ELECTRIC CIRCUITS
1.3 TOPOLOGY
1.3.4 Concept of Tree and Co-Tree
A tree is a connected subgraph of a network, which consists of all the nodes of the original graph but no closed path.
1.3.4.1 Properties of a Tree
A tree consists of all the nodes of graph. For an n-node graph, the number of tree branches is n − 1.
Number of possible trees = det [[A
r][A
r]T] where A is the reduced incidence matrix and [A]T is transpose of reduced incidence matrix.
1. Twig: The branch of a tree is known as twig.
Number of twigs (n - 1) is known as tree value of the graph. It is also called the rank of the tree.
2. Co-tree: The part of a directed graph that is not covered by the tree is known as co-tree. A co-tree is represented by dotted line.
3. Link or chord: The branch of tree is known as link or chord. Suppose a graph has n nodes and b branches, then the number of branches in a tree is (n − 1) and number of branches in co-tree (links) is l = b − (n − 1).
2
1
a b
c
V d
3
4
2
4
1 3
2
4 1
a b
c d
3
1 a
b c
d 4 2
3
Figure 1.31
|
Network graphs.1.3.2 Directed and Undirected Graph
If every branch of a graph has a direction, then the graph is called as directed (oriented). If the branches of a graph have no direction then it is called an undirected graph as shown in Figs. 1.32(a) and (b).
2
4
1
a b
c d
3
2
4
1
a b
c d
3 Figure 1.32
|
(a) Directed graph and (b) undirectedgraph.
Node can be incident to one or more elements. The number of branches incident at a node of a graph indi- cates the degree of the node. A graph is connected if and only if there is a path between every pair or nodes.
1.3.3 Planar and Non-Planar Graph
A graph is said to be planar if it can be drawn on a plane surface such that no two branches cross each other [Fig. 1.33(a)]. A non-planar graph cannot be drawn on a plane surface without a crossing over of two or more branches. [Fig. 1.133 (b)].
(a) (b)
Figure 1.33
|
(a) Planar graph. (b) Non-planar graph.Problem 1.25: Identify which among the following graphs is non-planar.
b
d
b
d
c c
a ≅ a
1 2 3 4
5
4 3
2 1 2
3 4
5 1
2
4
3 5
1 (a)
(b)
(c)
(d)
1.3 TOpOLOGY 23
nodes and b branches, the complete incidence matrix A is a rectangular matrix of order n × b, whose branches repre- sent the column and nodes represents the rows.
Formation of incidence matrix:
1. Obtain directed graph for the given network.
2. Assign `+1’ in the matrix if the arrow of a branch is oriented away from the node.
3. Assign `−1’ in the matrix if the arrow of a branch is oriented towards a node.
4. Assign `0’ in the matrix if the branch is not con- nected to a node.
Problem 1.26: What is the number of equations required to analyse the circuit given below?
R
R R
C C
C
Solution: (n − 1) = 3 equations, as n = 4.
Ans. (4) Solution: Graphs given in (a), (b) and (c) are planar while graph in (d) is non-planar. In graph (d):
Number of branches, b =10 Nodes n = 5
Number of tree branches = n − 1 = 4 Links = 6
Problem 1.27: If a graph of network has 10 branches and 6 nodes, then what is the number of mesh equa- tions or KVL equation required to solve the network?
Solution:
Required KVL equations = Links = b − n + 1 = 10
− 6 + 1 = 5
Required KCL equations = Twigs = n − 1 = 6 − 1 = 5 Ans. (5, 5) Problem 1.28: How many nodes and branches does the following circuit have?
10 Ω 5 Ω 50 Ω 4 Ω
30V 20V
2A
+− +−
Solution: Five nodes and seven branches.
Ans. (5, 7) 1.3.5 Incidence Matrix
A matrix representing the relation between number of branches and number of nodes in a directed graph is known as incidence matrix. For a given directed graph with n
Problem 1.29: Draw the incidence matrix of the fol- lowing graph.
