1. GRAPH THEORY 1.1 INTRODUCTION INTRODUCTION
An electrical circuit or network is an interconnection of electrical elements such as resistors, inductors, capacitors, transmission lines, voltage sources, current sources, and switches. The analysis of electrical circuit should results in knowing the voltage across and currents through all the circuit elements. In circuit analysis, all the elements in a network must satisfy Kirchhoff’s laws, besides their own characteristics. Network topology is a generic name that refers to all properties arising from the structure or geometry of a network. If each element or a branch of a network is represented on a diagram by a line irrespective of the characteristics of the elements, we get a graph. Hence network topology is network geometry. Thus topology deals with with the t he way in which the various elements are interconnected at their terminals without considering the properties and type of the elements connected. This method is considered to be a more systematic approach to the analysis of large electrical networks. For small circuit analysis based on nodal and mesh equation methods by using Kirchoff’s law and Ohm’s law are sufficient. But for complex networks these methods are difficult and take more time for solving the equations. The high speed digital computers has made it possible to use graph theory advantageously for larger network analysis. In order to describe the geometrical structure of the network, it is sufficient to replace the different power system co mponents mponents such as generators, transformers and transmission lines etc. by a single line element irrespective of the characteristics of the power system components. These line segments are called elements and their terminals are called nodes. 1.2 BASIC DEFINITIONS Node: A node is is a junction point or inter-connection point of two or more more network elements. Element or Branch or Edge:
An element is a line segment representing one network
element or a combination of network elements connected between two nodes. Degree of a node: It is the number of branches incident on a node. Graph: In a network, if the branches are represented by straight line segments and nodes by
dots, then the resultant diagrammatic representat ion is called as a graph. Oriented Graph: If each line segment of a graph is assigned with a direction, it is called as a
oriented graph. Connected Graph: When there exits at least one path between every pair of nodes, then the
graph is called as a connected co nnected graph. 1
Tree: The tree of a network connects all the nodes of the network but contains no closed
path. Twig: Each branch Each branch of a tree is called as a twig. Co-tree: All branches that are present in the network, but not in the tree of the network,
together constitute a Co-tree. Link: Each branch of a co tree is called as a link or chord. Path: A path is a traversal from one node to another node of a graph along the branches,
such that no node is encountered twice. Planar graph: If a graph contains no cross-overs i.e. if it can be represented on a single
plane, then it is called as a planar graph. Sub-graph: A graph Gs is said to be the sub graph of a graph G, if every node of G s is a
node of G and every branch of G s is also a branch of G. Rank: The rank of a connected connected graph is defined as (n-1)
Where n = no. of nodes in the graph Tie-set: Tie-set is a set of branches which forms a closed path or loop.
A basic tie-set or a fundamental tie-set is a tie-set having one and only one link branch, the other elements being tree branches Cut-set: A fundamental fundamental cut-set of a graph w.r.t a tree is a cut-set formed by one and only
one twig and and a set of links, which must must be cut to divide the network graph into into two parts. The conventional direction for the fundamental cut-set is taken to be the same as the direction of the tree branch defines the particular cut-set. Example: The following fig 1.1 shows all the above things clearly
Fig.(1.1) 2
In the above graph (i) Degree of node 1 is 2 (ii) The rank of the graph is 3 1.3 TREE OF A GRAPH
A tree is a sub-graph of a network which consists of all the nodes as in the graph, but has no closed path. Every graph has at least one tree. The branches of tree are called as twigs and the branches removed from the graph in forming the tree are called as links or chords. The number of twigs = the rank of the tree = (n-1) where n = number of nodes. If the connected graph has ‘b’ branches and ‘n’ nodes, then the number o f links corresponds to the graph, l = b-n +1 The set of all links corresponding to a given tree is called a co-tree. Example:
Fig.(1.2) Note: Since energy sources are replaced by their internal impedance, voltage and
current sources are replaced by short circuit and open circuit respectively. 1.3.1 Properties of a Tree
The tree of a graph has the following properties: i) It is a connected sub-graph. ii) It contains all the nodes of the original graph. iii) A graph having ‘n’ nodes will have (n-1) branches in its tree. iv) Tree does not contain any closed path. v) There may be many trees for a given graph.
