Electrical Network Topology

PATHAKAMURI NARESH, Asst. Prof.,
EEE Department, RAGHU Engg. College (A)
Course
Electric Circuits
Module
Network Topology

UNIT-V
Network Topology
Definitions of Graph and Tree
Network Incidence matrices
Basic cut-set and tie-set matrices for planar networks
Loop and nodal methods of analysis of networks with dependent
and independent voltage and current sources.
Duality and Dual networks

Network topology is a graphical representation of electric circuits.
It is useful for analyzing complex electric circuits by converting them into
network graphs.
Network topology is also called as Graph theory.
Graph
Network graph is simply called as graph.
It consists of a set of nodes connected by branches.
In a graph, node is a common point of two or more branches. Sometimes,
only a single branch may connect to the node.
A branch is a line segment that connects two nodes.

Branch:
A branch is a line segment representing one network element or
combination of element connected between two points.
Node:
Node is an end point of a line segment and exists at the junction
between two branch.
Degree of Node:
It is number of branches incident to it.

No. of Nodes in the graph : 4  A, B, C and D
No. of Branches in the graph: 6  (1), (2), (3), (4), (5) and (6).
Degree of Node-A is 3 (there are 3 branches incident to Node-A)
Similarly here all the remaining nodes have degree as 3.

Any electric circuit or network can be converted into its equivalent graph by
replacing the passive elements( R, L and C) and voltage sources(V) with
short circuits and the current sources (I) with open circuits.
So the line segments in the graph represent the branches corresponding to
either passive elements or voltage sources of electric circuit.
Representation of network elements to draw the graph:
SI NO ELEMENT RERESESNTAION
1 Resistor, R short circuit (line segment)
2 Inductor, L short circuit (line segment)
3 Capacitor, C short circuit (line segment)
4 Voltage Source short circuit (line segment)
5 Current Source open circuit

Element Nodes Representation Graph

Draw the graph for the following network shown in figure
Fig: Network
Step-1: Represent nodes in the network
Step-2: Takeout nodes to draw graph Step-3: connect nodes as per network to
complete graph

(a) Electrical Network
(b) Graph of network
In the network: Nodes are 4 and Branches
are 7 ( voltage source, current source and 5
resistances)
In the graph : 4 nodes and 6 branches;
since current source is made open.

For the following electrical network shown in the figure (a) obtain
the Graph?
Voltage sources and passive elements are
represented by lines (short circuit)
Current sources open circuit.
The Graph for the given network is
shown in fig (b)
Conclusions:
1. The number of nodes present in a graph
will be equal to the number of principal
nodes present in an electric circuit.
2. The number of branches present in a
graph will be less than or equal to the
number of branches present in an electric
circuit.

Types of Graphs
Following are the types of graphs
1. Connected and Unconnected Graphs
2. Oriented and Un-Oriented Graphs
3. Planar and Non-planar Graphs
1. Connected Graph
If there exists at least one branch between any of the two nodes of a graph, then it is called as
a connected graph.
Each node in the connected graph will be having one or more branches that are connected to
it. So, no node will present as isolated or separated.
Fig: Connected Graph

2. Unconnected Graph
If there exists at least one node in the graph that remains unconnected by even single branch,
then it is called as an unconnected graph.
So, there will be one or more isolated nodes in an unconnected graph.
Consider the graph shown in the following figure.
In this graph, the nodes 2, 3, and 4 are connected by two branches each. But, not even a single
branch has been connected to the node 1.
So, the node 1 becomes an isolated node. Hence, the above graph is an unconnected graph.

3. Oriented Graph
If all the branches of a graph are represented with arrows, then that graph is called as a
oriented graph.
These arrows indicate the direction of current flow in each branch.
Consider the graph shown in the following figure.
This graph is oriented graph, the direction of current flow is represented with an arrow in
each branch.

(a) Network (b) Oriented Graph
Fig: Network and Oriented Graph

4. Un-oriented Graph
If the branches of a graph are not represented with arrows, then that graph is called as an un-
oriented graph.
The graph that was shown in the following example is an un-oriented graph, because there
are no arrows on the branches of that graph.