1 c
a b
e 4 2 f
3 d
Solution: In the incidence matrix column represents branches, while rows represent nodes. Therefore,
Nodes
Branches
a b c d e f
1 1 0 1 0 0 1
2 −1 −1 0 1 0 0
3 0 1 0 0 1 −1
4 0 0 −1 −1 −1 0
Note: In matrix A with n rows and b columns, an entry aij in the ith row and jth column has the following values:
aij = 1, if jth branch is incident to and oriented away from the ith node.
aij = −1, if jth branch is incident to and oriented towards ith node.
aij = 0, if jth branch is not incident to the ith node.
Problem 1.30: Draw the graph of the following network and write the incidence matrix.
V
b
d
a R1 c
R2 R3 R4
C1
+−
1.3.6 Link Currents: Tie Set Schedules
In any tree of a graph, addition of a link results in a closed path or loop and a circulatory curve or the link current. The current in any branch of the graph may then be obtained by noting the
1. various link currents flowing through the branch;
2. direction of flow of the link currents through branch (Fig. 1.34).
Solution: The graph for the given network is
1
a b c
4 3 2
d
5 6
The incidence matrix A is given as
Nodes
Branches
1 2 3 4 5 6
a −1 0 0 0 1 0
b 0 −1 0 0 −1 1
c 0 0 −1 1 0 −1
d 1 1 1 −1 0 0
1.3.5.1 Properties of Incidence Matrix
1. The sum of entries in any column is zero.
2. The determinant of the incidence matrix of a closed loop is zero.
3. The rank of the incidence matrix of a connected graph is (n − 1).
1.3.5.2 Reduced Incidence Matrix [A
r]
The reduced incidence matrix is obtained by deleting any row in the incidence matrix. The number of possible trees of a graph is = det [A
r] [A
r]T where A
r is reduced incidence matrix and AT
r is the transpose of reduced incidence matrix.
Problem 1.31: Find the reduced matrix of the matrix in Problem 1.30, and find the number of possible trees.
Solution: Reduced incidence matrix is [ ]
Ar =
−
− −
− −
1 0 0 0 1 0
0 1 0 0 1 1
0 0 1 1 0 1
The number of possible trees of a graph = det {[Ar] × [Ar]T}
= det
−
− −
− −
−
−
−
×
1 0 0 0 1 0
0 1 0 0 1 1
0 0 1 1 0 1
1 0 0 0 1 0 0 0 1 0 0 1
3 6 1
−−
− ×
1 0 0 1 1
6 3
= det
2 1 0 1 3 1 0 1 3
3 3
−
− −
− ×
= 2(9 - 1) + (−3) + 0 = 16 - 3 = 13
Problem 1.32: If the reduced incidence matrix net- work is given as
A=
−
− −
−
1 0 0 1 1 0
0 1 1 1 0 0
0 0 1 0 1 1
then what are possible number of trees?
Solution:
AT =
−
− −
−
−
−
−
−
1 0 0 1 1 0
0 1 1 1 0 0
0 0 1 0 1 1
1 0 0
0 1 0
0 1 0
1 1 0
1 0 1
0 0
1 1
=
3 1 1
1 3 1
1 1 3
− −
− −
− −
The number of trees is det [A
rA
r
T] = 3(9 − 1) + 1(−3 − 1) − 1 = 24 − 4 − 4 = 16
Ans. (16) Problem 1.33: For a given network, the reduced inci- dence matrix is given by
Nodes
Branches
1 2 3 4 5 6
1 1 0 0 1 −1 0
2 0 1 0 −1 1 −1
3 0 1 0 −1 1 −1
4 0 0 1 0 0 1
Given that i
2 = 2 A, i
4 = 4 A, i
5 = 2 A, determine i
6. Solution: The network can be drawn as
1 1
6
3 4
2 4 5
2
3
i6 + i
4- i
5- i
2 = 0 i6 = −i
4 + i
5 + i
2 = −4 + 2 + 2 = 0 A
Ans. (0)
1.3 TOpOLOGY 25
Problem 1.34: If tree is as shown below, what is the tie set for this tree?
j3
j4 j6
j2 i1
i3
i2 i4 j1 Solution: The tie set is
j3 = i
1 j
6 = i
3
j4 = i
2 j
8 = i
4
j1 = i
2 + i
4 j
5 = i
1− i
2− i
3− i
− − 4
Problem 1.35: For the graph shown below, what is the tie-matrix corresponding to tree formed branches 4, 5, 6?
