.
The basic elements of this theory are nodes, branches, loops and meshes.
Node It is defined as a point at which two or more elements have a common connection.
Branch It is a line connecting a pair of nodes, the line representing a single element or series connected
elements.
Loop Whenever there is more than one path between two nodes, there is a circuit or loop.
Mesh It is a loop which does not contain any other loops within it.

9.2 GRAPH OF A NETWORK

A linear graph is a collection of nodes and branches. The nodes are joined together by branches.
The graph of a network is drawn by first marking the nodes and then joining these nodes by lines which
correspond to the network elements of each branch. All the voltage and current sources are replaced by their
internal impedances. The voltage sources are replaced by short circuits as their internal impedances are zero
whereas current sources are replaced by open circuits as their internal impedances are infinite. Nodes and
branches are numbered. Figure 9.1 shows a network and its associated graphs.
Each branch of a graph may be given an orientation or a direction with the help of an arrow head which
represents the assigned reference direction for current. Such a graph is then referred to as a directed or
oriented graph.
Branches whose ends fall on a node are said to be incident at that node. Degree of a node is defined as
the number of branches incident to it. Branches 2, 3 and 4 are incident at Node 2 in Fig. 9.1(c). Hence, the
degree of Node 2 is 3.
9.2 Network Analysis and Synthesis

(6) (6) (6)

2 2
2 1 3 1 3
1 3 (2) (3) (2) (3)
(2) (3)
(1) (5) (1) (4) (5) (1) (4) (5)
(4)
+
V −

4 4 4

(a) Network (b) Undirected graph (c) Directed or oriented graph

Fig. 9.1 Network and its graphs

9.3 DEFINITIONS ASSOCIATED WITH A GRAPH

1. Planar Graph A graph drawn on a two-dimensional plane is said to be planar if two branches do not
intersect or cross at a point which is other than a node. Figure 9.2 shows such graphs.
(2) (2) 2 (8)
1 3
2
1 3
(3) (4)
(1) (7) (9)
(3) (6)
(1) (5) (6)

4 6
(4) 5 (5)
4
Fig. 9.2 Planar graphs

2. Non-planar Graph A graph drawn on a two-dimensional plane is said to be non-planar if there is

intersection of two or more branches at another point which is not a node. Figure 9.3 shows non-planar
graphs.
1 (2) 2 (2) 2 (3)
1 3
(5) (6)
(1) (4) (1) (7) (4)
(8)
3 4 4 6
(3) (5) 5 (6)
Fig. 9.3 Non-planar graphs

3. Sub-graph It is a subset of branches and nodes of a graph. It is a proper sub-graph if it contains

branches and nodes less than those on a graph. A sub-graph can be just a node or only one branch of the
graph. Figure 9.4 shows a graph and its proper sub-graph.
 9.3 Definitions Associated with a Graph 9.3

(2) (2)
2 2 (5)
1 3 3
(3) (5) 1 3

(1) (4) (6) (4)

4 4
(a) (b)
Fig. 9.4 (a) Graph (b) Proper sub-graph
4. Path It is an improper sub-graph having the following properties:
1. At two of its nodes called terminal nodes, there is incident only one branch of sub-graph.
2. At all remaining nodes called internal nodes, there are incident two branches of a graph.
In Fig. 9.5, branches 2, 5 and 6 together with all the four nodes, constitute a path.
(3)

2
1 3
(2) (4)

(1) (5) (6)

Fig. 9.5 Path

5. Connected Graph A graph is said to be connected if there exists a path between any pair of nodes.
Otherwise, the graph is disconnected.
6. Rank of a Graph If there are n nodes in a graph, the rank of the graph is (n − 1).
7. Loop or Circuit A loop is a connected sub-graph of a connected graph at each node of which are incident
exactly two branches. If two terminals of a path are made to coincide, it will result in a loop or circuit.
Figure 9.6 shows two loops.
(2) 2 (3)
1 3
(2)

(1) (4) 1 2
(1)
4
Loops: {1, 2, 3, 4} Loops: {1, 2}

Fig. 9.6 Loops

Loops of a graph have the following properties:
1. There are at least two branches in a loop.
2. There are exactly two paths between any pair of nodes in a circuit.
3. The maximum number of possible branches is equal to the number of nodes.
9.4 Network Analysis and Synthesis
8. Tree A tree is a set of branches with every node connected to every other node in such a way that
removal of any branch destroys this property.
Alternately, a tree is defined as a connected sub-graph of a connected graph containing all the nodes of
the graph but not containing any loops.
Branches of a tree are called twigs. A tree contains (n − 1) twigs where n is the number of nodes in the
graph. Figure 9.7 shows a graph and its trees.
(6)

2 2 2 (2)
1 3 1 3 1 3
(1) (2) (1)

(3) (4) (5) (4) (3) (5)

(5)

4 4
4
(a) Graph (b) (c)
Twigs: {1, 4, 5} Twigs: {2, 3, 5}
Fig. 9.7 Graph and its trees

Trees have the following properties:

1. There exists only one path between any pair of nodes in a tree.
2. A tree contains all nodes of the graph.
3. If n is the number of nodes of the graph, there are (n − 1) branches in the tree.
4. Trees do not contain any loops.
5. Every connected graph has at least one tree.
6. The minimum terminal nodes in a tree are two.
9. Co-tree Branches which are not on a tree are called links or chords. All links of a tree together
constitute the compliment of the corresponding tree and is called the co-tree.
A co-tree contains b − (n − 1) links where b is the number of branches of the graph.
In Fig. 9.7 (b) and (c) the links are {2, 3, 6} and {1, 4, 6} respectively.

Example 9.1 Draw directed graph of the networks shown in Fig. 9.8.

R1 R2
C1 L
L1
v
L2 C1
R1
R2 C2
+ C3
V − C2 R3

(a) (b)

Fig. 9.8
 9.3 Definitions Associated with a Graph 9.5

Solution For drawing the directed graph,

1. replace all resistors, inductors and capacitors by line segments,
2. replace the voltage source by a short-circuit,
3. assume directions of branch currents, and
4. number all the nodes and branches.
The directed graph for the two networks are shown in Fig. 9.9.
1

1 (1) (3)
(2)

4 2
(5) 5 (4)
(1) (2) (3)
(7)
(6) (8)

2 3
(a) (b)
Fig. 9.9

Example 9.2 Figure 9.10 shows a graph of the network. Show all the trees of this graph.
(1) 2 (2)
1 3

(4)
(3) (5)

4
Fig. 9.10
Solution A graph has many trees. A tree is a connected sub-graph of a connected graph containing all the
nodes of the graph but not containing any loops. Figure 9.11 shows various trees of the given graph.
2 (2) 3 1 (1) 2 3 1 (1) 2 3 1 (1) 2 (2) 3 1 (1) 2 (2) 3
1

(4) (4) (5)

(3) (3) (5) (5) (4)

4 4 4 4 4
1 2 3 1 2 (2) 3 1 (1) 2 (2) 3

(3) (4) (5)

(3) (5)
(3)

4 4 4

Fig. 9.11
9.6 Network Analysis and Synthesis

9.4 INCIDENCE MATRIX

A linear graph is made up of nodes and branches. When a graph is given, it is possible to tell which branches
are incident at which nodes and what are its orientations relative to the nodes.

9.4.1 Complete Incidence Matrix (Aa)

For a graph with n nodes and b branches, the complete incidence matrix is a rectangular matrix of order n × b.
Elements of this matrix have the following values:
aij = 1, if branch j is incident at node i and is oriented away from node i.
= −1, if branch j is incident at node i and is oriented towards node i.
= 0, if branch j is not incident at node i.
For the graph shown in Fig. 9.12, branch 1 is incident at nodes 1 and 4. It is oriented away from Node 1 and
oriented towards Node 4. Hence, a11 = 1 and a41 = −1. Since branch 1 is not incident at nodes 2 and 3, a21 =
0 and a31 = 0. Similarly, other elements of the complete incidence matrix are written.
(6)
Nodes Branches →
2
1 3 ↓ 1 2 3 4 5 6
(2) (4)
1 1 1 0 0 0 1
(1) (3) (5)
2 0 −1 1 −1 0 0
3 0 0 0 1 1 −1
4 −1 0 −1 0 −1 0
4
Fig. 9.12 Graph
The complete incidence matrix is

⎡ 1 1 0 0 0 1⎤
⎢ 0 −1 1 −1 0 0 ⎥
Aa = ⎢ ⎥
⎢ 0 0 0 1 1 −1⎥
⎢⎣ −1 0 −1 0 −1 0 ⎥⎦
It is seen from the matrix Aa that the sum of the elements in any column is zero. Hence, any one row of the
complete incidence matrix can be obtained by the algebraic manipulation of other rows.

9.4.2 Reduced Incidence Matrix (A)

The reduced incidence matrix A is obtained from the complete incidence matrix Aa by eliminating one of the
rows. It is also called incidence matrix. It is of order (n − 1) × b.
Eliminating the third row of matrix Aa,

⎡ 1 1 0 0 0 1⎤
A = ⎢ 0 −1 1 −1 0 0 ⎥
⎢ ⎥
⎣ −1 0 −1 0 −1 0 ⎦

When a tree is selected for the graph as shown in Fig. 9.13, the incidence matrix is obtained by arranging
a column such that the first (n − 1) column corresponds to twigs of the tree and the last b − (n − 1) branches
corresponds to the links of the selected tree.
 9.4 Incidence Matrix 9.7

2
1 3 Twigs Links
(2) (4)
2 3 4 1 5 6
Twigs: {2, 3, 4}
(3) Links: {1, 5, 6} ⎡ 1 0 0 1 0 1⎤
A = ⎢ −1 1 1 0 0 0⎥
⎢ ⎥
4 ⎣ 0 −11 0 −11 1 0⎦
Fig. 9.13 Tree

The matrix A can be subdivided into submatrices At and Al.

A [ At : Al ]
Where At the is twig matrix and Al is the link matrix.

9.4.3 Number of Possible Trees of a Graph

Let the transpose of the reduced incidence matrix A be AT. It can be shown that the number of possible trees
of a graph will be given by
Number of possible trees = |AAT|
For the graphs shown in Fig. 9.12, the reduced incidence matrix is given by

⎡ 1 1 0 0 0 1⎤
A = ⎢ 0 −1 1 −1 0 0 ⎥
⎢ ⎥
⎣ −1 0 −1 0 −1 0 ⎦
Then transpose of this matrix will be
⎡ 1 0 −1⎤
⎢ 1 −1 0 ⎥
⎢ ⎥
0 1 −1⎥
AT = ⎢
⎢0 −1 0 ⎥
⎢0 0 −1⎥
⎢ ⎥
⎣ 1 0 0⎦

Hence, number of all possible trees of the graph

⎡ 1 0 −1⎤
⎢ 1 −1 0 ⎥
⎡ 1 1 0 0 0 1⎤ ⎢0 ⎥
1 −1⎥ ⎡
3 −1 −1⎤
AAT = ⎢ 0 −1 1 −1 0 0 ⎥ ⎢ = ⎢ −1 3 −1⎥
⎢ ⎥ ⎢0 −1 0 ⎥ ⎢ ⎥
⎣ −1 0 −1 0 −1 0 ⎦ ⎢ ⎥ ⎣ −1 −1 3⎦
0 0 −1
⎢ ⎥
⎣ 1 0 0⎦

3 −1 −1
T
| | = −1 3 −1 = 3(( ) + ( )( ) − 1(( ) = 16
−1 −1 3
Thus, 16 different trees can be drawn.
9.8 Network Analysis and Synthesis

9.5 LOOP MATRIX OR CIRCUIT MATRIX

When a graph is given, it is possible to tell which branches constitute which loop or circuit. Alternately, if a
loop matrix or circuit matrix is given, we can reconstruct the graph.
For a graph having n nodes and b branches, the loop matrix Ba is a rectangular matrix of order b columns
and as many rows as there are loops.
Its elements have the following values:
bij = 1, if branch j is in loop i and their orientations coincide.
= − 1, if branch j is in loop i and their orientations do not coincide.
= 0, if branch j is not in loop i.
A graph and its loops are shown in Fig. 9.14.
(6)

2 Loop 1: {1, 2, 3}
1 3 Loop 2: {3, 4, 5}
(2) (4) Loop 3: {2, 4, 6}
Loop 4: {1, 2, 4, 5}
(3) Loop 5: {1, 5, 6}
(1) (5)
Loop 6: {2, 3, 5, 6}
Loop 7: {1, 3, 4, 6}
4
Fig. 9.14 Graph
All the loop currents are assumed to be flowing in a clockwise direction.
Loops Branches →
↓ 1 2 3 4 5 6
1 −1 1 1 0 0 0
2 0 0 −1 −1 1 0
3 0 −1 0 1 0 1
4 −1 1 0 −1 1 0
5 −1 0 0 0 1 1
6 0 −1 −1 0 1 1
7 −1 0 1 1 0 1
⎡ −1 1 1 0 0 0⎤
⎢ 0 0 −1 −1 1 0⎥
⎢ ⎥
⎢ 0 −1 0 1 0 1⎥
Ba = ⎢ −1 1 0 −1 1 0⎥
⎢ −1 0 0 0 1 1⎥
⎢ 0 −1 −1 0 1 1⎥⎥
⎢
⎣ −1 0 1 1 0 1⎦

9.5.1 Fundamental Circuit (Tieset) and Fundamental Circuit Matrix

When a graph is given, first select a tree and remove all the links. When a link is replaced, a closed loop or
circuit is formed. Circuits formed in this way are called fundamental circuits or f-circuits or tiesets.
Orientation of an f-circuit is given by the orientation of the connecting link. The number of f-circuits is
same as the number of links for a graph. In a graph having b branches and n nodes, the number of f-circuits or
tiesets will be (b − n + 1). Figure 9.15 shows a tree and f-circuits (tiesets) for the graph shown in Fig. 9.14.
 9.5 Loop Matrix or Circuit Matrix 9.9

(6)
1 2 3

(2) (4) (2) (4) (2) (4) (2) (4)

(3) (1) (3) (3) (5) (3) tieset 1: {1, 2, 3}

tieset 5: {5, 3, 4}
tieset 6: {6, 2, 4}
4 tieset 1 tieset 5 tieset 6
(a) Tree (b) f-circuits (tiesets)

Fig. 9.15 Tree and f-circuits

Here, b = 6 and n = 4.
Number of tiesets = b − n + 1 = 6 − 4 + 1 = 3
f-circuits are shown in Fig. 9.15. The orientation of each f-circuit is given by the orientation of the
corresponding connecting link.
The branches 1, 2 and 3 are in the tieset 1. Orientation of tieset 1 is given by orientation of branch 1. Since
the orientation of branch 1 coincides with orientation of tieset 1, b11 = 1. The orientations of branches 2 and 3
do not coincide with the orientation of tieset 1, Hence, b12= − 1 and b13 = − 1. The branches 4, 5 and 6 are not
in tieset 1. Hence, b14 = 0, b15 = 0 and b16 = 0. Similarly, other elements of the tieset matrix are written.
Then, the tieset schedule will be written as

Tiesets Branches →
↓ 1 2 3 4 5 6
1 1 −1 −1 0 0 0
5 0 0 −1 −1 1 0
6 0 −1 0 1 0 1
Hence, an f-circuit matrix or tieset matrix will be given as

⎡ 1 −1 −1 0 0 0 ⎤
B = ⎢0 0 −1 −1 1 0 ⎥
⎢ ⎥
⎣0 −1 0 1 0 1⎦

Usually, the f-circuit matrix B is rearranged so that the first (n − 1) columns correspond to the twigs and
b − (n − 1) columns to the links of the selected tree.
Twigs
gs Links
2 3 4 1 5 6
⎡−1
1 1 0 1 0 0⎤
B=⎢ 0 1 −1 0 1 0⎥
⎢ ⎥
⎣−1 0 1 0 0 1⎦

The matrix B can be partitioned into two matrices Bt and Bl.

B [ Bt : Bl ] [ Bt : U ]
where Bt is the twig matrix, Bl is the link matrix and U is the unit matrix.
9.10 Network Analysis and Synthesis
9.5.2 Orthogonal Relationship between Matrix A and Matrix B
For a linear graph, if the columns of the two matrices Aa and Ba are arranged in the same order, it can be
shown that
Aa BaT = 0
or Ba AaT = 0
The above equations describe the orthogonal relationship between the matrices Aa and Ba.
If the reduced incidence matrix A and the f-circuit matrix B are written for the same tree, it can be shown
that
A BT = 0
or B AT = 0
These two equations show the orthogonal relationship between matrices A and B.

