UNIT III

Network Topology
3.1 Basic definitions
Network Topology
 Is another method of solving electric circuits
 Is generalized approach
Network
A combination of two or more network elements is called a network.
Topology
Topology is a branch of geometry which is concerned with the properties of
a geometrical figure, which are not changed when the figure is physically distorted,
provided that, no parts of the figure are cut open or joined together.

The geometrical properties of a network are independent of the types of

elements and their values.

Every element of the network is represented by a line segment with dots at

the ends irrespective of its nature and value.
Circuit
If the network has at least one closed path it is a circuit.

Note that every circuit is a network but every network is not a circuit.
Branch
Representation of each element (component) of a electric network by a line
segment is a branch.
Node
A point at which two or more elements are joined is a node. End points of
the branches are called nodes.
Graph
It is collection of branches and nodes in which each branch connects
two nodes.
Graph of a Network
The diagram that gives network geometry and uses lines with dots at
the ends to represent network element is usually called a graph of a given
network. For example,

Fig.3.1 Network

Fig.3.2 Graph

Sub-graph
A sub-graph is a subset of branches and nodes of a graph for example
branches 1, 2, 3 & 4 forms a sub-graph. The sub-graph may be connected or
unconnected. The sub- graph of graph shown in figure 2 is shown in figure
3.3.

Fig.3.3 Sub-graph
Connected Graph
If there exists at least one path from each node to every other node,
then graph is said to be connected. Example ,

Fig.3.4 Connected Graph

Un-connected Graph
If there exists no path from each node to every other node, the graph is
said to be un-connected graph. For example, the network containing a
transformer (inductively coupled parts) its graph could be un-connected.

Fig.3.5 Network

Fig.3.6 Un-connected
Graph

Path (Walk)
A sequence of branches going from one node to other is called path.
The node once considered should not be again considered the same node .

Loop (Closed Path)

Loop may be defined as a connected sub-graph of a graph, which has
exactly two branches of the sub-graph connected to each of its node.

For example, the branches1, 2 & 3 in figure 3.7 constitute a loop.

 Fig.3.7 Loop

Planar and Non-planar Graphs

A planar graph is one where the branches do not cross each other
while drawn on a plain sheet of paper. If they cross, they are non-planar .

Fig.3.8 Planar Graph Fig.3.9 Non-planar graph

Oriented Graph
The graph whose branches carry an orientation is called an oriented
graph

Fig.3.10 Oriented Graph

The current and voltage references for a given branches are selected
with a +ve sign at tail side and –ve sign at head
+
-
Tree
Tree of a connected graph is defined as any set of branches, which
together connect all the nodes of the graph without forming any loops. The
branches of a tree are called Twigs.
Co-tree
Remaining branches of a graph, which are not in the tree form a co-tree. The
branches of a co-tree are called links or chords.

The tree and co-tree for a given oriented graph shown in figure3.11 is shown
in figure 3.12 and figure 3.13.

Fig.3. 11 Oriented Graph

Fig.12 Trees
 Fig.3.13 Co-trees

Tree Twigs Links (Chords)

1 2, 4 & 5 1, 3 & 6
2 3, 4 & 5 1, 2 & 6
3 2, 5 & 6 1, 3 & 4
Properties of Tree
 It contains all the nodes of the graph.

 It contains (nt-1) branches. Where ‘nt’ is total number of nodes in the

given graph.

 There are no closed paths.

 Total number of tree branches, n = (nt-1)

 Where nt = Total number of nodes

 Total number of links, l = (b-n)

 Where b = Total number of branches in the graph.

Degree of Node
The number of branches attached to the node is degree of node.

3.2. Complete Incidence Matrix (Aa)

Incidence matrix gives us the information about the branches, which are
joined to the nodes and the orientation of the branch, which may be towards a node
or away from it.
Nodes of the graph form the rows and branches form the columns. If the
branch is not connected to node, corresponding element in the matrix is given the
value ‘0’. If a branch is joined, it has two possible orientations. If the orientation is
away from the node, the corresponding matrix element is written as ‘+1’. If it is
towards the node, the corresponding matrix element is written as ‘-1’.