4 1 5 2 2
3 6 4
1 3
5
6 4
0
Solution: The tie-set schedule or tie-matrix is Link
Number
Branch
1 2 3 4 5 6
1 1 0 0 1 −1 0
2 0 1 0 0 1 −1
3 0 0 1 −1 0 1
3
4 8 7 6 5 2
1
Graph Tree
j3
j4 j8 j7 i3
i4 i1 i2
j6
j5 j2
j1 3
4 8 7 6 5 2
1
Graph Tree
j3
j4 j8 j7 i3
i4 i1 i2
j6
j5 j2
j1
Figure 1.34
|
Direction of flow of the link currents.Here j
1, j
2, …, j
8 are branch currents and i
1, i
2, i
3, i
4 are link currents. The set of branches forming the closed loop on which the link circulates is called a tie set. The tie set for the tree shown in Fig. 1.34 is
j1 = i
1, j
2 = i
2, j
3 = i
3, j
4 = i
4
j5 = i
2− i
1, j6 = i
3− i
2, j7 = i
4− i
3, j8 = i
1− i
4
Tie-set schedule is as follows Link
number
Branch number
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
3 0 0 +1 0 0 +1 −1 0
4 0 0 0 +1 0 0 +1 −1
1.3.7 Cut Set and Tree Branch Voltages
A cut set is the minimum set of branches of a connected graph such that the removal of these branches causes the graph to be cut into exactly two parts.
1.3.7.1 Procedure
1. Select a tree and co-tree from the directed graph.
2. Determine the number of tree branches, that is, twigs.
3. The number of cut sets to be assigned equal to the number of twigs.
4. A cut set may contain any number of links, but should cover only one twig.
5. The cut set is oriented in the same direction as the twig it covers.
6. Assign +1 in matrix if the direction of cut sets and branches are the same.
7. Assign −1 in the matrix if the direction of cut sets and branches are different.
8. Assign `0’ in the matrix if cut sets do not cover the branch.
Problem 1.36: Formulate the cut-set matrix for the directed graph shown below.
a 1
4 b 5
3 d
2
c 6
Solution: The tree and cut set corresponding to the graph are shown as follows.
Branches Cut set
Links Twigs
1 2 3 4 5 6
1 −1 1 0 1 0 0
2 0 −1 −1 0 1 0
3 −1 0 −1 0 0 1
a
4 b 5
d Tree
c 6
a
b 1
4 5
3 d
2
c 6
Cut set
C1 C2
C3
Problem 1.37: Valid cut set of the following circuit is
1 2
4
5 3
6
(a) 1, 2, 3, 4 (b) 3, 4, 5, 2 (c) 1, 6, 5, 3 (d) 1, 6, 4, 3
Ans. (d)
Solution: The tree is as shown below:
8
5
6 7
3 2
1 4
The f-cut sets are: (1, 6, 8) (1, 2, 5) (1, 2, 3, 7, 8) (1, 2, 3, 4)
Fundamental Cut Set Matrix and KCL
We can obtain KCL equations at any node is terms of the Q-matrix.
QIb = 0
With reference to the figure given in Problem 1.36,
−
− −
− −
1 1 0 1 0 0 0 1 1 0 1 0 1 0 1 0 0 1
1 2 3 4 4 5
I I I I I I
= 0 0 0
Expanding the above equations, we get
−I
1 + I
2 + I
4 = 0
−I
2− I
3 + I
5 = 0
−I
1− I
3 + I
6 = 0
These equations are the KCL equation at nodes a, c and d. The branch voltages and fundamental cut set voltages are as follows. We can express the branch voltages in terms of tree branch voltage or twig voltages.
Vb = QTV
t
where V
b is the branch voltage and V
t is the twig voltage.
V V V V V V
1 2 3 4 5 6
1 0 1
1 1 0
0 1 1
1 0 0
0 1 0
0 0 1
=
− −
−
− −
V V V
1 2 3
These are the KVL equations of the cut set matrix.
1.3.7.2 Fundamental Cut Sets
A fundamental cut set is a cut set that cuts or contains one and only one tree branch (twig). Therefore, for a given tree the number of f-cut sets will be equal to the number of twigs (n − 1).
Problem 1.38: For the tree shown, what are the pos- sible f-cut sets?
8
5
6 7
3 2
1 4