3
1.4 INCIDENCE MATRICES OF A NETWORK
Matrix representation of a graph is very important in the development of the network equations. There are several incidence matrices that are important in developing the various network matrices such as bus impedance matrix and branch admittance matrix etc., using singular or non-singular transformation. We shall discuss the following matrix representation of the graph 1.4.1 Element node (branch node) Incidence matrix (
):
This matrix shows which branch is incident to which node. Each column of the matrix represents the corresponding node of the graph; each row represents the corresponding element. If there are ‘n’ nodes and ‘e’ elements in a graph, then the order of the element node incidence matrix is n e 1
The elements aij of the complete incidence matrix [A ] are found as follows aij = 1, if i th element is incident and oriented away from the jth node. aij = -1, if ith element is incident and oriented towards the jth node. aij = 0, if i
th
th
element is not incident to the j node.
Consider the graph as shown in fig.(1.3). The element node incidence matr ix is given by
1.4.2 Bus incidence matrix (A)
Any node of a connected graph can be selected as the reference node. Then, the variables of the other nodes, referred to as buses, can be measured with respect to the assigned reference. The matrix obtained from
deleting the column corresponding to the
reference node is the bus incidence matrix A. The order of the matrix is ex (n-1) and the rank is (n-1).
4
The matrix A can be partitioned into two sub-matrices (i) A t of dimension (n-1)x(n-1) corresponding to the branches which are twigs and (ii) A l of dimension lx(n-1) corresponding to links. The partitioned matrices are shown below
Since A gives the incidence of various branches on the node with their direction of incidence, the KCL equation for the nodes can be written as AT i = 0
--- (1.1)
T
Where A is the transpose of matrix A i is the vector of branch currents 1.4.3 Branch path incidence matrix (K)
The branch path incidence matrix shows the incidence of branches to paths in a tree, where a path is oriented from a node to the reference node. The elements k ij of the branch path incidence matrix [K] are found as follows k ij = 1, if the ith branch is in the path from the jth bus to reference and is oriented in the same direction. k ij = -1, if the ith branch is in the path from the jth bus to reference and is oriented in the opposite direction. k ij = 0, if the ith branch is not in the path from the jth bus to reference bus. The dimension of the matrix is (n-1)x(n-1). The matrix K associated with a tree of fig .(1.4) is
The branch path incidence matrix and the sub matrix A t relate the branches to paths and branches to buses respectively. So there is one to one correspondence between paths and buses.
5
At K T=U K T= At-1
--- (1.2)
1.4.4 Basic Cut-set incidence matrix (B)
A fundamental cut-set of a graph w.r.t a tree is a cut-set formed by one and only one twig and a set of links, which must be cut to divide the network graph into two parts. The conventional direction for the fundamental cut-set is taken to be the same as the direction of the tree branch defines the particular cut-set. The elements bij of the basis cut set matrix [B] are found as follows bij = 1, if the ith element is incident to and oriented in the same direction as the jth basic cut set. bij = -1, if the i
th
element is incident to and oriented in the opposite
direction as the jth basic cut set. th
bij =0, if the i
th
element is not incident to the j
basic cut set.
The basic cut set incidence matr ix for the example of fig. (1.5) is
The sub matrix B l can be obtained from the bus impedance matrix A. The incidence of links to buses is given by the sub matrix A l and the incidence of elements to buses is given by the sub matrix A b. Since there is a one to one correspondence of the elements and basic cut-sets, Bl A b gives the incidence of links to buses. i.e
Bl At= Al Bl = Al At-1 T
= Al K
-1
T
( Since At = K )
--- (1.3)
1
1.4.5 Augmented cut-set matrix (B )
Fictitious cutsets, called tie-cut sets can be introduced to get a square cutest matrix. Each tie cutset contains only one link of the connected graph and is oriented in the same direction as the link (Note that it is not a cutset in the actual sense, since removal of the cutset elements 6
does not divide the graph into two separate parts). We get a non-singular square matrix of dimension e×e. The augmented cutset matrix for the example o f fig.(1.5) is
1.4.6 Basic loop incidence matrix ( C )
The incidence of elements to basic loops of a connected graph is shown by the basic loop incidence matrix. Choose the loop direction always along the link direction. The dimension of the basic loop incidence matrix is e×l The elements Cij of the basic loop incidence matrix [C] are found as follows Cij = 1, if the i
th
element is incident to and oriented in the same direction as
the jth basic loop. Cij = -1, if the i
th
element is incident to and oriented in the opposite
direction as the jth basic loop Cij = 0, if the ith element is not incident in the jth basic loop. The basic loop incidence matrix for the example of fig. (1.6) is given by
Application to KVL to each fundamental loop, constitutes a set of l linearly independent equations, written as T
C V=0
--- (1.4)
1
1.4.7 Augmented loop incidence matrix (C )
The basic loop incidence matrix can be augmented by adding open-loops corresponding to the tree branches, to obtain a non-singular square matrix of dimensions e×e. An open loop is simply a path between adjacent nodes connected by a twig. 7
The augmented loop incidence matrix for fig.(1.6) is given by
Example -1
For the oriented graph as shown in figure, obtain the bus incidence matrix A, branch path incidence matrix K and basic cut set matrix B. [ JNTU, supplementary , Nov- 05] Solution:
Number of nodes, n=5
1
Number of elements, e=7 Number of tree branches or twigs, b (or t) = n-1 = 5-1 =4 Number of links, l= e-n+1 = 7-5+1 = 3
Bus incidence matrix A
Bus incidence matrix A is taken between number of elements e and (n-1) nodes. The tree of the given graph is as shown in the following figure. Here the reference node is not taken into consideration. Taking ‘0’ as reference node.