9.6 CUTSET MATRIX 1

(2)
2

Consider a linear graph. By removing a set of branches without affecting the nodes,
two connected sub-graphs are obtained and the original graph becomes unconnected. (1) (3) (5)
The removal of this set of branches which results in cutting the graph into two parts (4)
are known as a cutset. The cutset separates the nodes of the graph into two groups, 4 3
each being in one of the two groups. Fig. 9.16 Graph
Figure 9.16 shows a graph.
Branches 1, 3 and 4 will form a cutset. This set of branches separates the graph into two parts. One having
an isolated node 4 and other part having branches 2 and 5 and nodes 1, 2 and 3.
Similarly, branches 1 and 2 will form a cutset. Each branch of the cutset has one of its terminals incident
at a node in one part and its other end incident at other nodes in the other parts. The orientation of a cutset is
made to coincide with orientation of defining branch.
For a graph having n nodes and b branches, the cutset matrix Qa is a rectangular matrix of order b columns
and as many rows as there are cutsets. Its elements have the following values:
qij = 1, if the branch j is in the cutset i and the orientation coincide.
= −1, if the branch j is in the cutset i and the orientations do not coincide.
= 0, if the branch j is not in the cutset i.
Figure 9.17 shows a directed graph and its cutsets.
(6)

Cutsets Branches →
1
2
3 Cutset 1: {1, 2, 6} ↓ 1 2 3 4 5 6
Cutset 2: {2, 3, 4}
(2) (4) Cutset 3: {3, 1, 5} 1 1 1 0 0 0 1
Cutset 4: {4, 5, 6} 2 0 1 −1 1 0 0
Cutset 5: {5, 2, 3, 6} 3 1 0 1 0 1 0
(3)
(1) Cutset 6: {6, 1, 3, 4} 4 0 0 0 1 1 −1
(5)
5 0 −1 1 0 1 −1
6 1 0 1 −1 0 1
4

Fig. 9.17 Directed graph

For the cutset 2, which cuts the branches 2, 3 and 4 and is shown by a dotted circle, the entry in the cutset
schedule for the branch 2 is 1, since the orientation of this cutset is given by the orientation of the branch 2
and hence it coincides. The entry for branch 3 is −1 as orientation of branch 3 is opposite to that of cutset 2,
 9.6 Cutset Matrix 9.11

i.e., branch 2 goes into cutset while branch 3 goes out of cutset. The entry for branch 4 is 1 as the branch 2
and the branch 4 go into the cutset. Thus their orientations coincide.
Hence, the cutset matrix Qa is given as
⎡1 1 0 0 0 1⎤
⎢0 1 −1 1 0 0⎥
⎢ ⎥
1 0 1 0 1 0⎥
Qa = ⎢
⎢ 0 0 0 1 1 −1⎥
⎢0 −1 1 0 1 −1⎥
⎢ 1 0 1 −1 0 1⎥⎦
⎣

9.6.1 Fundamental Cutset and Fundamental Cutset Matrix

When a graph is given, first select a tree and note down its twigs. When a twig is removed from the tree, it
separates a tree into two parts (one of the separated part may be an isolated node). Now, all the branches
connecting one part of the disconnected tree to the other along with the twig removed, constitutes a cutset.
This set of branches is called a fundamental cutset or f-cutset. A matrix formed by these f-cutsets is called an
f-cutset matrix. The orientation of the f-cutset is made to coincide with the orientation of the defining branch,
i.e., twig. The number of f-cutsets is the same as the number of twigs for a graph.
Figure 9.18 shows a graph, selected tree and f-cutsets corresponding to the selected tree.
(6) (6)
1 2 3
(2) (4) (2) (4) (2) (4)

f-cutset 2: {2, 1, 6}
(1) (3) (5) (3) (1) (3) (5) f-cutset 3: {3, 1, 5}
f-cutset 4: {4, 5, 6}

4
(a) Graph (b) Tree (c) f-cutsets

Fig. 9.18 Graph, selected tree and f-cutsets

The branches 2, 1, and 6 are in the f-cutset 2. Orientation of f-cutset 2 is given by orientation of the branch
2 which is moving away from the f-cutset 2. Since the orientations of branches 2, 1 and 6 coincide with the
orientation of the f-cutset 2, q11 = 1, q12 = 1 and q16 = 0. The branches 3, 4 and 5 are not in the f-cutset 2.
Hence, q13 = 0, q14 = 0 and q15 = 0.
Similarly, other elements of the f-cutset matrix are written.
The cutset schedule is
f-cutsets Branches →
↓ 1 2 3 4 5 6
2 1 1 0 0 0 1
3 1 0 1 0 1 0
4 0 0 0 1 1 −1
Hence, the f-cutset matrix Q is given by
⎡ 1 1 0 1⎤
0 0
Q=⎢ 1 0 1 0⎥
1 0
⎢ ⎥
⎣ 0 0 1 −1⎦
0 1
The f-cutset matrix Q is rearranged so that the first (n − 1) columns correspond to twigs and b − (n − 1)
columns to links of the selected tree.
9.12 Network Analysis and Synthesis

Twigs Links
2 3 4 1 5 6
⎡ 1 0 0 1 0 1⎤
Q=⎢ 0 1 0 1 1 0⎥
⎢ ⎥
⎣ 0 0 1 0 1 −11⎦

The matrix Q can be subdivided into matrices Qt and Ql.

Q Qt : Q U : Ql ]
where Qt is the twig matrix, Ql is the link matrix and U is the unit matrix.

9.6.2 Orthogonal Relationship between Matrix B and Matrix Q

For a linear graph, if the columns of two matrices Ba and Qa are arranged in the same order, it can be shown
that
Q BaT = 0
or B QaT = 0
If the f-circuit matrix B and the f-cutset matrix Q are written for the same selected tree, it can be shown that
B QT = 0
or Q BT = 0
These two equations show the orthogonal relationship between matrices A and B.

9.7 RELATIONSHIP AMONG SUBMATRICES OF A, B AND Q

Arranging the columns of matrices A, B and Q with twigs for a given tree first and then the links, we get the
partitioned forms as
A [ At : Al ]
B [ Bt : Bl ] [ Bt : U ]
Q Qt : Q U : Ql ]

From the orthogonal relation, ABT = 0,

⎡ Bt T ⎤
AB = [ At : Al ] ⎢ … ⎥
T
⎢ ⎥
⎢⎣ Bl T ⎥⎦
At Bt T + Al Bl T = 0
At Bt T = − Al Bl T

Since At is non-singular, i.e., |A| ≠ 0, At−1 exists.

Premultiplying with At−1,
Bt T At−1 Al Bl T
Bt Bl ( At 1 Al )T
Since Bl is a unit matrix
Bt ( At−1. Al )T
 9.7 Relationship Among Submatrices of A, B and Q 9.13

Hence, matrix B is written as

B [ ( At−1 Al )T : U ] …(9.1)

We know that ABT = 0

Al Bl T = − At Bt T
T −1
Postmultiplying with ( ) ,
Al At Bt T ( Bl T ) −1 = − At Bt T (B
( Bl 1 )T At ( Bl 1 Bt )T
Hence matrix A can be written as
A [ Al : At ( Bl−1 Bt )T ]
…(9.2)
= At [U : − ( Bl 1 Bt )T ]
Similarly we can prove that
Q U : − ( Bl 1 Bt )T ] …(9.3)
From Eqs (9.2) and (9.3), we can write
A At Q
Q At−1 A At−1[ At : Al ] = [U : At−11 Al ]
We have shown that Bt ( At−1. Al )T
Bt T ( At−1 Al )
Hence, Q can be written as
Q U : − Bt T ]
Q Bt T

Example 9.3 For the circuit shown in Fig. 9.19, draw the oriented graph and write the (a) incidence
matrix, (b) tieset matrix, and (c) f-cutset matrix.
R5

R2 L1 R4 C

R1 R6 R3
+
V − I
(4)
Fig. 9.19
1 2
Solution For drawing the oriented graph, (3)
1. replace all resistors, inductors and capacitors by line segments,
(1) (2)
2. replace the voltage source by short circuit and the current source by an open
circuit,
3. assume the directions of branch currents arbitrarily, and
4. number all the nodes and branches. 3

The oriented graph is shown in Fig. 9.20. Fig. 9.20

9.14 Network Analysis and Synthesis
(a) Incidence Matrix (A)
Nodes Branches → ⎡ −1 0 −1 1⎤
↓ 1 2 3 4 Aa = ⎢ 0 1 1 −1⎥
1 −1 0 −1 1 ⎢ ⎥
⎣ 1 −1 0 0 ⎦
2 0 1 1 −1
3 1 −1 0 0
Eliminating the third row from the matrix Aa, we get the incidence matrix A.
⎡ −1 0 −1 1⎤
A= ⎢
⎣ 0 1 1 −1⎥⎦
(b) Tieset Matrix (B)
The oriented graph, selected tree and tiesets are shown in Fig. 9.21.
(4)

1 2 Links: {3, 4}
(3) Tieset 3: {3, 1, 2} 1 2 3 4
Tieset 4: {4, 1, 2}
(1) (2) 3 ⎡ −11 1 1 0⎤
B= ⎢
4⎣ 1 1 0 1⎥⎦

Fig. 9.21
(c) f-cutset Matrix (Q)
The oriented graph, selected tree and f-cutsets are shown in Fig. 9.22.
(4)
1 2
Twigs: {1, 2}
f-cutset 1: {1, 3, 4} 1 2 3 4
(3)
f-cutset 2: {2, 3, 4} 1⎡1 0 1 1⎤
Q= ⎢
(1) (2)
2 ⎣0 1 1 1⎥⎦

Fig. 9.22

Example 9.4 For the network shown in Fig. 9.23, draw the oriented graph and write the
(a) incidence matrix, (b) tieset matrix, and (c) f-cutset matrix.
2Ω
2Ω
4Ω 10 Ω I

1F 1H
2Ω 2Ω
4F
+ +
V1 − − V2

Fig. 9.23
 9.7 Relationship Among Submatrices of A, B and Q 9.15

Solution For drawing the oriented graph, (6) 5 (7)

1. replace all resistors, inductors and capacitors by line segments, 2
2. replace all voltage sources by short circuits and current source 1 3
(1) (2)
by an open circuit,
3. assume directions of branch currents arbitrarily, and (3)
(4) (5)
4. number all the nodes and branches.
The oriented graph is shown in Fig. 9.24.
4
(a) Incidence Matrix (A)
Fig. 9.24
Nodes Branches →
↓ 1 2 3 4 5 6 7 ⎡ 1 0 0 1 0 1 0⎤
⎢ −1 −1 1 0 0 0 0 ⎥
1 1 0 0 1 0 1 0 ⎢ ⎥
2 −1 −1 1 0 0 0 0 Aa = ⎢ 0 1 0 0 1 0 1⎥
3 0 1 0 0 1 0 1 ⎢ 0 0 −1 −1 −1 0 0 ⎥
4 0 0 −1 −1 −1 0 0 ⎢⎣ 0 0 0 0 0 −1 −1⎥⎦
5 0 0 0 0 0 −1 −1
Eliminating the last row from the matrix Aa, we get the incidence matrix A.

⎡ 1 0 0 1 0 1 0⎤
⎢ −1 −1 1 0 0 0 0⎥
A= ⎢ ⎥
⎢ 0 1 0 0 1 0 1⎥
⎢⎣ 0 0 −1 −1 −1 0 0 ⎥⎦
(b) Tieset Matrix (B)
The oriented graph, selected tree and tiesets are shown in Fig. 9.25.
(6) 5 (7)

2
1 3 Links: {2, 4, 7} 1 2 3 4 5 6 7
(1) (2) Tieset 2: {2, 3, 5}
Tieset 4: {4, 1, 3} 2⎡ 0 1 1 0 −1 0 0⎤
(4)
(3)
(5)
Tieset 7: {7, 6, 1, 3, 5} B = 4 ⎢ −1 0 1 1 0 0 0⎥
⎢ ⎥
7⎣ 1 0 1 0 −11 1 1⎦

4
Fig. 9.25
(c) f-cutset Matrix (Q)
The oriented graph, selected tree and f-cutsets are shown in Fig. 9.26.
5
(6) (7)
2 Twigs: {1, 3, 5, 6} 1 2 3 4 5 6 7
1 f-cutset 1: {1, 4, 7}
(1) (2) 3
f-cutset 3: {3, 4, 7, 2}
1⎡1 0 0 1 0 0 −1⎤
(3) f-cutset 5: {5, 2, 7} 3 ⎢0 −1 1 1 0 0 −1⎥
(4) (5)
Q= ⎢ ⎥
f-cutset 6: {6, 7} 5 ⎢0 1 0 0 1 0 1⎥
6 ⎢⎣0 0 0 0 0 1 1⎥⎦
4

Fig. 9.26
9.16 Network Analysis and Synthesis

Example 9.5 For the circuit shown in Fig. 9.27, draw the oriented graph and write (a) incidence
matrix, (b) tieset matrix, and (c) cutset matrix.

r2 L1 r4 r5

r1 r6
r3

+
V C1 L2
−

Fig. 9.27

1 (4)
Solution For drawing the oriented graph, 2

1. replace all resistors, inductors and capacitors by line segments,

2. replace voltage source by short circuit and current source by an open (1) (2) (3)
circuit,
3. assume directions of branch currents arbitrarily, and
4. number the nodes and branches. 3
The oriented graph is shown in Fig. 9.28. Fig. 9.28
(a) Incidence Matrix (A)
Nodes Branches →
↓ 1 2 3 4
1 −1 1 0 −1 ⎡ −1 1 0 −1⎤
2 0 0 1 1 Aa = ⎢ 0 0 1 1⎥
⎢ ⎥
3 1 −1 −1 0 ⎣ 1 −1 −1 0 ⎦

Eliminating the third row from the matrix Aa, we get the incidence matrix A.

⎡ −1 1 0 −1⎤
A= ⎢
⎣ 0 0 1 1⎥⎦
(b) Tieset Matrix (B)
The oriented graph, selected tree and tiesets are shown in Fig. 9.29.
1 (4)
2

Links: {1, 3}
(1) (2)
Tieset 1: {1, 2} 1 2 3 4
(3) Tieset 3: {3, 2, 4}
1⎡1 1 0 0⎤
B= ⎢
3 ⎣0 1 1 1⎥⎦
3
Fig. 9.29
(c) f-cutset Matrix (Q)
The oriented graph, selected tree and f-cutsets are shown in Fig. 9.30.
 9.7 Relationship Among Submatrices of A, B and Q 9.17

1 (4)
2 Twigs: {2, 4}
f-cutset 2: {2, 1, 3}
f-cutset 4: {4, 3} 1 2 3 4
(1) (2) (3)
2 ⎡ −1 1 1 0 ⎤
Q= ⎢
4 ⎣ 0 0 1 1⎥⎦

Fig. 9.30

Example 9.6 For the circuit shown in Fig. 9.31, (a) draw its graph, (b) draw its tree, and (c) write
the fundamental cutset matrix.
2Ω

1F 1F

1Ω
I 1Ω 1Ω
1H

(3)
Fig. 9.31
2
1 3
Solution (2) (4)
(a) For drawing the oriented graph,
(5)
1. replace all resistors, inductors and capacitors by line segments, (1) (6)
2. replace the current source by an open circuit,
3. assume directions of branch currents, and
4. number all the nodes and branches. 4

The oriented graph is shown in Fig. 9.32. Fig. 9.32

(b) Tree
The oriented graph and its selected tree are shown in Fig. 9.33.
(3)

2
1 3
(2) (4) Twigs: {2, 4, 5}
f-cutset 2: {2, 1, 3}
(5) f-cutset 4: {4, 3, 6}
(1) (6)
f-cutset 5: {5, 1, 6}

Fig. 9.33
(c) Fundamental Cutset Matrix (Q)
1 2 3 4 5 6
2 ⎡ 1 1 1 0 0 0⎤
Q = 4 ⎢0 0 −1 1 0 1⎥
⎢ ⎥
5 ⎣ 1 0 0 0 1 1⎦
9.18 Network Analysis and Synthesis

Example 9.7 The graph of a network is shown in Fig. 9.34. Write the (a) incidence matrix,
(b) tieset matrix, and (c) f-cutset matrix.