Example: 1 Obtain complete incidence matrix for the graph shown

Solution:

Branches
Nodes 1 2 3 4
Aa = 1 1 0 1 -1
2 0 1 -1 1
3 -1 -1 0 0
 1 0 1 -1
Aa = 0 1 -1 1
-1 -1 0 0

3.2.1Properties of Incidence Matrix

i) Each column has only two non-zero elements and all other elements are
zero.
ii) If all the rows of ‘Aa’ are added, the sum will be a row whose elements
equal zero.
If the graph has ‘b’ branches and ‘nt’ nodes, the complete incidence
matrix is of the order (nt x b).

3.3. Reduced Incidence Matrix (A)

x When one row is eliminated from the complete incidence matrix, the
remaining matrix is called reduced incidence matrix
If the graph has ‘b’ branches and ‘nt’ nodes, the reduced incidence
matrix is of the order (nt-1) b.
Example: 2 Write the complete and reduced incidence matrix for the
given graph shown

Solution:

Nodes Branches
1 2 3 4 5 6
1 1 1 0 0 0 0
Aa = 2 0 -1 1 1 0 0
3 0 0 0 -1 0 1
4 -1 0 -1 0 1 0
5 0 0 0 0 -1 -1
 1 1 0 0 0 0
0 -1 1 1 0 0
Complete Incidence Matrix, Aa = 0 0 0 -1 0 1
-1 0 -1 0 1 0
0 0 0 0 -1- 1

1 1 0 00 0
0 -1 1 1 0 0
Reduced Incidence Matrix, A= 0 0 0 -1 0 1
-1 0 -1 0 1 0

Example: 3 Draw the oriented graph of incidence matrix shown below

1 1 0 0 0 0 0
0 -1 1 1 0 0 0
A= 0 0 0 -1 1 1 0
0 0 0 0 0 -1 1

Solution: The given matrix is a reduced incidence matrix. Obtain the complete
incidence matrix in order to draw the oriented graph.

1 1 0 0 0 0 0
0 -1 1 1 0 0 0
Aa = 0 0 0 -1 1 1 0
0 0 0 0 0 –1 1
-1 0 -1 0 -1 0 -1

Total number of nodes = nt = 5

Total number of branches = b = 7

Oriented Graph
Example: 4 Draw the oriented graph of incidence matrix shown below

1 0 0 0 1 -1
-1 1 1 0 00
Aa = 0 -1 0 -1 0 1
Solution: 0 0 -1 1 -1 0

Total number of nodes = nt = 4

Total number of branches = b = 6

Oriented Graph

Example: 5) Draw the oriented graph of incidence matrix shown below

-1 -1 0 0 1 0 0
A= 0 1 1 0 0 1 0
0 0 -1 1 0 0 1

Solution: The given matrix is a reduced incidence matrix. Obtain the complete
incidence matrix in order to draw the oriented graph.
-1 -1 0 0 1 0 0
Aa = 0 1 1 0 0 1 0
0 0 -1 1 0 0 1
1 0 0 -1 -1 –1 -1
Total number of nodes = nt = 4
Total number of branches = b = 7

Oriented Graph
Example: 6 Show that determinant of the incidence matrix of a
closed loop is zero.

Proof: let us consider a closed path B

ABC 3

Total number of nodes = nt = 3 2

Total number of branches = b = 3
A 1 C

The complete incidence matrix is -1 0 -1

Aa = 0 1 1
1 -1 0

The determinant of complete incidence matrix of the closed loop is

-1 0 -1 = -1 (0 + 1) -0 –1 (0 –1) = -1+1 =
0 1 1 0
1 -1 0

3.4. Number of Possible Trees of a Graph

For the given network graph, it is possible to write several trees. The
t
number of possible trees is equal to determinant of [A] [A] . Where [A] is the
reduced incidence matrix obtained by removing any one row from complete
t
incidence matrix and [A] is the transpose of [A].

C Node Branches
Aa = No. 1 2 3 4
2 A 1 0 0 1
4
B -1 1 1 0
3 c 0 -1 -1 -1
2

A 1 B
 1 0 0 1
[Aa] = -1 1 1 0
0 -1 - 1 - 1

1 0 0 1
[A] = -1 1 1 0

t
[A] [A] = 1 0 0 1 1 -1 1+1 -1 2 -1
-1 1 1 0 0 1 -1 1 +1+1 = -1 3
0 1
1 0

[A] [A]t = 2 -1 = (6-1) = 5

-1 3
Node Pair Voltages
The voltage between any two nodes of a network is known as the
node-pair voltages. In general all branch voltages are node-pair voltages.
Network Variables
In Loop analysis, the loop currents are unknown parameters. Once,
they are evaluated, all branch currents can be determined in terms of these
loop currents.
Similarly, in nodal analysis, the node-pair voltages are the unknown
parameters. Once, they are evaluated, the voltages across any two nodes of
the network can be found. Hence, the node-pair voltages and loop currents
are called network variables.
The network variables are independent variables and all other
quantities depend on these values.