Branch path incidence matrix K
Branch path incidence matrix K
relates the no. of tree branches ‘b’ to paths in a tr ee. It
shows the incidence of branches to paths.
8
Basic cut set matrix B
Basic cut set matrix B is taken between number of elements ‘e’ and number of basic cut sets consists of only one branch, so number of basic cut sets are equal to number of branches.
Example -2 1
1
Determine the incidences matrices A, K, B, B , C, C for the figure as shown below and verify the following i) C b = -BTl
ii) C1 (B1) T = U
iii) At K T = U
Solution:
The oriented graph for the given network is
9
iv) Bl = Al K T
Let us assume node 1 as reference node number of nodes, n=4 number of tree branches, b= n-1 = 4-1 = 3 number of links, l = e-n+1 = 4-4+1 = 1
10
Verification:T i) Cb = -B l
from the matrices C b and Bl values, we have C b = -BTl
11
1.4.8 Interrelationships between the matrices A, B and C of the network graph
The matrix A can be partitioned into t wo sub-matrices i) A b of dimension (n-1)×(n-1) corresponding to the tree branches and ii) Al of dimension l×(n-1) corresponding to links, thus we can write as Ab
A
--- (1.5)
Al
The following properties are now stated without a rigorous proof and illustrated for some examples. Property 1: For a given tree of a graph each row of the fundamental loop matrix C is
orthogonal to each row of the fundamental cut set matrix B. Mathematically this relationship implies T
T
B C = C B = 0 Since B =
U B l
,
C b
and C =
T
[ U/ B l]
U C b U
, it follows
= 0
[C bU+BlTU]=0 Since U
--- (1.6)
[C b+BlT]U=0
C b+BlT = 0
0
[BTl] [ C b] = 0 T
C b = - Bl C b = - B
T
--- (1.7)
l
12
This is a very important result. It tells us that for a given tree of a graph, if the fundamental loop matrix C is known, the fundamental cut set matrix is also known and vice-versa. This relationship can be verified from eq. (1.7) Property 2: Let the incidence matrix A can be arranged in the order of tree branches and
links for a given tree i.e A
Ab
--- (1.8)
Al
It can be shown that A b is non-singular. Furthermore, the fundamental cut set matrix for a given tree is given by B = A A b =
=
Since We have
B=
Ab
-1
A b-1
Al
U Al Ab
--- (1.9)
1
U
--- (1.10)
Bl
Bl = [ Al A-1 b]
--- (1.11)
This important result tells us that, by choosing a tree and writing the incidence matrix by inspection we can obtain the fundamental cut set matrix B and also the fundamental loop matrix C from property 1. Example-3: For the network graph as shown in figure, choose a tree whose branches are
(1,2,3). Find the fundamental cut set and loop matrices B and C by using the bus incidence matrix.