5. Planar Graph:
A Planar Graph is a graph drawn on a two-dimensional plane so that no two branches
intersect at a point which is not a node.
The graphs shown in the figure are planar graphs.
Fig: Planar Graphs

6. Non- Planar Graph:
It is a graph drawn on a two dimensional plane such that two or more branches intersect at
a point other than node on a graph.
The graphs shown in the figure are planar graphs.
Fig: Non-Planar Graphs

Subgraph
A subgraph is a subset of branches and nodes of a graph.
A subgraph can be of two types (1) proper subgraph and (2) improper subgraph.
proper subgraph:
If the subgraph contains branches and nodes less than those on the graph, then such a
subgraph is called proper subgraph.
The graph and its proper subgraphs are shown in figure.
(a) Graph (b) Subgraphs

Improper subgraph:
If a subgraph contains all the nodes and less branches of a graph then a subgraph is
called improper subgraph.
A graph and its improper subgraphs are as shown in figure.
Fig: Graph and improper graph
(a) Graph (b) Improper Graph

Tree
Tree is a connected subgraph of a given graph, which contains all the nodes of a graph. But,
there should not be any loop in that subgraph.
The branches of a tree are called as twigs and represented with solid lines.
If there ‘n’ number of nodes in the graph then the number of branches in a Tree equal to n – 1.
Fig (a) : Graph Fig (b): one of Tree of Graph
The connected subgraph in fig(b) contains all the four nodes of the given graph but there is
no loop. Hence, it is a Tree.

Draw the different possible trees for the following graph in the figure.
Fig: Graph
Fig: possible trees of given Graph

Draw the different possible trees for the following graph in the figure.
Fig: Graph
Fig: Possible trees of Graph

Co-Tree
Co-Tree is a subgraph, which is formed with the branches that are removed while forming a
Tree.
It is called as Complement of a Tree.
The co-tree branches are called as links or chords.
In general, the links are represented with dotted lines.
The Co-Tree corresponding to the above Tree is shown in the following figure.
Fig (a) : Graph Fig (b): one of Tree of Graph
Fig (c) : Co-tree
corresponding to tree in fig (b)

Mathematically, it can be written as
Where,
l number of links
b number of branches in given graph
n number of nodes in given graph
The Tree branches d, e & f are represented with solid
lines. The Co-Tree branches a, b & c are represented with
dashed lines.
This Co-Tree has only three nodes instead of four nodes
of the given graph, because Node 4 is isolated from the
above Co-Tree. Therefore, the Co-Tree need not be a
connected subgraph.
This Co-Tree has three branches and they form a loop.
The number of branches that are present in a co-tree will
be equal to the difference between the number of
branches of a given graph and the number of twigs.
Fig: Graph with a tree and
corresponding co-tree
If we combine a Tree and its corresponding Co-Tree, then we will get the original graph as
shown below.

Draw the possible tree and the corresponding co-tree for the given graph shown below
Fig: Graph
No. of Nodes , n = 4 ( 1, 2, 3 and 4)
No. of Branches, b = 6 (a, b, c, d, e and f)
No. of Twigs = n-1 = 3
Therefore NO. of links, l = b-(n+1) = 3
Fig: Co-Tree
Fig: Selected Tree
For the selected tree
The twigs are a, c
and e
and then the links for
the co-tree are b, d and
f.

Matrices Associated with Network Graphs
Following are the three matrices that are used in Graph theory.
1. Incidence Matrix
2. Fundamental Loop (Tie-set) Matrix
3. Fundamental Cut set Matrix
Incidence Matrix
A graph consists of a set of nodes and those are connected by some branches. So, the
connecting of branches to a node is called as incidence. An Incidence Matrix represents the
graph of a given electric circuit or network.
It is possible to draw the graph of that same electric circuit or network from the incidence
matrix.
Incidence matrix is represented with the letter A. It is also called as node to branch incidence
matrix or node incidence matrix.
If there are ‘n’ nodes and ‘b’ branches are present in a directed graph, then the incidence
matrix will have ‘n’ rows and ‘b’ columns.
Rows and columns are corresponding to the nodes and branches of a directed graph. Hence,
the order of incidence matrix will be n × b.

Procedure to find Incidence Matrix
Follow these steps in order to find the incidence matrix of directed graph
1. Select a node at a time of the given directed graph and fill the values of the
elements of incidence matrix corresponding to that node in a row.
2. The elements of incidence matrix will be having one of these three values,
+1, -1 and 0.
If the branch current is leaving from a selected node, then the value of the
element will be +1.
If the branch current is entering towards a selected node, then the value of the
element will be -1.
If the branch current neither enters at a selected node nor leaves from a
selected node, then the value of element will be 0.
Repeat the above step for all the nodes of the given directed graph.