(2)
1 3
(4) (5)
2
(1) (6) (3)

Fig. 9.34
Solution
(a) Incidence Matrix (A)
1 2 3 4 5 6
1 ⎡ −1 1 0 1 0 0⎤
2⎢ 0 0 0 −1 1 1⎥
Aa = ⎢ ⎥
3⎢ 0 1 1 0 −1 0⎥
4 ⎢⎣ 1 0 −11 0 0 1⎥⎦

The incidence matrix A is obtained by eliminating any row from the matrix Aa.

⎡ −1 1 0 1 0 0⎤
A = ⎢ 0 0 0 −1 1 1⎥
⎢ ⎥
⎣ 0 −1 1 0 −1 0 ⎦
(b) Tieset, Matrix (B)
The oriented graph, selected tree and tiesets are shown in Fig. 9.35.
(2)
1 3
(4) (5) Links: {1, 2, 3} 1 2 3 4 5 6
2 Tieset 1: {1, 4, 6}
Tieset 2: {2, 4, 5} 1⎡1 0 0 1 0 1⎤
(1)
(6)
(3) Tieset 3: {3, 5, 6} B = 2 ⎢0 1 0 1 −1 0⎥
⎢ ⎥
3 ⎣0 0 1 0 1 1⎦

Fig. 9.35
(c) f-cutset Matrix (Q)
The oriented graph, selected tree and f-cutsets are shown in Fig. 9.36.
(2)
1 3
(4) (5)
2
1 2 3 4 5 6
Twigs: {4, 5, 6}
(3)
f-cutset 4: {4, 1, 2} 4 ⎡ −1 1 0 1 0 0 ⎤
(1)
(6) f-cutset 5: {5, 2, 3} Q = 5 ⎢ 0 1 −1 0 1 0 ⎥
f-cutset 6: {6, 1, 3} ⎢ ⎥
6 ⎣ −1 0 1 0 0 1⎦

Fig. 9.36
 9.7 Relationship Among Submatrices of A, B and Q 9.19

Example 9.8 For the graph shown in Fig. 9.37, write the incidence matrix, tieset matrix and f-cutset
matrix.
(5)

1 2 3 4
(2) (3) (4)

(6)
(1)
(7)

Fig. 9.37
Solution
(a) Incidence Matrix (A)
1 2 3 4 5 6 7
1⎡ 1 1 0 0 0 0 0⎤
2⎢ 0 1 1 0 1 0 0⎥
⎢ ⎥
Aa = 3 ⎢ 0 0 −1 1 0 0 0⎥
4⎢ 0 0 0 −11 1 1 1⎥
5 ⎢⎣ −1 0 0 0 0 −1 1⎥⎦

The incidence matrix A is obtained by eliminating any row from the matrix Aa.

⎡1 1 0 0 0 0 0⎤
⎢0 −1 1 0 1 0 0⎥
A= ⎢ ⎥
⎢0 0 −1 1 0 0 0⎥
⎢⎣0 0 0 −1 −1 1 −1⎥⎦

(b) Tieset Matrix (B) 1 2 3 4 5 6 7

Links: {5, 6, 7}
5⎡ 0 0 −1 −1 1 0 0 ⎤
Tieset 5: {5, 3, 4}
B = 6 ⎢ −1 1 1 1 0 1 0⎥
Tieset 6: {6, 1, 2, 3, 4} ⎢ ⎥
Tieset 7: {7, 1, 2, 3, 4} 7⎣ 1 1 1 1 0 0 1⎦

(c) f-cutset Matrix (Q)

The oriented graph, selected tree and f-cutsets are shown in Fig. 9.38.
(5)

1 2 3
4
(2) (3) (4) 1 2 3 4 5 6 7
Twigs: {1, 2, 3, 4}
1 ⎡1 0 0 0 0 1 1⎤
f-cutset 1: {1, 6, 7}
(6) f-cutset 2: {2, 6, 7} 2 ⎢0 1 0 0 0 −1 1⎥
(1) Q= ⎢ ⎥
f-cutset 3: {3, 5, 6, 7} 3 ⎢0 0 1 0 1 1 1⎥
(7) f-cutset 4: {4, 5, 6, 7} 4 ⎢⎣0 0 0 1 1 1 1⎥⎦
5

Fig. 9.38
9.20 Network Analysis and Synthesis
Example 9.9 For the graph shown in Fig. 9.39, write the incidence matrix, tieset matrix and
f-cutset matrix.
(1)
1 2
(5) (6)

(4) (2)
5
(8) (7)
4 3
(3)

Fig. 9.39
Solution
(a) Incidence Matrix (A)
1 2 3 4 5 6 7 8
1⎡ 1 0 0 1 1 0 0 0⎤
2 ⎢ −1 1 0 0 0 1 0 0⎥
⎢ ⎥
Aa = 3 ⎢ 0 1 1 0 0 0 1 0⎥
4⎢ 0 0 −1 1 0 0 0 1⎥
5 ⎢⎣ 0 0 0 0 −1 −1 −1 −1⎥⎦

The incidence matrix is obtained by eliminating any one row.

⎡ 1 0 0 −1 1 0 0 0⎤
⎢ −1 1 0 0 0 1 0 0⎥
A= ⎢ ⎥
⎢ 0 −1 1 0 0 0 1 0⎥
⎢⎣ 0 0 −1 1 0 0 0 1⎥⎦

(b) Tieset Matrix (B)

1 2 3 4 5 6 7 8
Links: {2, 4, 6, 8}
Tieset 2: {2, 7, 5, 1} 2⎡1 1 0 0 −1 0 1 0⎤
Tieset 4: {4, 5, 7, 3} 4 ⎢0 0 1 1 1 0 −1 0⎥
B= ⎢ ⎥
Tieset 6: {6, 5, 1} 6 ⎢1 0 0 0 −1 1 0 0⎥
Tieset 8: {8, 7, 3} 8 ⎢⎣0 0 1 0 0 0 −1 1⎥⎦
(c) f-cutset Matrix (Q)
The oriented graph, selected tree and f-cutsets are shown in Fig. 9.40.
(1)
1 2
(5) Twigs: {1, 3, 5, 7} 1 2 3 4 5 6 7 8
(6)
f-cutset 1: {1, 6, 2} 1⎡1 1 0 0 0 1 0 0⎤
f-cutset 3: {3, 4, 8}
(4) (2)
3 ⎢0 0 1 1 0 0 0 1⎥
5 f-cutset 5: {5, 4, 6, 2} Q= ⎢ ⎥
(8) (7)) f-cutset 7: {7, 2, 8, 4} 5 ⎢0 1 0 1 1 1 0 0⎥
4 3 7 ⎢⎣0 1 0 1 0 0 1 1⎥⎦
(3)

Fig. 9.40
 9.7 Relationship Among Submatrices of A, B and Q 9.21

Example 9.10 How many trees are possible for the graph of the network of Fig. 9.41.
2 3

+
1 −

Fig. 9.41
Solution To draw the graph, (2)
2 3
1. replace all resistors, inductors and capacitors by line segments,
2. replace voltage source by short circuit and current source by an open (1) (3)
circuit,
(4)
3. assume directions of branch currents arbitrarily, and
4. number all the nodes and branches. 1

The oriented graph is shown in Fig. 9.42. Fig. 9.42

The complete Incidence Matrix (Aa) is written as
3 4
1 ⎡ 1 0 −1 1⎤
Aa = 2 ⎢ −1 1 0 0⎥
⎢ ⎥
3 ⎣ 0 −1
1 1 1⎦
The reduced incidence matrix A is obtained by eliminating the last row from matrix Aa.
⎡ 1 0 −1 1⎤
A= ⎢
⎣ −1 1 0 0 ⎥⎦
⎡ 1 −1⎤
⎡ 1 0 − 1 1⎤⎢ 0 1⎥ ⎡ 3 −1⎤
AAT = ⎢ ⎢ ⎥=
⎣ −1 1 0 0 ⎦ ⎢ −1 0 ⎥ ⎢⎣ −1 2⎥⎦
⎥
⎢⎣ 1 0 ⎥⎦
3 −1
The number of possible trees = AAT = = 6 − 1 = 5.
−1 2

Example 9.11 Draw the oriented graph from the complete incidence matrix given below;
(8)
Nodes Branches →
(5) 2 (6) 3 (7)
1 4
↓ 1 2 3 4 5 6 7 8
1 1 0 0 0 1 0 0 1 (2) (3)
(1)
2 0 1 0 0 −1 1 0 0 (4)
3 0 0 1 0 0 −1 1 −1
4 0 0 0 1 0 0 −1 0 5
5 −1 −1 −1 −1 0 0 0 0
Fig. 9.43
Solution First, note down the nodes 1, 2, 3, 4, 5 as shown in Fig. 9.43. From the complete incidence matrix, it
is clear that the branch number 1 is between nodes 1 and 5 and it is going away from node 1 and towards node 5
as the entry against node 1 is 1 and that against 5 is −1. Hence, connect the nodes 1 and 5 by a line, point the arrow
towards 5 and call it branch 1 as shown in Fig. 9.43. Similarly, draw the other oriented branches.
9.22 Network Analysis and Synthesis

Example 9.12 The reduced incidence matrix of an oriented graph is given below. Draw the
graph.
⎡ 0 −1 1 1 0 ⎤
A = ⎢ 0 0 −1 −1 −1⎥
⎢ ⎥
⎣ −1 0 0 0 1⎦

Solution First, writing the complete incidence matrix from the matrix A such that the sum of all entries in
each column of Aa will be zero, we have
(4)
1 2 3 4 5 2
1 3
1 ⎡ 0 −1 1 1 0⎤ (3) (5)
2 ⎢ 0 0 −1
1 1 1⎥
Aa = ⎢ ⎥
3 ⎢ −1 0 0 0 1⎥ (1)
(2)
4 ⎢⎣ 1 1 0 0 0 ⎥⎦

Fig. 9.44
Now, the oriented graph can be drawn with matrix Aa as shown in Fig. 9.44.

Example 9.13 The reduced incidence matrix of an oriented graph is

⎡ 0 −1 1 0 0 ⎤
A = ⎢ 0 0 −1 −1 −1⎥
⎢ ⎥
⎣ −1 0 0 0 1⎦

(a) Draw the graph. (b) How many trees are possible for this graph? (c) Write the tieset and cutset matrices.
Solution
(a) First, writing the complete incidence matrix Aa such that the sum of all the entries in each column of Aa
is zero, we have
2
1 3
1 2 3 4 5 (3) (5)

1⎡ 0 1 1 0 0⎤
(4)
2⎢ 0 0 −1
1 1 1⎥ (2) (1)
Aa = ⎢ ⎥
3 ⎢ −1 0 0 0 1⎥
4 ⎢⎣ 1 1 0 1 0 ⎥⎦
4

Fig. 9.45
Now, the oriented graph can be drawn with the matrix Aa, as shown in Fig. 9.45.
(b) The number of possible trees = |AAT|
⎡ 0 0 −1⎤
⎡ 0 −1 1 0 0 ⎤ ⎢ −1 0 0 ⎥ ⎡ 2 −1 0 ⎤
⎢ ⎥
AAT = ⎢ 0 0 −1 −1 −1⎥ ⎢ 1 −1 0 ⎥ = ⎢ −1 3 −1⎥
⎢ ⎥ ⎢ ⎥
⎣ −1 0 0 0 1⎦ ⎢ 0 −1 0 ⎥ ⎣ 0 −1 2⎦
⎢⎣ 0 −1 1⎥⎦
 9.7 Relationship Among Submatrices of A, B and Q 9.23

2 −1 0
AAT = −1 3 −1 = 2(6 − 1) + 1( −2) = 8
0 −1 2

The number of possible trees = 8.

(c) Tieset Matrix (B)
The oriented graph, selected tree and tiesets are shown in Fig. 9.46.
2
1 3
(3) (5) 1 2 3 4 5
Links: {1, 2} 1 ⎡1 0 0 −1 1 ⎤
(4) B= ⎢
(2) (1) Tieset 1: {1, 4, 5}
2 ⎣0 1 1 −1 0 ⎥⎦
Tieset 2: {2, 3, 4}

Fig. 9.46
f-cutset Matrix (Q)
The oriented graph, selected tree and f-cutsets are shown in Fig. 9.47
2
1 3
(3) (5) Twigs: {3, 4, 5}
f-cutset 3: {3, 2} 1 2 3 4 5
f-cutset 4: {4, 2, 1} 3⎡ 0 1 1 0 0⎤
(4) f-cutset 5: {5, 1}
(2) (1)
Q = 4⎢ 1 1 0 1 0⎥
⎢ ⎥
5 ⎣ −1 0 0 0 1⎦

4
Fig. 9.47

Example 9.14 The fundamental cutset matrix of a network is given as follows;

Twigs
g Links
a c e b d f
1 0 0 1 0 1
0 1 0 0 1 1
0 0 1 1 1 1
Draw the oriented graph. (f)

Solution No. of links l = b − n + 1

No. of nodes n = b − l + 1 = 6 − 3 + 1 = 4 (a) (d) Twigs: {a, c, e}
f-cutsets are written as, Links: {b, d, f}
f-cutsets a: {a, b, f } (e)
(b) (c)
f-cutsets c: {c, d, f }
f-cutsets e: {e, b, d, f }
The oriented graph is drawn as shown in Fig. 9.48. Fig. 9.48
9.24 Network Analysis and Synthesis

Example 9.15 Draw the oriented graph of a network with the f-cutset matrix as shown below:

Twigs
g Links
k
1 2 3 4 5 6 7
1 0 0 0 −1 0 0
0 1 0 0 1 0 1
0 0 1 0 0 1 1
0 0 0 1 0 1 0

Solution No. of links l = b − n + 1

No. of nodes n = b − l + 1 = 7 − 3 + 1 = 5
f-cutsets are written as
(5) (2) (7)
f-cutset 1: {1, 5}
f-cutset 2: {2, 5, 7}
f-cutset 3: {3, 6, 7} (1) (3) Twigs: {1, 2, 3, 4}
f-cutset 4: {4, 6} Links: {5, 6, 7}
(4)
Then oriented graph can be drawn as shown in (6)
Fig. 9.49.
Fig. 9.49
9.8 KIRCHHOFF’S VOLTAGE LAW
KVL states that if vk is the voltage drop across the kth branch, then

∑ vk = 0
the sum being taken over all the branches in a given loop. If l is the number of loops or f-circuits, then there
will be l number of KVL equations, one for each loop. The KVL equation for the f-circuit or loop ‘l’ can be
written as
b
∑ bik k =0 (k = 1, 2, …, l)
k =1

where bik is the elements of the tieset matrix B, b being the number of branches. The set of l KVL
equations can be written in matrix form.
Vb = 0
BV

⎡ v1 ⎤
⎢v ⎥
where Vb = ⎢ 2 ⎥ is a column vector of branch voltages.
⎢ ⎥
⎢⎣ vb ⎥⎦
and B is the fundamental circuit matrix.

9.9 KIRCHHOFF’S CURRENT LAW

KCL states that if ik is the current in the kth branch then at a given node

∑ ik = 0
the sum being taken over all the branches incident at a given node. If there are ‘n’ nodes, there will ‘n’ such
equations, one for each node
 9.10 Relation Between Branch Voltage Matrix Vb, Twig Voltage Matrix Vt And Node Voltage Matrix Vn 9.25

b
∑ aik ik = 0 (k = 1, 2,… , n)
k =1
so that set of n equations can be written in matrix form.
Aa I b = 0 …(9.4)
⎡ i1 ⎤
⎢i ⎥
where I b = ⎢ 2 ⎥ is a column vector of branch currents.
⎢ ⎥
⎢⎣ib ⎥⎦
and Aa is the complete incidence matrix.
If one node is taken as reference node or datum node, we can write the Eq. (9.4) as,
AI b = 0 …(9.5)
where A is the incidence matrix of order (n − 1) × b.
We know that A At Q
Equation (9.5) can be written as
AQQII b = 0
Premultiplying with At−1,
At 1 At Q I b = At−1. 0
I Q Ib = 0
Q Ib = 0
where Q is the f-cutset matrix.