Solution:
The tree for the above graph can be drawn as
13
By choosing node 1 as reference node, we can write reduced incidence matr ix as
0
Therefore A b =
1
0
1
0
0
1
1
0 1
Ab
0
1
0
0
Adj A b =
0
Ab
Al Ab
1
1
0
1
0
Adj Ab
1
Ab
1
1(1 0)
0
1
1 0 0
1
1
0 1
0
1 0
0 1
Al =
,
0 1 0 1
-1
Bl = Al A b =
T
=
0
1
0
1
1
0 1
0
0
0
-1
0
-1 -1
1
0
0
0
-1
0
-1 -1
1
0
1 0
1 0
0
0 0
1
1
1
1
14
=
1 0
1
1
1
1
U b
B
=
Bl
1
0
0
0
1
0
0
0
1
1 0
1
1
1
1
T
Since C b= -Bl
1
= -
C=
C b U
0
=
1
0
1
1
1
1
1
0
0
1
1
1
1
1
1 0
T
=
1 1 1 1
1.5 PRIMITIVE NETWORK
The data obtained from electricity boards or the power companies is in the form of primitive network (or primitive network matrix). Primitive network is a set of uncoupled elements which gives information regarding the characteristics of individual elements only. There are two types of representation of primitive networks. 1. Impedance forms 2. Admittance form 1.5.1 Impedance form
Consider the network having two nodes ‘x’ and ‘y’ is shown in the fig .(1.7), the performance equations of primitive network in impedance form can be written as
Vxy = ex - e y Vxy – Zxy ixy + e xy = 0
15
Vxy + exy = Zxy ixy V + e = [Z] i
--- (1.12)
Where Vxy = voltage across element ‘x-y’ exy = voltage source in series with element ‘x-y’ ixy = current through the element ‘x-y’ Zxy = self impedance of element ‘x-y’ 1.5.2 Admittance form
Consider the network as shown in fig.(1.8), the performance equations of the primitive network in admittance form can be written as
ixy + jxy = yxy Vxy i + j = [y] V
--- (1.13)
Where jxy = cur rent source between nodes ‘x-y’ Yxy = self admittance of branch ’x-y’
1.6 NETWORK PERFORMANCE EQUATIONS 1.6.1 Bus frame of reference
A network is made up of an interconnected set of elements. In the bus frame of reference, the performance of an interconnected network is described by (n-1) independent nodal equations, where n is the number of nodes. In matrix notation, the performance equation is VBUS = ZBUS IBUS (impedance form)
--- (1.14)
IBUS = YBUS VBUS (admittance form)
--- (1.15)
Where VBUS = vector of bus voltages w.r.t reference bus IBUS = vector of impressed bus current matrix YBUS = bus admittance matrix
16
ZBUS = bus impedance matrix 1.6.2 Branch frame of reference
In the Branch frame of reference, the performance of an inter-connected network is described by ‘b’ independent branch equation, where ‘b’ is number of branches. In matrix notation, the performance equation is VBR = ZBR IBR (impedance form)
--- (1.16)
IBR = YBR VBR (admittance form)
--- (1.17)
Where VBR = vector of voltages across the branches IBR = vector of currents through the branches YBR = branch admittance matrix ZBR = branch impedance matrix 1.6.3 Loop frame of reference
In the loop frame of reference, the performance of an inter connected network is described by ‘l’ independent loop equation, where ‘l’ is no. of links or bas ic loops. In matrix notation, the performance equation is VLOOP = ZLOOP ILOOP (impedance form)
--- (1.18)
ILOOP = YLOOPVLOOP (admittance form)
--- (1.19)
Where VLOOP = vector of basic loop voltages ILOOP= vector of basic loop currents YLOOP = loop admittance matrix ZLOOP = loop impedance matrix
1.7 FORMATION OF NETWORK MATRICES
The admittance matrix Y and the impedance matrix Z of a network can be determined by using the following methods. 1. Based on incidence matrices a. Singular transformation method b. Non singular transformation method 2. Based on network analysis equations(by direct inspection method) 1.8 SINGULAR TRANSFORMATION METHODS