In matrix A with n rows and b columns an entry aij in the ith row and jth
column has the following values.
Fig: Graph of a network Fig: Incidence matrix of graph

Consider the following directed graph. Obtain the incidence matrix?
From the graph: No. of nodes = 4 and
No. of branches = 6
The rows and columns of the above
matrix represents the nodes and branches
of given directed graph. Therefore the
order of incidence matrix is 4 × 6.
Fig: Graph
If the branch current is leaving from a selected node, then the value of the element will be +1.
If the branch current is entering towards a selected node, then the value of the element will be -1.
If the branch current neither enters at a selected node nor leaves from a selected node, then the
value of element will be 0.

By observing the above incidence matrix, we can conclude that the summation of column
elements of incidence matrix is equal to zero. That means, a branch current leaves from one
node and enters at another single node only.
Note: If the given graph is an un-directed type, then convert it into a directed graph by
representing the arrows on each branch of it.
We can consider the arbitrary direction of current flow in each branch.
The incidence matrix corresponding to the above directed graph will be

For the following graph find the incidence matrix?
From the graph: No. of nodes = 4 and
No. of branches = 6
The rows and columns of the above
matrix represents the nodes and branches
of given directed graph. Therefore the
order of incidence matrix is 4 × 6.
Fig: graph
The incidence matrix corresponding to the above directed graph will be

Draw the oriented graph from the given incidence matrix?
Nodes 4  L, M, N and O
Branches 6  a, b, c, d, e and f
Form Matrix, branch ‘a’ is connected between nodes L and M and orientation
of branch ‘a’ at node L is -1 so it is coming towards node L .
while the orientation at node M is +1 so it is going away from node.
Similarly remaining can be drawn for the graph.

So, the fundamental loop matrix will have ‘b-n+1’ rows and ‘b’ columns.
Here, rows and columns are corresponding to the links of co-tree and
branches of given graph.
Hence, the order of fundamental loop matrix will be (b - n + 1) × b.
The elements of fundamental loop matrix will be having one of these three
values, +1, -1 and 0.
The value of elements will be 0 for the remaining links and twigs, which are
not part of the selected f-loop.
If the direction of twig current of selected f-loop is same as that of f-loop
link current, then the value of element will be +1.
If the direction of twig current of selected f-loop is opposite to that of f-loop
link current, then the value of element will be -1.

Procedure to find Fundamental Loop Matrix
Follow these steps in order to find the fundamental loop matrix of given
directed graph.
1. Select a tree of given directed graph.
2. By including one link at a time, we will get one f-loop.
3. Fill the values of elements corresponding to this f-loop in a row of
fundamental loop matrix.
4. Repeat the above step for all links.

RELATION BETWEEN LOOP CURRENTS AND BRANCH CURRENTS
Consider the graph shown below

Therefore the tie-set matrix is given from
the above three tie-sets

It is possible to relate branch currents as a linear combination of link currents using matrix B.
Since all the branches are connected to all the nodes in the graph.
Let the branch currents ia, ib, ic, id, ie & if.
There are three links and let the link currents be Ia, Ib, Ic for links b, d and f respectively.

Obtain the incidence matrix for the directed graph and its selected
tree shown in figure.
(a) Directed Graph
(b) Selected Tree

The given Tree contains three branches d, e & f.
Hence, the branches a, b & c will be the links of the Co-Tree
corresponding to the given Tree.
By including one link at a time to the above Tree, we will get one f-
loop.
So, there will be three f-loops, since there are three links.
These three f-loops are shown in the following figure.
Branches, which are represented with dotted lines (links) will
forms the fundamental-loops.

We will get the row wise element values of Tie-set matrix from each f-loop.
So, the Tie-set matrix of the above considered Tree will be
The rows and columns of the above matrix represents the links and branches of
given directed graph.
The order of this incidence matrix is 3 × 6.
Note: The number of Fundamental loop matrices of a directed graph will be
equal to the number of Trees of that directed graph. Because, every Tree will be
having one Fundamental loop matrix.