RELATION BETWEEN BRANCH VOLTAGE MATRIX Vb, TWIG

9.10
VOLTAGE MATRIX Vt AND NODE VOLTAGE MATRIX Vn
We know that Vb = 0
BV
⎡Vt ⎤
[ : Bl ] ⎢ ⎥ = 0
⎢ ⎥
⎣Vl ⎦
Bt Vt + Bl Vl = 0
Bl Vl = − Bt Vt
Premultiplying with Bl−1 .
Vl Bl−11 Bt Vt = − ( Bl 1 Bt )Vt
⎡Vt ⎤
Now Vb = ⎢ ⎥
⎢ ⎥
⎣Vl ⎦
⎡ ⎤ ⎡ ⎤
⎢ Vt ⎥ ⎢ U ⎥
=⎢ … ⎥=⎢ … ⎥ .Vt …(9.6)
⎢−
⎢⎣ ( )
Vt ⎥⎥ ⎢⎢ −
⎦ ⎣ ( ) ⎥
⎥⎦
9.26 Network Analysis and Synthesis

Also, V QT Vt
Q At−1 A
QT AT ( At 1 )T AT ( AtT ) −1
Hence Eq. (9.6) can be written as
Vb AT ( AtT ) 1Vt = AT {A T
t Vt } AT Vn

where Vn ( AtT ) −1Vt is node voltage matrix.

RELATION BETWEEN BRANCH CURRENT MATRIX Ib AND LOOP

9.11
CURRENT MATRIX Il
We know that, A I b = 0
⎡ It ⎤
[ : At ] ⎢…⎥ = 0
⎢ ⎥
⎣ Il ⎦
At I t + Al I l = 0
At I t = − Al I l
Premultiplying with At−1 ,

It At−11 Al I l = − ( At 1 Al ) I l
⎡ I t ⎤ ⎡ − ( At Al ) I l ⎤ ⎡ − ( At Al ) ⎤
1 1

I b = ⎢…⎥ = ⎢ … ⎥ = ⎢ … ⎥ .I l
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
Now ⎣ I l ⎦ ⎢⎣ Il ⎥⎦ ⎢⎣ U ⎥⎦
I b = BT I l

9.12 NETWORK EQUILIBRIUM EQUATION

9.12.1 KVL Equation
1. If there is a voltage source vsk in the branch k having impedance zk and carrying current ik, as shown in
Fig. 9.50,
Zk vsk
vk zk ik vsk (k = 1, 2,…, b) + −
−+
In matrix form, ik
Vb Zb I b Vs vk

where Zb is the branch impedance matrix, Ib is the column vector Fig. 9.50 Circuit diagram
of branch currents and Vs is the column vector of source voltages.
Hence, KVL equation can be written as
Vb = 0
BV
B ( Zb I b Vs ) = 0
B Zb I b BV Vs
 9.12 Network Equilibrium Equation 9.27

Also, Ib BT I l
B Zb BT I l = BV
Vs
Z Il = E
where E BV Vs
and Z B Z b BT
The matrix Z is called loop impedance matrix.
2. If there is a voltage source in series with an impedance and vk
a current source in parallel with the combination as shown in ik + Zk
−
Fig. 9.51, −+
(v + vsk ) vsk
ik = k − isk
zk
vk = zk ik zk isk vsk isk
In matrix form,
Fig. 9.51 Circuit diagram
Vb Zb I b Zb I s Vs
Vb = 0.
KVL equation is BV
Vb = B ( Zb I b Zb I s Vs ) = 0
BV
B Zb I b BV
Vs B Zb I s
Now Ib BT I l
B Zb BT I l = BV
Vs − B Zb I s
Z I l = BV
Vs − B Zb I s
where Z B Zb BT is the loop impedance matrix. This is the generalised KVL equation.

9.12.2 KCL Equation

vk
1. If the branch k contains an input current source isk and an yk
ik + −
admittance yk as shown in Fig. 9.52,
i yk vk isk (k = 1, 2,…, b)
In the matrix form,
Ib Yb Vb Is isk
where Yb is the branch admittance matrix.
Fig. 9.52 Circuit diagram
Hence KCL equation is given by,
A Ib = 0
A (Yb Vb I s ) = 0
AYYb Vb A I s
Also Vb AT Vn
Yb AT Vn = A I s
AY
Y Vn = I
where Y Yb AT
AY
and I A Is
9.28 Network Analysis and Synthesis
The matrix Y is called admittance matrix. This is the KCL equation in matrix form.
In terms of f-cutset matrix, the KCL equation can be written as
Q Ib = 0
Q (Yb Vb Vs ) = 0
QY
Yb V Q I s

Also V QT Vt

Yb QT Vt = Q I s
QY
Y Vt = I
where Y QY Y QT
and I Q Is
This is the KCL equation in matrix form.
2. If there is a voltage source in series with an impedance and a vk
current source in parallel with the combination as shown in zk
ik + −
Fig. 9.53, −+
vsk
1
yk =
zk
isk
i yk v + yk vsk isk
In matrix form, Fig. 9.53 Circuit diagram
I b Yb Vb Yb Vs I s
KCL equation will be given by,
A Ib = 0
A (Yb Vb Yb Vs I s ) = 0
AY
Yb Vb A I s AY
Yb Vs

Also Vb AT Vn

Yb AT Vn = A I s − AY
AY Yb Vs
Y Vn = A I s − AV
Vb Vs
where Y AY Yb AT is the node admittance matrix. This is a generalised KCL equation.
In terms of f-cutset matrix, the KCL equation can be written as
Q Ib = 0
Q (Yb Vb Yb Vs I s ) = 0
QYYb V Q I QY
Yb Vs

Also V QT Vt

Yb QT Vt = Q I s − QY
QY Yb Vs
Y Vt = Q I s − QY
Yb Vs
This is a generalised KCL equation.
 9.12 Network Equilibrium Equation 9.29

Note
(i) For a graph having b branches, the branch impedance matrix Zb is a square matrix of order b, having
branch impedances as diagonal elements and the mutual impedances between the branches as non-
diagonal elements. For a network having no mutual impedances, only diagonal elements will be present
in the branch impedance matrix.
(ii) For a graph having b branches, the branch admittance matrix Yb is a square matrix of order b, having
branch admittances as diagonal elements and the mutual admittances between the branches as non-
diagonal elements. For a network having no mutual admittances, only diagonal elements will be present
in the branch admittance matrix.
(iii) For a graph having b branches, the voltage source matrix or vector Vs is a rectangular matrix of order
b × 1, having the value of the voltage source in the particular branch. The value will be positive if there
is a voltage rise in the direction of current and will be negative if there is a voltage fall in the direction
of current.
(iv) For a graph having b branches, the current source matrix or vector Is is a rectangular matrix of order
b × 1, having the value of the current source in the particular parallel branch. The value will be positive if
the direction of the current source and the corresponding parallel branch current are not same. The value
will be negative if the directions of the current source and corresponding parallel branch current are same.

Example 9.16 Write the incidence matrix of the graph of Fig. 9.54 and express branch voltages in
terms of node voltages. Write the tieset matrix and express branch currents in terms of loop currents.
(8)

(5) 2 (6)
1 3

(2) (7)
(1) (3)

5 (4)

Fig. 9.54
Solution
(a) Incidence Matrix
1 2 3 4 5 6 7 8
1⎡ 1 0 0 0 1 0 0 1⎤
2⎢ 0 1 0 0 −1 1 0 0⎥
⎢ ⎥
Aa = 3 ⎢ 0 0 1 0 0 −11 1 1⎥
4⎢ 0 0 0 1 0 0 1 0⎥
5 ⎢⎣ −1 −1 −1 −1 0 0 0 0 ⎥⎦
The incidence matrix is obtained by eliminating any one row.

⎡1 0 0 0 1 0 0 1⎤
⎢0 1 0 0 −1 1 0 0 ⎥
A= ⎢ ⎥
⎢0 0 1 0 0 −1 1 −1⎥
⎢⎣0 0 0 1 0 0 −1 0 ⎥⎦
9.30 Network Analysis and Synthesis
(b) Branch voltages in terms of node voltages
Vb AT Vn
⎡V1 ⎤ ⎡ 1 0 0 0⎤
⎢V2 ⎥ ⎢0 1 0 0⎥
⎢ ⎥ ⎢ ⎥
⎢V3 ⎥ ⎢0 0 1 0 ⎥ ⎡Vn1 ⎤
⎢V4 ⎥ = ⎢0 0 0 1⎥ ⎢Vn2 ⎥
⎢V5 ⎥ ⎢ 1 ⎢ ⎥
1 0 0 ⎥ ⎢Vn3 ⎥
⎢V ⎥ ⎢0 1 −1 0 ⎥⎥ ⎢⎣Vn4 ⎥⎦
⎢ 6⎥ ⎢
⎢V7 ⎥ ⎢0 0 1 1⎥
⎢⎣V8 ⎥⎦ ⎢⎣ 1 0 −1 0 ⎥⎦

(c) Tieset Matrix

Selected tree and tiesets are shown in Fig. 9.55.
(8)

(5) 2 (6) 1 2 3 4 5 6 7 8
1 3
Links: {1, 4, 6, 8} 1 ⎡1 1 0 0 −1 0 0 0⎤
4 ⎢0 0⎥
Tieset 1: {1, 2, 5}
0 −1 1 0 0 1
(2) (7)
Tieset 4: {4, 3, 7} B= ⎢ ⎥
(1) (3) Tieset 6: {6, 2, 3} 6 ⎢0 1 1 0 0 1 0 0⎥
Tieset 8: {8, 2, 3, 5} 8 ⎢⎣0 1 1 0 −1 0 0 1⎥⎦
4
5 (4)

Fig. 9.55
(d) Branch currents in terms of loop currents

Ib BT I l
⎡ I1 ⎤ ⎡ 1 0 0 0 ⎤
⎢ I 2 ⎥ ⎢ −1 0 −1 −1⎥
⎢ ⎥ ⎢ ⎥ ⎡I ⎤
⎢ I 3 ⎥ ⎢ 0 −1 1 1⎥ ⎢ l1 ⎥
⎢I4 ⎥ = ⎢ 0 1 0 0 ⎥ I l4
⎢ I 5 ⎥ ⎢ −1 0 0 −1⎥ ⎢⎢ I l6 ⎥⎥
⎢I ⎥ ⎢ 0 0 1 0 ⎥⎥ ⎢⎣ I l8 ⎥⎦
⎢ 6⎥ ⎢
⎢ I7 ⎥ ⎢ 0 1 0 0⎥
⎢⎣ I 8 ⎥⎦ ⎢⎣ 0 0 0 1⎥⎦

Example 9.17 Branch current and loop-current relationships are expressed in matrix form as

⎡ I1 ⎤ ⎡ 1 0 0 1⎤
⎢I2 ⎥ ⎢ 0 1 0 1⎥
⎢ ⎥ ⎢ ⎥
I
⎢ ⎥ ⎢3 0 1 1 0 ⎥ ⎡ I l1 ⎤
⎢I4 ⎥ = ⎢ 0 1 1 0 ⎥ ⎢ I l2 ⎥
⎢ I 5 ⎥ ⎢ 1 −1 0 ⎢ ⎥
0 ⎥ ⎢ I l3 ⎥
⎢ I ⎥ ⎢ 0 0 −1 0 ⎥⎥ ⎢⎣ I l4 ⎥⎦
⎢ 6⎥ ⎢
⎢ I 7 ⎥ ⎢ −1 0 0 0⎥
⎢⎣ I 8 ⎥⎦ ⎢⎣ 0 0 0 1⎥⎦
Draw the oriented graph.
 9.12 Network Equilibrium Equation 9.31

Solution Writing the equation in matrix form,

Ib BT I l
⎡ 1 0 0 −1⎤
⎢ 0 1 0 −1⎥ 1 2 3 4 5 6 7 8
⎢ ⎥
⎢ 0 1 1 0⎥ ⎡ 1 0 0 0 1 0 −1 0⎤
⎢ 0 1 −1 0⎥
BT = ⎢⎢
0 1 1 0⎥ 1 1 0 0
∴B = ⎢ ⎥
1 −1 0 0 ⎥ ⎢ 0 0 1 1 0 1 0 0⎥
⎢ 0 0 −1 0 ⎥
⎢ ⎥ ⎢⎣ −1
1 1 0 0 0 0 0 1⎥⎦ (8)

⎢ −1 0 0 0 ⎥
⎢⎣ 0 0 0 1⎥⎦ 1 (1) 2 (2) 3

(5) (3)
(7)
From tieset matrix,
No. of links l=4 4
5 (4)
No of branches b = 8 (6)
No of nodes n= b − l + 1 = 8 − 4 + 1 = 1
The oriented graph is shown in Fig. 9.56. Fig. 9.56

Example 9.18 For the given graph shown in Fig. 9.57, write down the basic tieset matrix and
taking a tree of branches 2, 4, 5, write down KVL equations from the matrix.

(2) 2
1

(5)
(1) (3)

4 3
(4)

(6)

Fig. 9.57
Solution Selecting branches 2, 4, and 5 as the tree as shown in Fig. 9.58,

(2) 2
1
1 2 3 4 5 6
(5)
(1) (3) Links: {1, 3, 6} 1⎡1 0 0 1 1 0⎤
Tieset 1: {1, 5, 4} B = 3 ⎢0 1 1 0 1 0⎥
Tieset 3: {3, 2, 5} ⎢ ⎥
Tieset 6: {6, 2, 5, 4}
6 ⎣0 1 0 1 1 1⎦
4 3
(4)

(6)

Fig. 9.58
9.32 Network Analysis and Synthesis
The KVL equation in matrix form is given by
Vb = 0
BV
⎡V1 ⎤
⎢V ⎥
⎡ 1 0 0 −1 1 0 ⎤ ⎢ 2 ⎥
⎢0 1 1 0 1 0 ⎥ ⎢V3 ⎥ = 0
⎢ ⎥ ⎢V4 ⎥
⎣0 1 0 −1 1 1⎦ ⎢V ⎥
5
⎢V ⎥
⎣ 6⎦
V1 V4 + V5 = 0
V2 V3 + V5 = 0
V2 V4 + V5 V6 = 0

Example 9.19 Obtain the f-cutset matrix for the graph shown in Fig. 9.59 taking 1, 2, 3, 4 as tree
branches. Write down the network equations from the f-cutset matrix.
2

(5) (6)
(3)

3
1
(2) (7)

(8)
(1)
(4)

Fig. 9.59
1 2 3 4 5 6 7 8
Solution Twigs: {1, 2, 3, 4}
f-cutset 1 : {1, 6, 7, 8} 1 ⎡1 0 0 0 0 1 1 1⎤
f-cutset 2 : {2, 5, 6, 7, 8} 2 ⎢0 1 0 0 1 1 1 1⎥
Q= ⎢ ⎥
f-cutset 3 : {3, 5, 6} 3 ⎢0 0 1 0 1 1 0 0⎥
f-cutset 4 : {4, 6, 7} 4 ⎢⎣0 0 0 1 0 1 1 0 ⎥⎦
The KCL equation in matrix form is given by
Q Ib = 0
⎡ I1 ⎤
⎢I2 ⎥
⎢ ⎥
⎡1 0 0 0 0 1 −1 1⎤ ⎢ I 3 ⎥
⎢0 1 0 0 1 −1 1 −1⎥ ⎢ I 4 ⎥
⎢ ⎥ =0
⎢0 0 1 0 1 −1 0 0⎥ ⎢ I 5 ⎥
⎢⎣0 0 0 1 0 1 −1 0 ⎥⎦ ⎢⎢ I 6 ⎥⎥
⎢ I7 ⎥
⎢⎣ I 8 ⎥⎦
 9.12 Network Equilibrium Equation 9.33

I1 I6 − I7 I8 = 0
I2 I5 − I6 I 7 − I8 = 0
I3 I5 − I6 = 0
I4 I6 − I7 = 0

Example 9.20 The reduced incidence matrix of a graph is given as

⎡ 1 0 0 0 −1⎤
A = ⎢ −1 −1 −1 0 0 ⎥
⎢ ⎥
⎣ 0 0 1 −1 0 ⎦
Express branch voltages in terms of node voltages.
Solution For the given graph,
No of branches b = 5
No of nodes n = 3
Branch voltages can be expressed in terms of node voltages by
T
Vb Vn
⎡V1 ⎤ ⎡ 1 −1 0 ⎤
⎢V2 ⎥ ⎢ 0 −1 0 ⎥ ⎡Vn1 ⎤
⎢ ⎥ ⎢ ⎥⎢ ⎥
⎢V3 ⎥ = ⎢ 0 −1 1⎥ ⎢Vn2 ⎥
⎢V4 ⎥ ⎢ 0 0 −1⎥ ⎣Vn3 ⎦
⎢⎣V5 ⎥⎦ ⎢⎣ −1 0 0 ⎥⎦
V1 Vn1 − Vn2
V2 Vn2
V3 Vn2 + Vn3
V4 Vn3
V5 Vn1

Example 9.21 The fundamental cutset matrix of a graph is given as

⎡ −1 1 0 0 −1⎤
Q = ⎢ 0 0 1 0 −1⎥
⎢ ⎥
⎣1 0 0 1 0⎦
Express branch voltages in terms of twig voltages.
Solution For the given graph,
No. of branches b = 5
No. of twigs = 3
Branch voltages are expressed in terms of twig voltages by
V QT Vt
⎡V1 ⎤ ⎡ −1 0 1⎤
⎢V2 ⎥ ⎢ 1 0 0 ⎥ ⎡Vt1 ⎤
⎢ ⎥ ⎢ ⎥
⎢V3 ⎥ = ⎢ 0 1 0 ⎥ ⎢Vt2 ⎥
⎢ ⎥
⎢V4 ⎥ ⎢ 0 0 1⎥ ⎣Vt3 ⎦
⎢⎣V5 ⎥⎦ ⎢⎣ −1 −1 0 ⎥⎦
9.34 Network Analysis and Synthesis

V1 Vt1 + Vt3
V2 Vt1
V3 Vt2
V4 Vt3
V5 Vt1 − Vt2

Example 9.22 For this network shown in Fig. 9.60, write down the tieset matrix and obtain the
network equilibrium equation in matrix form using KVL. Calculate the loop currents and branch currents.