The primitive impedance matrix is the most basic matrix and depends purely on the impedance of the individual elements. However, it contains no information about the behaviour of the interconnected network variables. Hence, it is to transform the primitive 17
network matrices into more meaningful matrices which can relate variables of the interconnected network. 1.8.1 Bus admittances and impedance matrices
The bus admittance matrix Y bus and bus impedance matrix Z bus can be determined by using bus incidence matrix ‘A’ which relates the variables and parameters of the primitive network to the bus quantities. The performance equation of the primitive network in admittance form is given by i+j = [y]V pre multiplying both sides with A T, we obtain AT i + AT j = AT [y] V
--- (1.20)
According to KCL, the algebraic sum of currents meet ing at any node is equal to zero. i.e AT i = 0
--- (1.21)
T
similarly algebraic sum of A j gives the algebraic sum of all source currents incident at each bus and this is nothing but the total current injected at the bus. Hence AT j = IBUS
--- (1.22)
Substituting the equations (1.21) and (1.22) in (1.20), we get IBUS = AT [y] V
--- (1.23)
Power into the network is [I BUS*]T VBUS and the sum of powers in the primitive network is * T
[j ] V. the power in the primitive and interconnected networks must be equal, i.e the transformation of the variables must power invariant. * T
* T
[IBUS ] VBUS = [j ] V
--- (1.24)
Taking the conjugate transpose of eqn.(1.22), we get T *T
* T
* T
[A ] [j ] = [IBUS ]
--- (1.25)
But A is a real matrix, so A* = A From the matrix property [AT]T = A * T
* T
[IBUS ] = A [j ]
--- (1.26)
Substituting eq. (1.26) in (1.24), we get A [j*]T VBUS = [j *]T V A VBUS = V
--- (1.27)
Substituting eq. (1.27) in (1.23), we get T
IBUS = A [y] A V BUS
--- (1.28)
But the performance equation of the network is IBUS = YBUS VBUS
--- (1.29)
18
From eq. (1.28) and (1.29) YBUS = AT [y] A
--- (1.30)
The bus impedance matrix is given by ZBUS = YBUS-1 = [ AT [y] A] -1
--- (1.31)
1.8.2 Branch admittance and impedance matrices:
The branch admittance matrix Y BR and The branch impedance matrix Z BR can determined by using basic cut set incidence matrix ‘B’ which relates the variables and parameters of the primitive network to the branch quantities of the interconnected network. The performance equation of the primitive network in admittance form is given by, i + j = [y] V T
Pre multiplying both sides with B , we get BTi + BT j = BT [y] V
--- (1.32)
T
B i is the algebraic sum of the currents through the elements inside the basic cut set and this is zero. BT i = 0
--- (1.33)
T
Similarly, B j is a vector in which each element is the algebraic sum of t he source currents of the elements to the basic cut set and can be treated as a single current source in parallel with the unique tree-branch in the basic cut set. It is denoted by I BR and given by BT j = IBR
--- (1.34)
Substituting eq. (1.33) and (1.34) in (1.32), we get T
IBR = B [y] V
--- (1.35)
Power into the network is given by [I BR *]T VBR , and this is equal to the sum of powers in the * T
primitive network, [j ] V since the power is invariant [IBR *]T VBR = [j*]T V
--- (1.36)
Taking the unique transpose of eq.(1.34), we get * T
T * T
* T
[IBR ] = [(B ) ] (j )
--- (1.37)
Since B is a real matrix, B* = B T T
From the matrix property [B ] = B [IBR *]T = B [j*]T
--- (1.38)
Substituting eq. (1.38) in (1.36), we get * T
* T
B [j ] VBR = [j ] V
--- (1.39)
Substituting eq. (1.39) and (1.34) in (1.35), we get T
IBR = B [y] B V BR
--- (1.40)
But the performance equation of the network is 19