Fundamental Cut-set Matrix
Fundamental cut set or f-cut set is the minimum number of branches that
are removed from a graph in such a way that the original graph will
become two isolated subgraphs.
The f-cut set contains only one twig and one or more links.
So, the number of f-cut sets will be equal to the number of twigs.
Fundamental cut set matrix is represented with letter C.
This matrix gives the relation between branch voltages and twig voltages.
If there are ‘n’ nodes and ‘b’ branches are present in a directed graph,
then the number of twigs present in a selected Tree of given graph will be
n-1.
So, the fundamental cut set matrix will have ‘n-1’ rows and ‘b’ columns.

Here, rows and columns are corresponding to the twigs of selected tree
and branches of given graph. Hence, the order of fundamental cut set
matrix will be (n-1) × b.
The elements of fundamental cut set matrix will be having one of
these three values, +1, -1 and 0.
The value of elements will be 0 for the remaining twigs and links,
which are not part of the selected f-cut-set.
If the direction of link current of selected f-cut set is same as that of f-
cut-set twig current, then the value of element will be +1.
If the direction of link current of selected f-cut set is opposite to that of
f-cut-set twig current, then the value of element will be -1.

Procedure to find Fundamental Cut-set Matrix
Follow these steps in order to find the fundamental cut set matrix of
given directed graph.
1. Select a Tree of given directed graph and represent the links with the
dotted lines.
2. By removing one twig and necessary links at a time, we will get one
f-cut set.
3. Fill the values of elements corresponding to this f-cut set in a row of
fundamental cut set matrix.
4. Repeat the above step for all twigs.

Selected the branches d, e & f of the directed graph as twigs. So, the
remaining branches a, b & c of this directed graph will be the links.
The twigs d, e & f are represented with solid lines and links a, b & c are
represented with dotted lines in the following figure.
Obtain the cut-set matrix for the following tree of a graph

By removing one twig and necessary links at a time, we will get one f-
cut set.
So, there will be three f-cut sets, since there are three twigs. These
three f-cut sets are shown in the following figure.

The rows and columns of the above matrix represents the twigs and branches
of given directed graph. The order of this fundamental cut set matrix is 3 × 6.
Note: The number of Fundamental cut set matrices of a directed graph
will be equal to the number of Trees of that directed graph. Because, every
Tree will be having one Fundamental cut set matrix.
We will be having three f-cut sets by removing a set of twig and links of
C1, C2 and C3. We will get the row wise element values of fundamental cut
set matrix from each f-cut set.
So, the fundamental cut set matrix of the above considered Tree will be

In the following network, the numerical values of resistances also indicate the
branch numbers. Write the oriented graph of the network. Select a tree with
brnches1, 2, 3 as the tree branches, write tie-set and cut-set matrices.
Fig: Given Network Fig: Oriented graph

Dual networks:
Two networks are said to be dual networks of each other if the mesh equations of given
network are the node equations of other network.
Dual networks are based on Kirchhoff’s voltage and current laws.
Consider simple series RLC circuit excited by AC voltage as shown in figure.
Applying KVL equation,
……..(1)

Consider parallel RLC circuit
Thus equations (1) and (2) are similar.
The voltage source V of the series RLC circuit corresponds to current source I in the
parallel RLC circuit.
Also the current in the first circuit corresponds to voltage in second network.
Applying KCL to the circuit
……..(2)

Sr. No. Element Dual Element
1 Resistance Conductance
2 Capacitance Inductance
3 Inductance Capacitance
4 Series Branch Parallel Branch
5 Mesh Node
6 Closed Switch Open Switch
List of element and its dual

Rules to Construct Dual Circuits
1. Place a node at the center of each mesh of the circuit.
2. Place a reference node (ground) outside of the circuit.
3. Draw lines between nodes such that each line crosses an
element.
4. Replace the element by its dual pair.
5. Determine the polarity of the voltage source and direction of
the current source.

Draw the dual of the following network shown in the fig.
Step-1: Place a node at the center of each mesh of the circuit.

Step 2: Place a reference node (ground) outside of the circuit.
Step 3: Draw lines between nodes such that each line crosses an element.

Step 4: Replace the element by its dual pair.
Step 5: Determine the polarity of the voltage source and direction of the current source.
Fig: Draw Dual circuit

Draw the dual network for the given network shown in figure.
Fig: Network

Place Dots in each mesh and also place a reference (datum) node
outside of the network

Draw the dual network by replacing the elements of the network with
their dual elements

Draw the dual network for the given network shown in figure.

Electrical Network Topology

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