2Ω
1Ω
1Ω

2V +
− 2Ω 2Ω
1Ω

Fig. 9.60

Solution The oriented graph and one of its trees are shown in Fig. 9.61.

1 1

Links: {1, 2, 3}
Tieset 1: {1, 4, 5}
1 2 3 4 5 6
(1) (4) (2) (1) (4) (2) Tieset 2: {2, 4, 6} 1⎡1 0 0 1 1 0⎤
Tieset 3: {3, 5, 6}
B = 2 ⎢0 1 0 1 0 1⎥
(5) (6) (5) (6) ⎢ ⎥
2 2 3 ⎣0 0 1 0 −1 1⎦

3 (3) 4 3 (3) 4

Fig. 9.61

The KVL equation in matrix form is given by

B Zb BT I l = BV Vs − B Zb I s
Here, I s = 0,
B Zb BT I l = BV
Vs
⎡1 0 0 0 0 0⎤ ⎡ 1 0 0⎤ ⎡ 2⎤
⎢0 1 0 0 0 0⎥ ⎢0 1 0⎥ ⎢0 ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
Zb ⎢0 0 1 0 0 0⎥ T ⎢0 0
;B =
1⎥ 0
; Vs = ⎢ ⎥
⎢0 0 0 2 0 0⎥ ⎢ 1 −1 0 ⎥ ⎢0 ⎥
⎢0 0 0 0 2 0⎥ ⎢ 1 0 −1⎥ ⎢0 ⎥
⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥
⎣ 0 0 0 0 2⎦ ⎣0 −1 1⎦ ⎣ ⎦
 9.12 Network Equilibrium Equation 9.35

⎡1 0 0 0 0 0⎤
⎢0 1 0 0 0 0⎥
⎡1 0 0 1 1 0⎤ ⎢ ⎥ ⎡1 0 0 2 2 0⎤
0 0 1 0 0 0⎥ ⎢
B Zb = ⎢0 1 0 −1 0 −1⎥ ⎢ = 0 1 0 2 0 2⎥
⎢ ⎥ ⎢0 0 0 2 0 0 ⎥ ⎢ ⎥
⎣0 0 1 0 −1 1⎦ ⎢
0 0 0 0 2 0⎥ ⎣
0 0 1 0 −2 2⎦
⎢ 0 0 0 0 0 2⎥
⎣ ⎦
⎡ 1 0 0 ⎤
⎢0 1 0⎥
⎡1 0 0 2 2 0⎤ ⎢ ⎥ ⎡ 5 −2 −2⎤
0 0 1⎥ ⎢
B Zb BT = ⎢0 1 0 −2 0 −2⎥ ⎢ = −2 5 −2⎥
⎢ ⎥ 1 −1 0 ⎥ ⎢ ⎥
⎣0 0 1 0 −2 2⎦ ⎢⎢ −2 −2 5⎦
1 0 −1⎥ ⎣
⎢0 −1 1⎥
⎣ ⎦
⎡ 2⎤
⎢0 ⎥
⎡1 0 0 1 1 0 ⎤ ⎢ ⎥ ⎡ 2⎤
0
Vs = ⎢0 1 0 −1 0
BV −1⎥ ⎢ ⎥ = ⎢0 ⎥
⎢ ⎥ 0 ⎢ ⎥
⎣0 0 1 0 −1 1⎦ ⎢⎢ ⎥⎥ ⎣0 ⎦
0
⎢0 ⎥
⎣ ⎦
The KVL equation in matrix form is given by

⎡ 5 2 −2⎤ ⎡ I l1 ⎤ ⎡ 2⎤
⎢ −2 5 −2⎥ ⎢ I l2 ⎥ = ⎢0 ⎥
⎢ ⎥⎢ ⎥ ⎢ ⎥
⎣ −2
2 2 5⎦ ⎣ I l3 ⎦ ⎣0 ⎦

Solving this matrix equation,

6
I l1 = A
7
4
I l2 = A
7
4
I l3 = A
7
The branch currents are given by

Ib BT I l
⎡6⎤
⎢7⎥
⎢4⎥
⎡ I1 ⎤ ⎡ 1 0 0⎤ ⎡ 6 ⎤ ⎢ ⎥
⎢ I 2 ⎥ ⎢0 1 0⎥ ⎢ 7 ⎥ ⎢ 7 ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢4⎥
I
⎢ 3⎥ = ⎢ 0 0 1⎥ ⎢ 4 ⎥ ⎢ ⎥
=
⎢ I 4 ⎥ ⎢ 1 −1 0⎥ ⎢ 7 ⎥ ⎢ 7 ⎥
2
⎢ I5 ⎥ ⎢ 1 0 −1⎥ ⎢ 4 ⎥ ⎢ ⎥
⎢ I ⎥ ⎢0 −1 ⎥ ⎢ ⎥ ⎢7⎥
⎣ 6⎦ ⎣ 1⎦ ⎣ 7 ⎦ ⎢ ⎥
2
⎢ ⎥
⎢7⎥
⎢⎣ 0 ⎥⎦
9.36 Network Analysis and Synthesis

Example 9.23 For the network shown in Fig. 9.62, write down the tieset matrix and obtain the
network equilibrium equation in matrix form using KVL. Calculate loop currents.
2Ω 8V
+−

4Ω Il 3 6Ω

6Ω 2Ω 4Ω

12 V + +
Il1 Il2
− − 6V

Fig. 9.62
Solution The oriented graph and its selected tree are shown in Fig. 9.63.
(3) (3)
2 2 Links: {1, 2, 3}
1 3 1 3
(4) (5) (4) (5) Tieset 1: {1, 4, 6} 1 2 3 4 5 6
Tieset 2: {2, 5, 6}
Tieset 3: {3, 5, 4} 1⎡1 0 0 1 0 1⎤
(1) (6) (2) (1) (6) (2) B = 2 ⎢0 1 0 0 1 −1⎥
⎢ ⎥
3 ⎣0 0 1 −1 −1 0 ⎦

4 4
Fig. 9.63
The KVL equation in matrix form is given by
B Zb BT I l = BV
Vs − B Zb I s
Here, I s = 0,
T
B Zb B I l = BV
Vs
⎡6 0⎤0 ⎡1 0
0 0 0 0⎤ ⎡ 12⎤
⎢0 0 ⎥
4 ⎢0
0 1 0 0 0 ⎥ ⎢ −6 ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢0 0
0⎥ T ⎢0 0 2 0 0 1⎥ −8
Zb ;B = ; Vs = ⎢ ⎥
⎢0 0⎥0 ⎢1 0
0 4 0 −1⎥ ⎢ 0⎥
⎢0 0⎥0 ⎢0
0 1 0 6 −1⎥ ⎢ 0⎥
⎢0 2⎥⎦ ⎢ 1 −1 0 ⎥⎦ ⎢ 0⎥
⎣ 0 ⎣0 0 0 ⎣ ⎦
⎡6 0 0 0 0 0⎤
⎢0 4 0 0 0 0⎥
⎡1 0 0 1 0 1⎤ ⎢ ⎥ ⎡6 0 0 4 0 2⎤
0 0 2 0 0 0⎥ ⎢
B Zb = ⎢0 1 0 0 1 −1⎥ ⎢ = 0 4 0 0 6 2⎥
⎢ ⎥ ⎢0 0 0 4 0 0⎥ ⎢ ⎥
⎣0 0 1 −1 −1 0 ⎦ ⎢0 0 0 0 6 0⎥ ⎣
0 0 2 −4 −6 0⎦
⎢0 0 0 0 0 2⎥⎦
⎣
⎡1 0 0⎤
⎢0 1 0⎥
⎡6 0 0 4 0 2⎤ ⎢ ⎥ ⎡ 12 −2 −4 ⎤
0 0 1⎥ ⎢
BZb BT = ⎢0 4 0 0 6 −2⎥ ⎢ = −1 12 −6 ⎥
⎢ ⎥ ⎢1 0 −1⎥ ⎢ ⎥
⎣0 0 2 −4 −6 0 ⎦ ⎢0 1 −1⎥ ⎣
−4 −6 12⎦
⎢ 1 −1 0 ⎥⎦
⎣
 9.12 Network Equilibrium Equation 9.37

⎡ 12⎤
⎢ −6 ⎥
⎡1 0 0 1 0 1⎤ ⎢ ⎥ ⎡ 12⎤
−8
Vs = ⎢0 1 0
BV 0 1 −1⎥ ⎢ ⎥ = ⎢ −6 ⎥
⎢ ⎥ 0 ⎢ ⎥
⎣0 0 1 1 1 0 ⎦ ⎢⎢ ⎥⎥ ⎣ −8⎦
0
⎢ 0⎥
⎣ ⎦
Hence, the KVL equation in matrix form is given by
⎡ 12 −2
2 4 ⎤ ⎡ I l1 ⎤ ⎡12 ⎤
⎢ −2
2 12 6 ⎥ ⎢ I l2 ⎥ = ⎢ −6 ⎥
⎢ ⎥⎢ ⎥ ⎢ ⎥
⎣ −4
4 6 12⎦ ⎣ I l3 ⎦ ⎣ −8⎦
Solving this matrix equation,
I l1 = 0 55A
I l2 = −0.866 A
I l3 = −0.916 A

Example 9.24 For the network shown in Fig. 9.64, draw the oriented graph. Write the tieset
schedule and hence obtain the equilibrium equation on loop basis. Calculate the values of branch currents
and branch voltages.

+ 1V
1A −
1Ω
1Ω 1Ω

1Ω 1Ω

Fig. 9.64
Solution The oriented graph and one of its trees are shown in Fig. 9.65.
1 1

(4) (4)
(1) (3) (1) (3) Links: {1, 2, 3}
(2) (2) Tieset 1: {1, 4, 5, 6}
Tieset 2: {2, 4, 5}
Tieset 3: {3, 4}

2 (6) 3 (5) 4 2 (6) 3 (5) 4

Fig. 9.65

1 2 3 4 5 6
1⎡1 0 0 1 1 1⎤
B = 2 ⎢0 1 0 1 1 0⎥
⎢ ⎥
3 ⎣0 0 1 1 0 0⎦
9.38 Network Analysis and Synthesis
The KVL equation in matrix form is given by
B Zb BT I l = BV
Vs − B Zb I s
⎡1 0 0 0⎤0 0 ⎡ 1 0 0⎤ ⎡0 ⎤ ⎡ −1⎤
⎢0 1 0 0⎥0 0 ⎢ 0 1 0⎥ ⎢0 ⎥ ⎢ 0⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
0 0 1 0 0
0⎥ T ⎢ 0 0 1⎥ 0 0
Zb = ⎢ ;B = ; Vs = ⎢ ⎥ ; I s =⎢ ⎥
⎢0 0 0 0⎥0 0 ⎢ −1 −1 1⎥ ⎢1 ⎥ ⎢ 0⎥
⎢0 0 0 0⎥0 1 ⎢ 1 1 0⎥ ⎢0 ⎥ ⎢ 0⎥
⎢0 1⎥⎦ ⎢ −1 0 ⎥⎦ ⎢0 ⎥ ⎢ 0⎥
⎣ 0 0 0 0 ⎣ 0 ⎣ ⎦ ⎣ ⎦
⎡1 0 0 0 0 0⎤
⎢0 1 0 0 0 0⎥
⎡ 1 0 0 −1 1 −1⎤ ⎢ ⎥ ⎡1 0 0 0 1 1⎤
0 0 1 0 0 0⎥ ⎢
B Zb = ⎢0 1 0 −1 1 0 ⎥ ⎢ = 0 1 0 0 1 0⎥
⎢ ⎥ ⎢0 0 0 0 0 0⎥ ⎢ ⎥
⎣0 0 1 1 0 0 ⎦ ⎢0 0 0 0 1 0⎥ ⎣
0 0 1 0 0 0⎦
⎢0 0 0 0 1 ⎥⎦
⎣ 0
⎡ 1 0 0⎤
⎢ 0 1 0⎥
⎡1 0 0 0 1 −1⎤ ⎢ ⎥ ⎡3 1 0⎤
0 0 1⎥ ⎢
BZb BT = ⎢0 1 0 0 1 0⎥ ⎢ = 1 2 0⎥
⎢ ⎥ ⎢ −1 −1 1⎥ ⎢ ⎥
⎣0 0 1 0 0 0⎦ ⎢
1 1 0⎥ ⎣
0 0 1⎦
⎢ −1 0 0 ⎥
⎣ ⎦
⎡0 ⎤
⎢0 ⎥
⎡1 0 0 −1 1 −1⎤ ⎢ ⎥ ⎡ −1⎤
0
Vs = ⎢0 1 0
BV −1 1 0 ⎥ ⎢ ⎥ = ⎢ −1⎥
⎢ ⎥ ⎢1 ⎥ ⎢ ⎥
⎣0 0 1 1 0 0 ⎦ ⎢ ⎥ ⎣ 1⎦
0
⎢0 ⎥
⎣ ⎦
⎡ −1⎤
⎢ 0⎥
⎡ 1 0 0 0 1 −1⎤ ⎢ ⎥ ⎡ −1⎤
0
B Zb I s = ⎢0 1 0 0 1 0 ⎥ ⎢ ⎥ = ⎢ 0 ⎥
⎢ ⎥ ⎢ 0⎥ ⎢ ⎥
⎣0 0 1 0 0 0 ⎦ ⎢ 0 ⎥ ⎣ 0 ⎦
⎢ 0⎥
⎣ ⎦
Hence, the KVL equation in matrix form is given by
⎡ 3 1 0 ⎤ ⎡ I l1 ⎤ ⎡ −1⎤ ⎡ −1⎤ ⎡ 0 ⎤
⎢1 2 0 ⎥ ⎢ I l ⎥ = ⎢ −1⎥ − ⎢ 0 ⎥ = ⎢ −1⎥
⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣0 0 1 ⎦ ⎣ I l3 ⎦ ⎣ 1⎦ ⎣ 0 ⎦ ⎣ 1⎦
Solving this matrix equation,
1
I l1 = A
5
3
I l2 = − A
5
I l3 = 1 A
 9.12 Network Equilibrium Equation 9.39