IBR = YBR VBR
--- (1.41)
From eq. (1.40) and (1.41), we get T
YBR = B [y] B
--- (1.42)
The branch impedance matrix is given by ZBR = [YBR ]-1 = [BT [y] B]T
--- (1.43)
1.8.3 Loop admittance and impedance matrices:
The loop admittance matrix Y LOOP and loop impedance matrix Z LOOP can be determined by using basic loop incidence matrix ‘C’ which relates the variables and parameters of the primitive network to the loop quantities of the interconnected network. The performance equation of the primitive network in impedance form is given by, V + e = [Z] i Pre multiplying both sides with C T, we get T
T
T
C V + C e = C [Z] i
--- (1.44)
CT V is the algebraic sum of the all branch voltages in the loop a nd is equal to zero. CT V = 0
--- (1.45)
T
Similarly C e gives the algebraic sum of the source voltage around each basic loop. VLOOP = CT e
--- (1.46)
Substituting eq. (1.45) and (1.46) in (1.44), we get VLOOP = CT [Z] i
--- (1.47)
Power into the network is given by [I LOOP*]T VLOOP, and this is equal to the sum of powers in * T
the primitive network, [i ] e since the power is invariant [ILOOP*]T VLOOP = [i*]T e
--- (1.48)
From eq. (1.46) and (1.48), we get [ILOOP*]T CT e = [i*]T e [ILOOP*]T CT = [i*]T
--- (1.49)
[i*]T = [C I LOOP*]T i* = C I LOOP * *
i = C ILOOP
--- (1.50)
*
Since C is a real matrix, C = C i = C ILOOP
--- (1.51)
Substituting eq. (1.51) in (1.47), we get VLOOP = CT [Z] C ILOOP
--- (1.52)
But the performance equation of the network is
20
VLOOP = ZLOOP ILOOP
--- (1.53)
From eq. (1.52) and (1.53), we get T
ZLOOP = C [Z] C
--- (1.54)
The loop admittance matrix is given by YLOOP = [ZLOOP]-1 = {CT [Z] C}-1
--- (1.55)
Example 4: Derive an expression for ZLOOP for the orient graph as shown below [JNTU regular Nov-2005]
Solution:
No. of nodes, n = 5 No. of tree branches, b = n-1 = 5-1 = 4 No. of links, l = e-b = 7-4 = 3
For the graph shown, let us assume 1,2,3,7 are tree branches. So the tree and corresponding basic loop incident matrix for the given graph as
Let us assume the primitive impedance matrix for the given system is Z 11
Z 12
Z 13
Z 14
Z 15
Z 16
Z 17
Z 21 Z 22 Z 23 Z 24 Z 25 Z 26 Z 27 Z 31 Z 32
Z 33 Z 34
Z 35
Z 36
Z 37
z = Z 41 Z 42 Z 43 Z 44 Z 45 Z 46 Z 47 Z 51 Z 52
Z 53 Z 54
Z 55
Z 56
Z 57
Z 61 Z 62
Z 63 Z 64
Z 65
Z 66
Z 67
Z 71 Z 72
Z 73 Z 74
Z 75
Z 76
Z 77
21
Note:- if there is no mutual coupling between the elements, [Z] is a diagonal matrix, whose
elements are impedances of the branches. If mutual coupling exists, then the corresponding off-diagonal elements of [Z] will have a non-zero entry. Let the mutual coupling between the elements is zero Z 11
0
0
0
0
0
0
0
Z 22
0
0
0
0
0
0
0
Z 33
0
0
0
0
0
0
0
Z 44
0
0
0
0
0
0
0
Z 55
0
0
0
0
0
0
0
Z 66
0
0
0
0
0
0
0
Z 77
[z] =
T
Zloop = C [z] C
=
Z 11
0
0
0
0
0
0
0
1
1
0
Z 22
0
0
0
0
0
1
1
1
0
1
1
1
0
0
0
0
0
Z 33
0
0
0
0
1
0
0
1
1
0
0
1
0
0
0
0
0
Z 44
0
0
0
1
0
0
0
0
0
1
0
0
0
0
Z 55
0
0
0
1
0
0
0
0
0
0
Z 66
0
0
0
1
0
0
0
0
0
0
Z 77
0
0
0
1
1
1
1
1
1
1
=
0
Z 22
Z 11
Z 22
Z 11
1
Z 33 Z 44
Z 22
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
0
Z 55
0
0
0
0
Z 66
Z 22 Z 33 Z 44 =
0
Z 22 Z 22
Z 77
Z 22 Z 11 Z 22
1
1
Z 22 Z 55
Z 11 Z 22
Z 22 Z 11 Z 11 Z 22 Z 66 Z 77
Example 5: For the system as shown in figure, construct YBUS by singular transformation
method. The parameters of various elements are given in table. Take node 1 as reference node.
22
Element
Reactance in p.u
1-2 (1)
0.04
1-6 (2)
0.02
2-4 (3)
0.0.
2-3 (4)
0.02
3-4 (5)
0.08
4-5 (6)
0.06
5-6 (7)
0.05
Solution: The oriented graph and tree for the given system can be drawn as shown below.