3.5 Tie-set
A tie-set is a set of branches contained in a loop such that each loop
contains one link or chord and remainder are tree branches.
Or
The set of branches forming the closed loop in which link or loop
current circulate is called a Tie-set.

Oriented Graph Tree

Let the branch currents in the network graph denoted by the symbol ‘j’ and
various loop currents by symbol ‘i’.The orientation of a closed loop will be
chosen to be the same as that of its connecting link.

For the given network graph,

Number of branches, b = 8
Number of nodes, nt= 5
Number of closed loops = [b–(nt - 1)]
Where (nt-1) = Number of tree branches.

3.5.1Tie-set Schedule
For a given network tree, a systematic way of indicating the links is
through use of a schedule called Tie-set Schedule
The tie-set schedule for the given network oriented graph is shown below

Link Branches
Curre nt
or 1 2 3 4 5 6 7 8
Number
1 1 0 0 0 -1 0 0 1
2 0 1 0 0 1 -1 0 0
3 0 0 1 0 0 1 -1 0
4 0 0 0 1 0 0 1 -1

The tie-set schedule can be written in matrix form is known as Tie-set matrix
(B).
 1 0 0 0 -1 0 0 1
B= 0 1 0 0 1 -1 0 0
0 0 1 0 0 1 -1 0
0 0 0 1 0 0 1 -1
Example: 7
For the given resistive network, write a tie-set schedule and equilibrium
equations on the current basis. Obtain values of branch current and branch
voltages. Given that R1=5 Ω; R2=5 Ω; R3=R4=R6= 10 Ω and R5= 2 Ω.

Solution:
 Total number of branches, b = 6;
Total number of nodes nt = 4;
Total number of tree branches, n = (nt-1) = (4-1) = 3;
Total number of links, l = (b– n) = (6 – 3) = 3

Tie-set Schedule:
Link Branches
Curre nts 1 2 3 4 5 6
i1 1 0 1 0 0 -1
i2 0 1 0 1 0 1
i3 0 0 -1 -1 1 0

Example: 8
(a) For the given network shown. Draw the graph, select a tree with branches 9,
4, 7, 5, & 8 and write the tie-set matrix. The number inside the brackets
indicates branch numbers.
Solution: Total number of branches, b = 9;
Total number of nodes nt =6;
Total number of tree branches, n = (nt-1) = (6-1) = 5;
Total number of links, l = (b– n) = (9 – 5) = 4

Oriented Graph

i1, i2, i3 and i4 are the loop currents

1 0 0 -1 0 0 -1 0 1
B= 0 1 0 0 -1 0 1 -1 0
0 0 1 1 1 0 0 1 -1
0 0 0 1 1 1 0 0 0
3.6 Cut-set
Tree branches connect all the nodes in the network graph. Hence, it is
possible to trace the path from one node to any other node by traveling along the
tree branch only. Therefore, potential difference between any two nodes called
node-pair voltage can be expressed in terms of tree branch voltages.
The cut set is a minimal set of branches of the graph, removal of which
cuts the graph into two parts. It separates the nodes of the graph into two groups.
The cut-set consists of only one tree branch and remainders are links. Each
branch of the cut-set has one of its terminal incident at a node in one group and
its other end at a node in the other group and its other end at a node in the other
group. The orientation of the cut-set is same as orientation of tree branch.
The number of cut-sets is equal to number of tree branches [i.e. (nt-1) = n
where nt is total number of nodes in the network graph.

Oriented Graph

Tree and Cut-sets

3.6.1Cut-set schedule

For a given network tree, a systematic way of indicating the tree

branch voltage through use of a schedule called cut-set schedule
To write the cut-set schedule for network graph,
(i) Consider an oriented network graph
(ii) Write any one possible tree of the network graph
(iii)Assume tree branch voltages as (e1, e2…en) independent variables.
(iv)Assume the independent voltage variable is same direction as that of a tree
branch voltage
(v) Mark the cut-sets (recognize) in the network graph.