The branch currents are given by

I b = BT I l Is
⎡ 4⎤
⎢− 5 ⎥
⎢ 3⎥
⎡ I1 ⎤ ⎡ 1 0 0⎤ ⎡ −1⎤ ⎢ − ⎥
⎢I2 ⎥ ⎢ 0 ⎡ 1⎤ ⎢ 5⎥
1 0⎥ ⎢ ⎥ ⎢ 0⎥ ⎢ 1 ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ I3 ⎥ = ⎢ 0 0 1⎥ ⎢ 5 ⎥ + ⎢ 0 ⎥ = ⎢ ⎥
3 ⎢ 7⎥
⎢ I 4 ⎥ ⎢ −1 −1 1⎥ ⎢ − ⎥ ⎢ 0 ⎥ ⎢ ⎥
⎢ I5 ⎥ ⎢ 1 1 ⎢ 5⎥
0⎥ ⎣ 1 ⎦ ⎢ 0⎥ ⎢ 5 ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 2
⎣ I 6 ⎦ ⎣ −1 0 0⎦ ⎣ 0⎦ ⎢ − ⎥
⎢ 5⎥
⎢ 1⎥
⎢− ⎥
⎣ 5⎦
The branch voltages are given by
Vb Zb I b Vs
⎡ 4⎤
⎢− 5 ⎥ ⎡ 4⎤
⎢ 3⎥ ⎢− 5 ⎥
⎢ ⎥
⎡V1 ⎤ ⎡1 0 0 0 0 0⎤ ⎢ − ⎥ ⎡0 ⎤ ⎢ − 3 ⎥
⎢V2 ⎥ ⎢0 ⎥ ⎢ 5⎥ ⎢ ⎥
1 0 0 0 0 ⎢ ⎥ 0 ⎢ 5⎥
⎢ ⎥ ⎢ ⎥ 1 ⎢ ⎥ ⎢ ⎥
⎢V3 ⎥ = ⎢0 0 1 0 0 0⎥ ⎢ 7 ⎥ − ⎢0 ⎥ = 1
⎢ ⎥
⎢V4 ⎥ ⎢0 0 0 0 0 0 ⎥ ⎢ ⎥ ⎢1 ⎥ ⎢ −1⎥
⎢V5 ⎥ ⎢0 ⎢ 5⎥
0 0 0 1 0 ⎥ ⎢ 2 ⎥ ⎢0 ⎥ ⎢ − 2 ⎥
⎢ ⎥ ⎢ ⎥ − ⎢ ⎥
⎣V6 ⎦ ⎣0 0 0 0 0 1 ⎦ ⎢ ⎥ ⎣0 ⎦ ⎢ 5 ⎥
⎢ ⎥ 5 ⎢ 1⎥
⎢ − 1⎥ ⎢− ⎥
⎢⎣ 5 ⎥⎦ ⎣ 5⎦

Example 9.25 For the network shown in Fig. 9.66, obtain the loop equation in matrix form.
R1

R2 I

V R4
R3

Fig. 9.66
Solution The oriented graph and one of its trees are shown in Fig. 9.67.
1 1

(2) (2) Links: {1, 4} 1 2 3 4

(1) 2 (4) (1) 2 (4) Tieset 1: {1, 2, 3} 1 ⎡1 1 1 0 ⎤
B= ⎢
4 ⎣0 1 1 1 ⎥⎦
Tieset 4: {4, 2, 3}
(3) (3)

3 3
Fig. 9.67
9.40 Network Analysis and Synthesis
The KVL equation in matrix form is given by

B Zb BT I l = BV
Vs − B Zb I s
⎡ R1 0 0 0⎤ ⎡1 0⎤ ⎡V ⎤ ⎡0 ⎤
⎢ 0 R2 0 0⎥ T ⎢1 1⎥ ⎢0⎥ ⎢I ⎥
Zb = ⎢ ⎥; B =⎢ ⎥ ; Vs = ⎢ ⎥ ; I s = ⎢ ⎥
⎢ 0 0 R 3 0⎥ ⎢ 1 1 ⎥ 0
⎢ ⎥ ⎢0 ⎥
⎢⎣ 0 0 0 R4 ⎦⎥ ⎢⎣ 0 1 ⎥⎦ ⎢⎣ ⎥⎦
0 ⎢⎣0 ⎥⎦
⎡ R1 0 0 0⎤
⎡1 1 1 0 ⎤ ⎢ 0 R2 0 0 ⎥ ⎡ R1 R2 R3 0⎤
B Zb = ⎢ ⎢ ⎥=
⎣0 1 1 1 ⎥⎦ ⎢ 0 0 R3 0 ⎥ ⎢⎣ 0 R2 R3 R4 ⎥⎦
⎢⎣ 0 0 0 R4 ⎥⎦
⎡1 0⎤
⎡ R R2 R3 0 ⎤ ⎢1 1 ⎥ ⎡ R1 + R2 + R3 R2 + R3 ⎤
B Z b BT = ⎢ 1 ⎢ ⎥=
⎣ 0 R2 R4 ⎥⎦ ⎢1 1 ⎥ ⎢⎣
R3 R2 R3 R2 R3 + R4 ⎥⎦
⎢⎣ 0 1 ⎥⎦
⎡V ⎤
⎡1 1 1 0 ⎤ ⎢ 0 ⎥ ⎡V ⎤
Vs = ⎢
BV ⎢ ⎥=
⎣0 1 1 1 ⎥⎦ ⎢ 0 ⎥ ⎢⎣ 0 ⎥⎦
⎢⎣ 0 ⎥⎦
⎡0 ⎤
⎡ R1 R2 R3 0 ⎤ ⎢ I ⎥ ⎡ R2 I ⎤
B Zb I s = ⎢ ⎢ ⎥=
⎣ 0 R2 R3 R4 ⎥⎦ ⎢0 ⎥ ⎢⎣ R2 I ⎥⎦
⎢⎣0 ⎥⎦
Hence, the KVL equation in matrix form is given by

⎡ R1 R2 + R3 R2 + R3 ⎤ ⎡ I l1 ⎤ ⎡V ⎤ ⎡ R2 I ⎤ ⎡V R2 I ⎤
= − =
⎢⎣ R2 R3 R2 R3 + R4 ⎥⎦ ⎢⎣ I l4 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ R2 I ⎥⎦ ⎢⎣ − R2 I ⎥⎦

Example 9.26 For the network shown in Fig. 9.68, write down the tieset matrix and obtain the
network equilibrium equation in matrix form using KVL.

2A

5Ω

5Ω j5 Ω

2Ω
−4Ω
−j 2Ω
+
10 V
−

Fig. 9.68
 9.12 Network Equilibrium Equation 9.41

Solution The oriented graph and its selected tree are shown in Fig. 9.69.
(3) (3)
2 2
1 3 1 3 Links: {1, 2, 3}
(4) (5) (4) (5) Tieset 1: {1, 4, 6} 1 2 3 4 5 6
Tieset 2: {2, 5, 6}
Tieset 3: {3, 5, 4}
1⎡1 0 0 1 0 1⎤
(1) (6) (2) (1) (6) (2) B = 2 ⎢0 1 0 0 1 1⎥
⎢ ⎥
3 ⎣0 0 1 −1 −1 0⎦

4 4

Fig. 9.69
The KVL equation in matrix form is given by

B Zb BT I l = BV
Vs − B Zb I s
⎡2 0 00⎤ 0 0 ⎡1 0 0⎤ ⎡10 ⎤ ⎡ 0⎤
⎢0 0 20⎥ 0 0 ⎢0 1 0⎥ ⎢ 0⎥ ⎢ 0⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
0 0 0 5 0
0 ⎥ T ⎢0 0 1⎥ 0 −2
Zb = ⎢ ;B = ; Vs = ⎢ ⎥ ; I s =⎢ ⎥
⎢0 0 00⎥ 0 5 ⎢1 0 −1⎥ ⎢ 0⎥ ⎢ 0⎥
⎢0 j5 00⎥ 0 0 ⎢0 1 −1⎥ ⎢ 0⎥ ⎢ 0⎥
⎢0 0 − j 4 ⎥⎦ ⎢1 1 0 ⎥⎦ ⎢ 0⎥ ⎢ 0⎥
⎣ 0 0 0 ⎣ ⎣ ⎦ ⎣ ⎦
⎡2 0 0 0 0 0⎤
⎢0 2 0 0 0 0⎥
⎡1 0 0 1 0 1⎤ ⎢ ⎥ ⎡2 0 0 5 0 j 4⎤
0 0 5 0 0 0⎥ ⎢
B Zb = ⎢0 1 0 0 1 −1⎥ ⎢ = 0 2 0 0 j5 j 4⎥
⎢ ⎥ ⎢0 0 0 5 0 0⎥ ⎢ ⎥
⎣0 0 1 −1 −1 0 ⎦ ⎢0 0 0 0 j5 0⎥ ⎣
0 0 5 5 − j5 0⎦
⎢0 0 0 j 4 ⎥⎦
⎣ 0 0
⎡1 0 0⎤
⎢0 1 0⎥
⎡2 0 0 5 0 − j 4⎤ ⎢ ⎥ ⎡7 − j 4 j4 −5⎤
0 0 1⎥ ⎢
B Z b BT = ⎢ 0 2 0 0 j5 j 4⎥ ⎢ = j 4 2 + j1 − j 5⎥
⎢ ⎥ 1 0 −1⎥ ⎢ ⎥
⎣0 0 5 −5 − j 5 0 ⎦ ⎢⎢ −5 − j5 10 + j 5⎦
0 1 −1⎥ ⎣
⎢ 1 −1 ⎥
⎣ 0⎦
⎡10 ⎤
⎢0⎥
⎡1 0 0 1 0 1⎤ ⎢ ⎥ ⎡100 ⎤
0
Vs = ⎢0 1 0 0
BV 1 −1⎥ ⎢ ⎥ = ⎢ 0 ⎥
⎢ ⎥⎢0 ⎥ ⎢ ⎥
⎣0 0 1 −1 −1 0 ⎦ ⎢ 0 ⎥ ⎣ 0 ⎦
⎢0⎥
⎣ ⎦
⎡ 0⎤
⎢ 0⎥
⎡2 0 0 5 0 j 4⎤ ⎢ ⎥ ⎡ 0⎤
−2
B Zb I s = ⎢0 2 0 0 j5 j 4⎥ ⎢ ⎥ = ⎢ 0⎥
⎢ ⎥ ⎢ 0⎥ ⎢ ⎥
⎣0 0 5 5 − j5 0 ⎦ ⎢ ⎥ ⎣ −10 ⎦
0
⎢ 0⎥
⎣ ⎦
9.42 Network Analysis and Synthesis
Hence, the KCL equation in matrix form is given by
⎡7 j 4 j4 5⎤ ⎡ I l1 ⎤ ⎡10 ⎤ ⎡ 0 ⎤ ⎡10 ⎤
⎢ j 4 2 + j1 j 5⎥ ⎢ I l2 ⎥ = ⎢ 0 ⎥ − ⎢ 0 ⎥ = ⎢ 0 ⎥
⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ −5 5 j 5 10 + j 5⎦ ⎣ I l3 ⎦ ⎣ 0 ⎦ ⎣ −10 ⎦ ⎣10 ⎦

Example 9.27 For the network shown in Fig. 9.70, write down the tieset matrix and obtain the
network equilibrium equation in matrix form using KVL.
j5.66 Ω
j5 Ω j10 Ω

+ 3Ω
50 ∠0° V − 5Ω
I1 −4Ω
−j I2

Fig. 9.70
Solution The branch currents are so chosen that they assume the direction out of the dotted terminals.
Because of this choice of current direction, the mutual inductance is positive. The oriented graph and its
selected tree are shown in Fig. 9.71.
1 1

Links: {1, 3} 1 2 3
(2) (3) (1) (2) (3) Tieset 1: {1, 2}
(1)
Tieset 3: {3, 2} ⎡1 1 0⎤
B=⎢
⎣0 1 1⎥⎦

2 2

Fig. 9.71
The KVL equation in matrix form is given by

B Zb BT I l = BV Vs − B Zb I s
Here, I s = 0,
B Zb BT I l = BV
Vs
⎡ j5 0 j 5 66 ⎤ ⎡ 1 0⎤ ⎡50 ∠0°⎤
Zb = ⎢ 0 3 − j4 0 ⎥ ; BT ⎢ 1 1⎥ ; Vs = ⎢ 0 ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ j 5 66 0 5 j10 ⎦ ⎣0 1⎦ ⎣ 0 ⎦
⎡ j5 0 j 5 66 ⎤
⎡ 1 1 0⎤ ⎢ ⎡ j5 3 − j4 j 5.66 ⎤
B Zb = ⎢ ⎥ 0 3 − j 4 0 ⎥=⎢
⎣ − ⎢
⎦ j 5 666 ⎥ + + j10 ⎥⎦
5 + j10 ⎦ ⎣
0 1 1 j 5 66 3 j 4 5
⎣ 0
⎡ 1 0⎤
⎡ j5 3 j4 j 5.66 ⎤ ⎢ ⎡ 3 + j1 −3 + j 9.66 ⎤
B Z b BT = ⎢ ⎥ 1 −1⎥ = ⎢
3 j 4 5 + j10 ⎦ ⎢
⎣ j 5 66 ⎥ −3 + j 9 66 8 + j 6 ⎥⎦
⎣0 1⎦ ⎣
⎡50 ∠0°⎤
⎡ 1 1 0⎤ ⎢ ⎡50 ∠0°⎤
Vs = ⎢
BV ⎥ 0 ⎥=⎢
⎣ 0 −1 1⎦ ⎢ ⎥ ⎣ 0 ⎥⎦
⎣ 0 ⎦
 9.12 Network Equilibrium Equation 9.43

Hence, the KVL equation in matrix form is given by

⎡ 3 j1 −3 j 9.66 ⎤ ⎡ I l1 ⎤ ⎡50 ∠0°⎤
=
⎢⎣ −3
3 j 9 66 8 j 6 ⎥⎦ ⎢⎣ I l2 ⎥⎦ ⎢⎣ 0 ⎥⎦

Example 9.28 For the network shown in Fig. 9.72, write down the tieset matrix and obtain the
network equilibrium equation in matrix form using KVL.
3Ω j4 Ω

+ j3 Ω
50 ∠45° V j5 Ω −j8 Ω
−
I1 I2

Fig. 9.72
Solution The branch currents are so chosen that they assume the direction out of the dotted terminals.
Because of this choice of current direction, the mutual inductance is positive. The oriented graph and its
selected tree are shown in Fig. 9.73.
1 1

Links: {1, 3} 1 2 3
(2) (3) (1) (2) (3) Tieset 1: {1, 2}
(1)
Tieset 3: {3, 2} ⎡1 1 0⎤
B=⎢
⎣0 1 1⎥⎦

2 2

Fig. 9.73
The KVL equation in matrix form is given by
B Zb BT I l = BV
Vs − B Zb I s
Here, I s = 0,
B Zb BT I l = BV
Vs
⎡3 + j 4
j3 0 ⎤ ⎡ 1 0⎤ ⎡50 ∠45°⎤
Zb = ⎢ j 3
j 5 0 ⎥ ; BT = ⎢ 1 −1⎥ ; Vs = ⎢ 0 ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ 0 0 − j 8⎦ ⎣0 1⎦ ⎣ 0 ⎦
⎡3 + j 4 j 3 0 ⎤
⎡ 1 1 0⎤ ⎢ ⎡3 + j 7 j 8 0 ⎤
B Zb = ⎢ ⎥ j3 j5 0 ⎥ = ⎢
⎣ − ⎦ ⎢ ⎥ − − − j8⎥⎦
0 − j 8⎦ ⎣
0 1 1 j 3 j 5
⎣ 0
⎡ 1 0⎤
⎡3 + j 7 j8 0⎤ ⎢ ⎡3 + j15 j 8⎤
B Z b BT = ⎢ 1 −1⎥ = ⎢
⎣ − j3 j 5 − j8⎥⎦ ⎢ ⎥ ⎣ − j8 j 3⎥⎦
⎣0 1⎦
⎡50 ∠45°⎤
⎡ 1 1 0⎤ ⎢ ⎡50 ∠45°⎤
Vs = ⎢
BV ⎥ 0 ⎥=⎢
⎣ 0 −1 1⎦ ⎢ ⎥ ⎣ 0 ⎥⎦
⎣ 0 ⎦
9.44 Network Analysis and Synthesis
Hence, the KVL equation in matrix form is given by,

⎡3 j15 j8⎤ ⎡ I l1 ⎤ ⎡50 ∠45°⎤

=
⎢⎣ − j8 j 3⎥⎦ ⎢⎣ I l3 ⎥⎦ ⎢⎣ 0 ⎥⎦

Example 9.29 For the network shown in Fig. 9.74, obtain branch voltages using KCL equation on
node basis.

8Ω

3Ω 4Ω
4Ω 5Ω

12 V 24 V

Fig. 9.74

Solution The oriented graph is shown in Fig. 9.75.