Cut-set schedule
Tre e Branches
Bra n c h 1 2 3 4 5 6 7 8
Voltag e s
e1 1 0 0 0 1 -1 0 0
e2 0 1 0 0 0 1 -1 0
e3 0 0 1 0 0 0 1 -1
e4 0 0 0 1 -1 0 0 1

The tree branch voltages e 1, e2, e3, & e4 entered in the first column of the
schedule correspond to 4 branches 1, 2, 3 & 4. In order to fill the first row
corresponding to the tree branch voltages e1, by looking into the direction of
currents in the branches connected to the cut-set under consideration. If the
direction of current in the cut-set branch is towards the cut-set node, write ‘+1’
in the branch column of concerned cut-set branch. If the direction of current in
the cut-set branch is away from the cut-set node, write ‘-1’ in that particular
cut-set branch column. Write ‘0’ in the branch columns, which are not in that
particular, cut-set.

The cut-set schedule can be written in matrix form is known as cut-set matrix.
This matrix can be represented by Q.

1 0 0 0 1 -1 0 0
0 1 0 0 0 1 -1 0
Q= 0 0 1 0 0 0 1 -1
0 0 0 1 -1 0 0 1
Example: 9
For the given resistive network, write a cut-set schedule and obtain
equilibrium equations on the voltage basis. Solve these equations and
hence calculate values of branch voltages and branch currents.

Solution:

Oriented Graph
Tree and Cut-sets

Cut-set Schedule:
C u t - set s or Branches
Tre e bra n c h
1 2 3 4 5 6
Volta g e s
1 or (e 1 ) -1 0 1 1 0 0
2 or (e 2 ) 1 -1 0 0 1 0
3 or (e 3 ) 0 1 -1 0 0 1
Example: 10
Draw the graph for the network shown in figure below. Write the cut-set
schedule & obtain equilibrium equations and hence calculate values of
branch voltages and branch currents

Solution:

Oriented Graph
Tree and Cut-sets
Cut-set Schedule:
C u t - sets or Branches
Tr e e br a n c h 1 2 3 4 5 6
Voltages
1 or (e1) 1 -1 0 1 -1 0
2 or (e2) 0 -1 1 0 -1 1

Columns of the above schedule give branch voltages in terms of tree branch
voltages

v1 = e1
v2 = -e1 -e2
v3 = e2
v4= e1 (1)
v5= -e1- e2
 v6= e2

Rows of the above schedule give KCL equations

i1- i2 + i4-i5 = 0
-i2 + i3 - i5 + i6= 0 (2)

We have from given network, (vk  vg) = jk x rk

Branch currents ik = (vk  vg) /rk. Hence,
i1= (v1 + 10)/1 = (e1+10)/1 =e1+ 10
i2= v2/1 = (-e1 -e2)/1 = -e1 – e2
i3= v3/1 = e2/1 = e2 (3)
i4= v4/2 = e1/2 = 0.5e1
i5= (v5 + 40) /2 = [(-e1 –e2) + 40]/2 = -0.5e1-0.5e2 +20
i6= v6/2 = e2/2 = 0.5e2
Substituting set of equations (3) in (2), we get
3e1 + 1.5e2 = 10 (4)
1.5e1 + 3e2 = 20
The set of equations (4) are called Equilibrium Equations

3 1.5 e1 10
1.5 3 e2 = 20
3 1. 5 10 1. 5
e 1 = ∆1 / ∆ where ∆= 1.5 3 = 6.75 and∆1 = 20 3 =0
e1 = 0/6.75 = 0 V.

3 1.5 10 1.5
e2 = ∆2 / ∆ where ∆ = 1.5 3 = 6.75 and∆2 = 20 3 = 45
e2 = 45/6.75 = 6.667 V.

The branch voltages are The branch currents are

v1 = e1 = 0 V i1= (v1 + 10)/1 = 10 A
v2 = -e1 -e2 = 0 - 6.667 = -6.667 V i2= v2/1 = 6.667 A
v3 = e2 = 6.667 V i3= v3/1 = 6.667 A
v4= e1= 0 V i4= v4/2 = 0 A
v5= -e1- e2 = -0 – 6.667 = -6.667Vj5= (v5 + 40) /2 = (-6.667 + 40)/2 = 16.66 A
v6= e2 = 6.667 V i6= v6/2 = 3.33 A
Example: 11
For the given network, draw the oriented graph and a tree. Select suitable tree
branch voltages and write the cut-set schedule and also write the equations for
the branch voltages in terms of the tree branch voltages.