1 (3) 2

(2)
(4)
(1) (5)

Fig. 9.75
The complete incidence matrix for the graph is
1 2 3 4 5
1 ⎡ 1 −1 1 0 0⎤
Aa = 2 ⎢ 0 0 −1 1 1⎥
⎢ ⎥
3 ⎣ −1 1 0 −1 1⎦

Eliminating the last row from the matrix Aa, we get the incidence matrix A.
1 2 3 4 5
1 ⎡ 1 −1 1 0 0 ⎤
A= ⎢
2 ⎣0 0 −1 1 1⎥⎦

The KCL equation in matrix form is given by

Yb AT Vn = AI s − AY
AY Yb Vs
Is = 0 ,
 9.12 Network Equilibrium Equation 9.45

Here, Yb AT Vn = AY
AY Yb Vs
⎡1 ⎤
⎢4 0 0 0 0⎥
⎢ ⎥
⎢0 1 0 0 0⎥ ⎡ 1 0⎤ ⎡ 0⎤
⎢ 3 ⎥ ⎢ −1 0 ⎥ ⎢ 12⎥
⎢ 1 ⎥ T ⎢ ⎥ ⎢ ⎥
Yb = ⎢ 0 0 0 0⎥ ; A = ⎢ − ⎥ ; Vs = ⎢ 0 ⎥
⎢ 8 ⎥
1 ⎢ 0 1⎥ ⎢ 24 ⎥
⎢0 0 0 0⎥ ⎢⎣ 0 1⎥⎦ ⎢⎣ 0 ⎥⎦
⎢ 4 ⎥
⎢ 1⎥
⎢0 0 0 0 ⎥
⎣ 5⎦
⎡1 ⎤
⎢4 0 0 0 0⎥
⎢ 1 ⎥
⎢0 0 0 0⎥
⎢ 3 ⎥ ⎡1 −
1 1
0 0⎥
⎤
⎡ 1 −1 1 0 0 ⎤ ⎢ 1 ⎥ ⎢
Yb = ⎢ 0⎥ = ⎢ 4 3 8
1 ⎥⎥
AY ⎢0 0 0
⎣0 0 −1 1 1⎥⎦ ⎢ 8 ⎥ ⎢0 0 −
1 1
0⎥ ⎣ 5⎦
⎢0 1 8 4
0 0
⎢ 4 ⎥
⎢ 1⎥
⎢0 0 0 0 ⎥
⎣ 5⎦
⎡ 1 0⎤
⎡1 1 1 ⎤ ⎡ 17 1⎤
⎢ − 0 0 ⎥ ⎢ −1 0 ⎥ ⎢ − ⎥
⎢ ⎥
Yb AT = ⎢ 4 3 8 1 −1⎥ = ⎢ 24 8
1 1 ⎥⎥ ⎢ 23 ⎥⎥
AY
⎢0 1 1
0 − ⎢ 0 1⎥ ⎢ −
⎣ 8 4 5⎦ ⎢ 0 1⎥⎦ ⎣ 8 40 ⎦
⎣
⎡0⎤
⎡1 1 1 ⎤⎢ ⎥
⎢4 − 3 8
0 0 ⎥ 12
⎢ ⎥ ⎡ −4 ⎤
Yb Vs = ⎢ 0 =
1 1 ⎥⎥ ⎢ ⎥ ⎢⎣ 6 ⎥⎦
AY
⎢0 1
0 − ⎢ 24 ⎥
⎣ 8 4 5⎦ ⎢ 0 ⎥
⎣ ⎦
Hence, the KCL equation in matrix form is given by
⎡ 17 1⎤
⎢ 24 − ⎥ ⎡V ⎤
8 n1 ⎡ −4 ⎤ ⎡ 4 ⎤
⎢ 1 = −⎢ ⎥ = ⎢ ⎥
⎢− 23 ⎥⎥ ⎢⎣Vn2 ⎥⎦ ⎣ 6 ⎦ ⎣ −6 ⎦
⎣ 8 40 ⎦
Solving this matrix equation,
Vn1 = 3 96 V
Vn2 = −9 57 V
9.46 Network Analysis and Synthesis
Branch voltages are given by,
Vb AT Vn
⎡V1 ⎤ ⎡ 1 0 ⎤ ⎡ 3 96 ⎤
⎢V2 ⎥ ⎢ −1 0 ⎥ ⎢ −3 96 ⎥
⎢ ⎥ ⎢ ⎥ ⎡ 3 96 ⎤ ⎢ ⎥
⎢V3 ⎥ = ⎢ 1 −1⎥ ⎢⎣ −9 57⎥⎦ = ⎢ 13.53⎥
⎢V4 ⎥ ⎢ 0 1⎥ ⎢ −9 57⎥
⎢⎣V5 ⎥⎦ ⎢⎣ 0 1⎥⎦ ⎢⎣ −9 57⎥⎦

Example 9.30 For the network shown in Fig. 9.76, write down the f-cutset matrix and obtain the
network equilibrium equation in matrix form using KCL.
1Ω 1Ω

1Ω
10 V 1Ω

2A

Fig. 9.76
Solution The oriented graph and its selected tree are shown in Fig. 9.77.
1 (3) 2 1 (3)
2
Twigs: {2, 4}
f-cutset 2: {2, 1, 3} 1 2 3 4
f-cutset 4: {4, 3}
(1) (2) (4) (1) (2) (4) 2 ⎡ −1 1 1 0 ⎤
Q= ⎢
4 ⎣ 0 0 −1 1⎥⎦

3 3

Fig. 9.77
The KCL equation in matrix form is given by
Yb QT Vt = Q I s − QY
QY Yb Vs
⎡1 0⎤
0 0 ⎡0 ⎤ ⎡10 ⎤ ⎡ −1 0 ⎤
⎢0 0⎥
1 0 ⎢0 ⎥ ⎢ 0⎥ T ⎢ 1 0⎥
Yb ⎢ ⎥ I s = ⎢ ⎥ ; Vs = ⎢ ⎥;Q = ⎢ ⎥
⎢0 0
0⎥ 1 ⎢0 ⎥ ⎢ 0⎥ ⎢ 1 1⎥
⎢⎣0 1 ⎥⎦
0 0 ⎢⎣ 2⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ 0 1⎥⎦
⎡1 0 0 0⎤
⎡ −1 1 1 0 ⎤ ⎢0 1 0 0 ⎥ ⎡ −1 1 1 0 ⎤
Yb = ⎢
QY ⎢ ⎥=
⎣ 0 0 −1 1⎥⎦ ⎢0 0 1 0 ⎥ ⎢⎣ 0 0 −1 1⎥⎦
⎢⎣0 0 0 1 ⎥⎦
⎡ −1 0⎤
⎡ −1 1 1 0 ⎤ ⎢ 1
T 0⎥ ⎡ 3 1⎤
Yb Q = ⎢
QY ⎢ ⎥=
⎣ 0 0 −1 1⎥⎦ ⎢ 1 1⎥ ⎢⎣ −1 2⎥⎦
⎢⎣ 0 1⎥⎦
 9.12 Network Equilibrium Equation 9.47

⎡0 ⎤
⎡ −1 1 1 0 ⎤ ⎢0 ⎥ ⎡0 ⎤
Q Is = ⎢ ⎢ ⎥=
⎣ 0 0 −1 1⎥⎦ ⎢0 ⎥ ⎢⎣ 2⎥⎦
⎢⎣ 2⎥⎦
⎡10 ⎤
⎡ −1 1 1 0 ⎤ ⎢ 0 ⎥ ⎡ −10 ⎤
QYYb Vs = ⎢ ⎢ ⎥=
⎣ 0 0 −1 1⎥⎦ ⎢ 0 ⎥ ⎢⎣ 0 ⎥⎦
⎢⎣ 0 ⎥⎦
Hence, the KCL equation is given by
⎡ 3 1⎤ ⎡Vt2 ⎤ ⎡0 ⎤ ⎡ −10 ⎤ ⎡10 ⎤
= − =
⎢⎣ −1 2⎥⎦ ⎢⎣Vt4 ⎥⎦ ⎢⎣ 2⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ 2 ⎥⎦
Solving this matrix equation,
Vt2 = 4.4 V
Vt4 = 3 2 V

Example 9.31 Calculate the twig voltages using KCL equation for the network shown in Fig. 9.78.
2Ω

5Ω 10 Ω
5Ω
10 Ω 5Ω

+
910 V −

Fig. 9.78
Solution The oriented graph and one of the trees are shown in Fig. 9.79.
(5) (5)

Twigs: {1, 2, 3}
(3) (4) (3) (4) f-cutset 1: {1, 4, 5, 6}
f-cutset 2: {2, 4, 5} 1 2 3 4 5 6
(6))
f-cutset 3: {3, 4, 6} 1 ⎡ 1 0 0 −1 −1 1⎤
Q = 2 ⎢0 1 0 −1 −1 0 ⎥
(1) (6) (2) (1) (2)
⎢ ⎥
3 ⎣0 0 1 1 0 1⎦

Fig. 9.79
The network equilibrium equation on node basis can be written as

Yb QT Vt = QI s − QY
QY Yb Vs
Here, I s = 0,
Yb QT Vt = QY
QY Yb Vs
9.48 Network Analysis and Synthesis

⎡0 2 0 0 0 0 0⎤ ⎡ 1 0 0⎤ ⎡9110 ⎤
⎢ 0 0.2 0 0 0 0⎥ ⎢ 0 1 0⎥ ⎢ 0⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
0 0 02 0 0 0⎥ T ⎢ 0 0 1⎥ 0⎥
Yb = ⎢ ;Q = ; Vs = ⎢
⎢ 0 0 0 01 0 0⎥ ⎢ −1 −1 −1⎥ ⎢ 0⎥
⎢ 0 0 0 0 0.5 0⎥ ⎢ −1 −1 0 ⎥ ⎢ 0⎥
⎢ 0 ⎥ ⎢ 1 1⎥⎦ ⎢ 0⎥
⎣ 0 0 0 0 0 1⎦ ⎣ 0 ⎣ ⎦
⎡0 2 0 0 0 0 0⎤
⎢ 0 0.2 0 0 0 0⎥
⎡1 0 0 1 −1 1⎤ ⎢ ⎥ ⎡0.2 0 0 −0.1 −0.5 0.1⎤
0 0 0 2 0 0 0⎥ ⎢
Yb = ⎢0 1 0
QY 1 −1 0 ⎥ ⎢ = 0 0.2 0 −0.1 −0 5 0⎥
⎢ ⎥⎢ 0 0 0 01 0 0⎥ ⎢ ⎥
⎣0 0 1 1 0 1⎦ ⎢ 0 0 0 0.2 −0.1 0 0.1⎦
0 0 0 0.5 0⎥ ⎣
⎢ 0 0 0 1⎥⎦
⎣ 0 0 0
⎡ 1 0 0⎤
⎢ 0 1 0⎥
⎡0.2 0 0 −0.1 −0.5 0.1⎤ ⎢ ⎥ ⎡ 0.9 0.6 0 2⎤
0 0 1⎥ ⎢
Yb QT = ⎢ 0 0.2
QY 0 −0.1 −0 5 0⎥ ⎢ = 0.6 0.8 0 1⎥
⎢ ⎥ ⎢ −11 1 1⎥ ⎢ ⎥
⎣ 0 0 0.2 −0.1 0 0.1⎦ ⎢ 0.2 0.1 0 3⎦
−11 1 0 ⎥ ⎣
⎢ 1 0 1⎥⎦
⎣
⎡910
1 ⎤
⎢ 0⎥
⎡0.2 0 0 −0.1 −0.5 0.1⎤ ⎢ ⎥ ⎡182⎤
0⎥ ⎢ ⎥
Yb Vs = ⎢ 0 0.2
QY 0 −0.1 −0 5 0⎥ ⎢ = 0
⎢ ⎥ 0⎥ ⎢ ⎥
⎣ 0 0 0.2 −0.1 0 0.1⎦ ⎢⎢ 0
0⎥ ⎣ ⎦
⎢ 0⎥
⎣ ⎦
Hence, KCL equation can be written as,
⎡ 0.9 0.6 0 2⎤ ⎡ vt1 ⎤ ⎡ −182⎤
⎢0.6 0.8 0 1⎥ ⎢ vt ⎥ = ⎢ 0 ⎥
⎢ ⎥⎢ 2⎥ ⎢ ⎥
⎣ 0.2 0.1 0 3⎦ ⎣ vt3 ⎦ ⎣ 0 ⎦
Solving this matrix equation,
vt1 = −460 V
vt2 = 320 V
vt3 = 200 V

Example 9.32 For the network shown in Fig. 9.80, obtain equilibrium equation on node basis.
Ω
5

Ω Ω
10 A 5 5 10
Ω

Fig. 9.80
 9.12 Network Equilibrium Equation 9.49

Solution The oriented graph and its selected tree are shown in Fig. 9.81.
(2) (2)
1 2 1 2 Twigs: {1, 3}
f-cutset 1: {1, 2} 1 2 3 4
(3) (3) f-cutset 3: {3, 2, 4}
(1) (1) 1 ⎡1 −1 0 0 ⎤
(4) (4) Q= ⎢
3 ⎣0 −1 1 1 ⎥⎦
3 3

Fig. 9.81
The KCL equation in matrix form is given by
Yb QT Vt = Q I s − QY
QY Yb Vs
Here, Vs = 0,
Yb QT Vt = Q I s
QY
⎡5 0 0 0⎤ ⎡ 1 0⎤ ⎡ −10 ⎤
⎢0 5 0 0 ⎥ T
⎢ −1 −1 ⎥ ⎢ 0⎥
Y ⎢ ⎥ Q =⎢ ⎥ ; Is = ⎢ ⎥
⎢0 0 5 0⎥ ⎢ 0 1⎥ ⎢ 0⎥
⎢⎣0 0 0 10 ⎥⎦ ⎢⎣ 0 1⎥⎦ ⎢⎣ 0 ⎥⎦
⎡ 5 0 0 0⎤
⎡1 −1 0 0⎤ ⎢0 5 0 0 ⎥ ⎡ 5 5 0 0 ⎤
Yb = ⎢
QY ⎢ ⎥=
⎣0 −1 1 1 ⎥⎦ ⎢0 0 5 0 ⎥ ⎢⎣0 5 5 10 ⎥⎦
⎢⎣0 0 0 10 1 ⎥⎦
⎡ 1 0⎤
⎡ 5 −5 0 0 ⎤ ⎢ −1 −1⎥ ⎡10 5⎤
Yb QT = ⎢
QY ⎢ ⎥=
⎣0 −5 5 10 ⎥⎦ ⎢ 0 1⎥ ⎢⎣ 5 20 ⎥⎦
⎢⎣ 0 1⎥⎦
⎡ −10 ⎤
⎡1 −1 0 0 ⎤ ⎢ 0 ⎥ ⎡ −10 ⎤
Q Is = ⎢ ⎢ ⎥=
⎣0 −1 1 1 ⎥⎦ ⎢ 0 ⎥ ⎢⎣ 0 ⎥⎦
⎢⎣ 0 ⎥⎦
Hence, KCL equation will be written as
⎡10 5⎤ ⎡ vt1 ⎤ ⎡ −10 ⎤
⎢⎣ 5 20 ⎥⎦ ⎢ vt ⎥ = ⎢⎣ 0 ⎥⎦
⎣ 3⎦
Solving this matrix equation,
8
vt1 = − V
7
2
vt3 = V
7

Example 9.33 For the network shown in Fig. 9.82, write down the f-cutset matrix and obtain the
network equilibrium equation in matrix form using KCL and calculate v.
9.50 Network Analysis and Synthesis
2v

2Ω

2Ω
+
+
2V − v 2Ω 2Ω
−

Fig. 9.82
Solution The oriented graph and its selected tree are shown in Fig. 9.83
Since voltage v is to be determined, Branch 2 is chosen as twig,
1 (4) 2 1 (4) 2
Twigs: {2, 4}
f-cutset 2: {2, 1, 3} 1 2 3 4
(2) (3) (2) (3) f-cutset 3: {4, 3}
⎡ −1 1 1 0 ⎤
Q=⎢
(1) (1) ⎣ 0 0 −1 1⎥⎦
3 3

Fig. 9.83
The KCL equation in matrix form is given by

Yb QT Vt = Q I s − QY
QY Yb Vs
⎡0 5 0 0 0⎤ ⎡ −1 0 ⎤ ⎡ 0⎤ ⎡ 2⎤
⎢ 0 0.5 0 0⎥ T ⎢ 1 0⎥ ⎢ 0⎥ ⎢0 ⎥
Y ⎢ ⎥ Q =⎢ ⎥ ; Is = ⎢ ⎥ ; Vs = ⎢ ⎥
⎢ 0 0 05 0⎥ ⎢ 1 −1⎥ ⎢ 0⎥ ⎢0 ⎥
⎢⎣ 0 0 0 0.5⎥⎦ ⎢⎣ 0 1⎥⎦ ⎢⎣ −2v ⎥⎦ ⎢⎣0 ⎥⎦
⎡0 5 0 0 0⎤
⎢
⎡ −1 1 1 0 ⎤ 0 0.5 0 0 ⎥ ⎡ −0.5 0.5 0.5 0⎤
QYYb = ⎢ ⎥ ⎢ ⎥=
⎣ 0 0 −1 1⎦ ⎢ 0 0 05 0 ⎥ ⎢⎣ 0 0 −0.5 0.5⎥⎦
⎢⎣ 0 0 0 0 5⎥⎦
⎡ −1 0 ⎤
T ⎡ −0.5 0.5 0 5 0 ⎤ ⎢ 1 0 ⎥ ⎡ 1 5 −0 5⎤
Yb Q = ⎢
QY ⎢ ⎥=
⎣ 0 0 −0.5 0.5⎥⎦ ⎢ 1 −1⎥ ⎢⎣ −0 5 1⎥⎦
⎢⎣ 0 1⎥⎦
⎡ 0⎤
⎡ −1 1 1 0 ⎤ ⎢ 0 ⎥ ⎡ 0 ⎤
Q Is = ⎢ ⎢ ⎥=
⎣ 0 0 −1 1⎥⎦ ⎢ 0 ⎥ ⎢⎣ −2v ⎥⎦
⎢⎣ −2v ⎥⎦
⎡ 2⎤
⎡ −0.5 0.5 0 5 0 ⎤ ⎢0 ⎥ ⎡ −1⎤
Yb Vs = ⎢
QY ⎢ ⎥=
⎣ 0 0 −0.5 0.5⎥⎦ ⎢0 ⎥ ⎢⎣ 0 ⎥⎦
⎢⎣0 ⎥⎦
⎡ 1⎤
Q I s − QY
Yb Vs = ⎢
⎣ −2v ⎥⎦
 9.12 Network Equilibrium Equation 9.51

Hence, the KCL equation can be written as

⎡ 1.5 −0.5⎤ ⎡ vt2 ⎤ ⎡ 1⎤
=
⎢⎣ −0 5 1⎥⎦ ⎢⎣ vt4 ⎥⎦ ⎢⎣ −2v ⎥⎦

From Fig. 9.82, vt2 v

Solving this matrix equation,
vt2 = 0 44 V
vt4 = 0 66 V
v vt2 = 0 44 V

Example 9.34 For the network shown in Fig. 9.84, write down the f-cutset matrix and obtain the
network equilibrium equation in matrix form using KCL and calculate v.

4Ω
0.5 A
2Ω
+ v −

1V +
− 4Ω 2Ω

Fig. 9.84
Solution The voltage and current sources are converted into accompanied sources by source–shifting
method as shown in Fig. 9.85.
4Ω

2Ω
+ v −
1V +
− 0.5 A
+ 4Ω 2Ω
1V −
0.5 A

Fig. 9.85
The oriented graph and its selected tree are shown in Fig. 9.86.
2
2
Twigs: {1, 2}
f-cutset 1: {1, 4}
1 1 f-cutset 2: {2, 3} 1 2 3 4
(1) (4) (1) (4)
⎡ 1 0 0 −1⎤
Q=⎢
(2) (3) (2) (3) ⎣0 1 −1 0 ⎥⎦

3 3

Fig. 9.86
9.52 Network Analysis and Synthesis
The KCL equation in the matrix form is given by

Yb QT Vt = Q I s − QY
QY Yb Vs
⎡0 25 0 0 0⎤ ⎡ 1 0⎤ ⎡ 0 ⎤ ⎡1 ⎤
⎢ 0 0.5 0 0⎥ T ⎢ 0 1⎥ ⎢ 0 ⎥ ⎢1 ⎥
Yb = ⎢ ⎥;Q ⎢ ⎥; I = ⎢ ⎥; V = ⎢ ⎥
⎢ 0 0 0 25 0 ⎥ ⎢ 0 −1⎥ s ⎢ 0.5 ⎥ s ⎢0 ⎥
⎢⎣ 0 0 0 0.5⎥⎦ ⎢⎣ −1 0 ⎥⎦ ⎢⎣ −0.5⎥⎦ ⎢⎣0 ⎥⎦
⎡0 25 0 0 0⎤
⎡ 1 0 0 −1⎤ ⎢ 0 0.5 0 0 ⎥ ⎡0.25 0 0 −0.5⎤
Yb = ⎢
QY ⎢ ⎥=
⎣0 1 −1 0 ⎥⎦ ⎢ 0 0 0 25 0 ⎥ ⎢⎣ 0 0.5 −0.25 0 ⎥⎦
⎢⎣ 0 0 0 0.5⎥⎦
⎡ 1 0⎤
⎡ 0 .25 0 0 − 0 . 5 ⎤ ⎢ 0 1⎥ ⎡0 75 0 ⎤
QYYb QT = ⎢ ⎥ ⎢ ⎥=⎢
⎣ 0 0. 5 −0 . 25 0 ⎦⎢ 0 −1 ⎥ ⎣ 0 0. 75⎥⎦
⎢⎣ −1 0 ⎥⎦
⎡ 0 ⎤
⎡1 0 0 −1⎤ ⎢ 0 ⎥ ⎡ 0 5⎤
Q Is = ⎢ ⎢ ⎥=
⎣0 1 −1 0 ⎥⎦ ⎢ 0 5 ⎥ ⎢⎣ −0 5⎥⎦
⎢⎣ −0 5⎥⎦
⎡1 ⎤
⎡0.25 0 0 −0.5⎤ ⎢1 ⎥ ⎡0 25⎤
Q Yb Vs = ⎢ ⎢ ⎥=
⎣ 0 0.5 −0.25 0 ⎥⎦ ⎢0 ⎥ ⎢⎣ 0 5 ⎥⎦
⎢⎣0 ⎥⎦
⎡0 25⎤
Q I s − Q Yb Vs = ⎢
⎣ −1 ⎥⎦
Hence, the KCL equation can be written as

⎡0 75 0 ⎤ ⎡ vt1 ⎤ ⎡0 25⎤
=
⎢⎣ 0 0.75⎥⎦ ⎢⎣ vt2 ⎥⎦ ⎢⎣ −1 ⎥⎦

Solving this matrix equation,

vt1 = 0 33 V
vt2 = −1 33 V
From Fig. 9.85, v 1 + vt2 = −0.33 V

9.13 DUALITY
Two networks are said to be the dual of each other when the mesh equations of one network are the same
as the node equations of the other. Kirchhoff ’s voltage law and current law are same, word for word, with
voltage substituted for current, independent loop for independent node pair, etc. Similarly, two graphs are
said to be dual of each other if the incidence matrix of any one of them is equal to the circuit matrix of the
other. Only planar networks have duals.
 9.13 Duality 9.53

Table 9.1 Conversion for dual electrical circuits

Loop basis Node basis
Current Voltage
Resistance Conductance
Inductance Capacitance
Branch current Branch voltage
Mesh Node
Short circuit Open circuit
Parallel path Series path

The following steps are involved in constructing the dual of a network:

1. Place a node inside each mesh of the given network. These internal nodes correspond to the
independent nodes in the dual network.
2. Place a node outside the given network. The external node corresponds to the datum node in the dual
network.
3. Connect all internal nodes in the adjacent mesh by dashed lines crossing the common branches.
Elements which are the duals of the common branches will form the branches connecting the
corresponding independent node in the dual network.
4. Connect all internal nodes to the external node by dashed lines corresponding to all external branches.
Duals of these external branches will form the branches connecting independent nodes and the datum
node.
5. A clockwise current in a mesh corresponds to a positive polarity (with respect to the datum node) at
the dual independent node.
6. A voltage rise in the direction of a clockwise mesh current corresponds to a current flowing towards
the dual independent node.

Example 9.35 Draw the dual of the network shown in Fig. 9.87.
R1 C

V +
− L R2

Fig. 9.87
Solution The following steps are involved in constructing the dual of the network as shown in Fig. 9.88.
1. Place a node inside each mesh.
2. Place a node outside the mesh which will correspond to the datum node.
3. Connect two internal nodes through a dashed line. The element which is dual of the common branch
(here capacitance) will form the branch connecting the corresponding independent node in the dual
network.
4. Connect all internal nodes to the external node by dashed lines crossing all the branches. The dual of
these branches will form the branches connecting the independent node and datum.
9.54 Network Analysis and Synthesis
R1 C

V + L R2
− 1 2

Fig. 9.88
The dual network is shown in Fig. 9.89.
1 C 2

G1 L G2
I

Fig. 9.89

Example 9.36 Draw the dual of the network of Fig. 9.90.

C2 C4

R1 L3 R5
I

Fig. 9.90
Solution For drawing the dual network, proceed in the same way as in Example 9.35 as shown in Fig. 9.91.
C2 C4

R1 L3
I
1 2 3 R5

Fig. 9.91
 9.13 Duality 9.55

The dual network is shown in Fig. 9.92.

1 G1 2 C3 3

V + L2 L4 G5
−

Fig. 9.92

Example 9.37 Draw the dual of the network shown in Fig. 9.93.
R1 V R2
+−
R3 R4

C
L R5

Fig. 9.93
Solution For drawing the dual network, proceed in the same way as in Example 9.35, as shown in Fig. 9.94.
R1 V R2
+−
1
R3 R4

2 3

C
L R5

Fig. 9.94
The dual network is shown in Fig. 9.95.
G4

G3 2 L
1 3

G1 G2 C G5
I

Fig. 9.95
9.56 Network Analysis and Synthesis

Example 9.38 Draw the dual of the network shown in Fig. 9.96.

L2

C
+
V
−
R1 L1 R2 R3

Fig. 9.96

Solution For drawing the dual network, proceed in the same way as in Example 9.35 as shown in Fig. 9.97.

C L2
1 2

+
V
− R3
R1 L1 R2

Fig. 9.97

The dual network is shown in Fig. 9.98.

C1

L 2 G2
1 3

I G1 −
C2 G3 V
+

Fig. 9.98
 9.13 Duality 9.57

Example 9.39 Draw the dual of the network shown in Fig. 9.99.

3Ω 8F

4H
10 sin w t 5Ω

Fig. 9.99

Solution For drawing the dual network, proceed in the same way as in Example 9.35 as shown in Fig. 9.100.
3Ω 8F

1 4H 2
10 sin w t 5Ω

Fig. 9.100

The dual network is shown in Fig. 9.101.

4F
1 2

Ω
10 sin w t 3 8H 5
Ω

Fig. 9.101

Example 9.40 Draw the dual of the network shown in Fig. 9.102.

2H 4F

5V + 1A 1H
−

1F 2Ω

Fig. 9.102
9.58 Network Analysis and Synthesis
Solution For drawing the dual network, proceed in the same way as in Example 9.35 as shown in
Fig. 9.103.
2H 4F

1 2
5V + 1A 1H
−

1F 2Ω

Fig. 9.103

The dual network is shown in Fig. 9.104.

1V
1 2
−+

Ω
5A 2F 1H 4H 1F 2

Fig. 9.104

Exercises
9.1 For the networks shown in Fig. 9.105–9.108, (iii)
write the incidence matrix, tieset matrix and
f-cutset matrix. 1Ω

(i) R2 C R4 L2 2Ω

R1 1Ω
2Ω 20 A 2Ω 1Ω
R3 +
L1 10 V −

V +
− Fig. 9.107
Fig. 9.105
(iv) I

(ii) L R1 C1

C R1
+ L1
V − L2
I1 R2 + V R2
− 2

Fig. 9.108
Fig. 9.106
 Exercises 9.59

9.2 For the graph shown in Fig. 9.109, write the 9.6 Draw the dual networks for the circuits shown
incidence matrix, tieset matrix and f-cutset in Fig. 9.111–9.115.
matrix. (i)
(4)
1 2 3 4
4A 4Ω 3H 2F
(1) (2) (3)

(6)
(5) (7)
Fig. 9.111
(ii)
5 R1 R3

Fig. 9.109
+
9.3 The incidence matrix is given as follows: v (t ) R2 C
−
Branches →
1 2 3 4 5 6 7 8
−1 −1 0 0 0 0 1 0 Fig. 9.112
0 1 1 0 1 0 0 0 (iii)
0 0 −1 −1 0 1 0 0 R L

1 0 0 1 0 0 0 1
Draw oriented graph and write tieset matrix. V + C
−
9.4 The incidence matrix is given below:
Branches →
1 2 3 4 5 6 7 8 9 10 Fig. 9.113
(iv)
0 0 1 1 1 1 0 1 0 0 L1 L2
0 −1 −1 0 0 0 −1 0 0 −1
−1 1 0 0 0 0 0 −1 −1 1 R1 R2
1 0 0 0 −1
1 1 1 0 0 0
Draw the oriented graph. V
+ C R3
−
9.5 For the network shown in Fig. 9.110, draw the
oriented graph and obtain the tieset matrix.
Use this matrix to calculate the current i.
Fig. 9.114
i 1Ω 2Ω (v)

2Ω 1V
1Ω 2H
2V 3F
+
1Ω V − 3Ω
3Ω
3Ω 4F 3Ω

Fig. 9.110
1A
[0.91 A]
Fig. 9.115
9.60 Network Analysis and Synthesis
9.7 Using the principles of network topology, R1
write the loop/node equation in matrix form
for the network shown in Fig. 9.116. R2 ig

Vg +
− R4
R3

Fig. 9.116

Objective-Type Questions
9.1 The number of independent loops for a (a) 3 (b) 4
network with n nodes and b branches is (c) 6 (d) 7
(a) n − 1 9.5 Consider the network graph shown in Fig.
(b) b − n 9.119.
(c) b − n + 1
(d) independent of the number of nodes
9.2 A network has 7 nodes and 5 independent
loops. The number of branched in the network
is
(a) 13 (b) 12 Fig. 9.119
(c) 11 (d) 10
Which one of the following is NOT a tree of
9.3 Identify which of the following is NOT a tree this graph?
of the graph shown in Fig. 9.117. (a)
(b)
a

2
1 3
b c

(c) (d)
d e f g

4 h 5
Fig. 9.117
9.6 Figure 9.120 below shows a network and its
(a) begh (b) defg
graph is drawn aside. A proper tree chosen for
(c) adfg (d) aegh
analyzing the network will contain the edges.
9.4 The minimum number of equations required
to analyze the circuit shown in Fig. 9.118 is
C C a b c
+
−
R R
d

V R C R Fig. 9.120
(a) ab, bc, ad (b) ab, bc, ca
Fig. 9.118 (c) ab, bd, ca? (d) ac, bd, ad
 Answers to Objective-Type Questions 9.61

9.7 The graph of an electrical network has n nodes (2) (4)

and b branches. The number of links with
respect to the choice of a tree is given by (3)
(1) (5)
(a) b − n + 1 (b) b + n
(c) n − b + 1 (d) n − 2b − 1.
9.8 In the graph shown in Fig. 9.121, one Fig. 9.122
possible tree is formed by the branches 4, 5,
6, 7. Then one possible fundamental cutset is (a) 4 (b) C
(c) 5 (d) 8
8
9.10 Which one of the following is a cutset of the
6 7
graph shown in Fig. 9.123?
1 2 5 4 (3)

3 (2) (4)

Fig. 9.121

(a) 1, 2, 3, 8 (b) 1, 2, 5, 6 (5)

(1) (6)
(c) 1, 5, 6, 8 (d) 1, 2, 3, 7, 8
9.9 Which one of the following represents the
total number of trees in the graph given in Fig. 9.123
Fig. 9.122?
(a) 1, 2, 3 and 4 (b) 2, 3, 4 and 6
(c) 1, 4, 5 and 6 (d) 1, 2, 4 and 5

Answers to Objective-Type Questions

9.1. (c) 9.2. (c) 9.3. (c) 9.4. (b) 9.5. (b)
9.6. (d) 9.7. (a) 9.8. (d) 9.9. (d) 9.10. (d)