Solution:
Oriented Graph

For the given network graph,

Number of branches, b = 12
Number of nodes, nt= 6
Number of Cut-sets = [nt - 1]
= [6-1] =5

Tree and Cut-sets

 Cut-set Schedule

Tree Branch Branches

Voltages or
1 2 3 4 5 6 7 8 9 10 11 12
Cut-set
e1 1 0 -1 1 0 0 0 0 -1 0 0 0
e2 1 0 -1 0 1 0 0 0 -1 1 -1 0
e3 0 1 -1 0 0 1 0 0 -1 1 -1 0
e4 0 1 -1 0 0 0 1 -1 0 0 0 0
e5 0 0 0 0 0 0 0 -1 1 -1 0 1

3.7 DUALITY AND DUAL NETWORK

The network is said to be dual network of each other if the mesh equations of
given network are the node equations of other network. The property of
duality is a mutual property. If network A is dual network B, then the network
B is also dual of network A.

Some of the dual pairs are given in the following table:

3.7.1 Methods of drawing the dual of a network

The following steps are followed to draw the dual of given electrical network:

1. A dot is placed in each independent loop of the original network.

These dots placed inside the loops correspond to the independent
nodes in dual network.

2. A dot is placed outside the given network. This corresponds to the

reference node of the dual network.

3. All the dots are connected by dotted lines crossing all the branches.
The dotted lines should cross only one branch at a time. The dual
elements will form the branches connecting the corresponding nodes
in the dual network.

Note A: The voltage rise in the clockwise direction corresponds to a

current flowing towards the independent network.

Note B: A clockwise current in a loop corresponds to positive polarity for

the at the dual independent node.

Example 12: Draw a dual network for the given network below.

The procedure for drawing the dual network is given below:

The dual network is given below:

Example 13: Obtain the dual network for network shown below.

Solution:

Putting nodes inside identified independent loops and considering one datum
node outside the network , the elements are traced as follows:

The dual network of the above figure is as follows,

Questions

Part A

1 What is a graph of a network?

2 What is tree of a network?

3 Properties of a tree in a graph

4. What is a Dual Network?

5. Steps to draw a Dual Network.

6. List out 4 pairs of dual quantities.

7. State the dual elements for inductance and mesh current.

8. State the dual elements for resistance and capacitance.

9. Explain the terms oriented graph and basic loop.

10. What are tie set and cut set?

11. How will you determine the number of nodes and loop equations of a
network from the graph?

12. Define Link.

13. Methods of drawing the dual of a network.

14. Define Network.

15. Define connected network.

16. Define Node.

17. What are the advantages of tree concept?

PART-B

1. Obtain the Tie set matrices for the following networks & write KVL
equations.
 10

5 15
10 20 5

100V 50V

2. Obtain the Cut set matrices for the following networks & write KCL

equations.

50

10 30V
15 20 15

10V 20V

3. Obtain dual for the network shown.

10mho

50 ohm 10H

50micro F

10V

4. Obtain dual for the network shown.

 R1

C1 C2
R2
R1
V 1
V1 R3
1
L1
C2
L R2 2

5 .Draw the dual network for the given networks:

(a) (b)
5F

1 2 1 4H 2

10 H
10 A 5F
3Ω 6Ω 6 MHO

c)

20 Ω 10 Ω

20 V
10 A 30 Ω 15 A

6. Find the Cut-set and Tie-set matrix of the given network.

C4
L1 R3
A 1 2
C
B

R1
R2 R4
E

7. For the given network find the tie set matrix.

 A 1Ω B 1Ω C

1Ω 5 V2 Ω 2Ω

8. Find the cut set matrix of the network shown:

1Ω

1Ω
2Ω 2Ω
30 V

9. Draw the dual network for the circuit shown

30 F

20 Vc

2 cos 6t
10 H 5

10. For the network given, draw the graph and a tree. Show the link Currents. Write
the tie-set schedule for the tree, the equation for branch currents in terms of link
currents. Also write independent equations.

5 6

3 2
4