No. of elements, e = 7 No. of nodes, n = 6 No. of tree branches, b = n-1 = 6-1 = 5 No. of links, l = e-b = 7-5 = 2 Let us assume the node 1 as reference node. The bus incidence matrix for the given syste m is
23
Primitive impedance matrix for the given system is j 0.04
0
0
0
0
0
0
0
j 0.02
0
0
0
0
0
0
0
j 0.03
0
0
0
0
0
0
0
j 0.02
0
0
0
0
0
0
0
j 0.08
0
0
0
0
0
0
0
j 0.06
0
0
0
0
0
0
0
j 0.05
[Z] =
Primitive admittance matrix is [y] = [Z] -1 The bus admittance matrix for the given system is = a YBUS = AT [y] A
[y] A =
0
0
0
0
0
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
j 20
0
0
0
j 25
0
0
0
0
0
0
0
j 50
0
0
0
0
0
0
0
0
0
0
0
0
0
j 50
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
=
j33.3
j12.5
j16.67 0
1
1
j 25
0
0
0
0
0
0
0
0
j50
j33.3
0
j33.3
0
0
j 50
j50
0
0
0
0
j12.5
j12.5
0
0
0
0
0
0
j16.67 j16.67 0
j 20
0
j 20
The bus admittance matrix for the given system is YBUS = AT [y] A 24
1 0 1
0 1
0
0
0
0
0
0
1 1
0 1
1
=
0
1
0
0
0
0
0
0
0
0
1
1 1
1
0
0
0
0
1
0
0
1
0
1
0
0
0
0
0
0
1 0
0
0
0
0
0
0
0
j50
j33.3
0
j33.3
0
0
j 50
j50
0
0
0
0
j12.5
j12.5
0
0
1
j 50 =
0
1
j108.33 YBUS
j 25
0
0
0
0
j16.67 j16.67
0
j 20
0
j 20
j 50
j 33.3
0
0
j 62.5
j12.5
0
0
j 33.3
j12.5
j 62.5
j 36.67
0
0
0
j16.67
j 36.67
j 20
0
0
0
j 20
j 70
OBJECTIVE TYPE QUESTIONS
1. If the number of branches in a network is B, the number of nodes is N and the number of dependent loops is L, then the number of independent node equations will be a) N + 1
b) N-1
2
N – 1
2
c) N -1
d) N +1
2. The number of independent loops for a network with N nodes and B branches is B - N + 1 a) B+ N + 1
b) B+ N-1
c) BN-1
d) B- N+1
3. The graph of an electrical network has N nodes, B branches. The number of links L with respect to the choice of tree is given by B - N + 1 a) B+ N - 1
b) B+ N-1
c) BN+1
d) B- N+1
4. The diagonal elements of bus admittance matrix is called Self admittances a) Self admittances b) Mutual admittance
c) both
d) none
5. The off diagonal elements of bus admittance matr ix is called a) Self admittances b) Mutual admittance
c) both
Mutual admittance
d) none
6. The matrix consisting of the self and mutual admittances o f the network 1of the power system is called Bus admittance matrix a) loop admittance b) node admittance c) Bus admittance 7. If a graph having N number of nodes, the rank of the graph is a) N + 1
b) N-1
c) N2-1
8. The terminal of an element is called a a) Vertex
b) node
c)both
d)none n-1
d) N2+1 vertex
d)none
9. In a graph, if there are 4 nodes and 7 elements, the number of links is 4 a) 11
b) 4
c) 7
d) 3
25
10. The dimension of the bus incidence matrix is a) e × (n-1)
b) e × n
e (n-1)
c) (e-1) × (n-1)
d) (e-1) × n
11. A tree has ------- No closed paths a) one closed path
b) no closed path
c) infinity closed paths d) none
12 The number of branches in a tree is ---------- the number of branches in a graph a) Less than
b) greater than
c) qual to
d) none
13. If a network contains B branches and N nodes, t hen the number of mesh current equations would be b-(n-1) a) B+ N + 1
b) B+ N-1
c) BN-1
d) B- N+1
14. The meeting of various components in a power system is called---------------. a) node
b) bus
d) none
c) both
15. If Y bus matrix is symmetrical then the corresponding Z bus matrix is a) symmetrical
b) unsymmetrical
c) both
symmetrical
d) none
16. Tree is a sub-graph containing all the _____ of the original graph vertices a) vertices
b) elements
c) both
d) none
17. Co-tree is a complement of a tree a) tree
b) branch
c) link
d)all
18. The incidence of element to nodes in a connected graph is given by -- ---- incidence matrix a) bus
element node
b) element node c) loop
d)all
19. The number of loops in a connected graph is equal to the number of links a) vertices
b) elements
c) links
d) none
20. The elements of the co-tree are called links a) vertices
b) branches
c) links
d) none
21. A graph consisting 'n' number of nodes, the number of tree branches is n-1 a) n
2
b) n
c) n-1
d)2n
22. A network has 7 nodes and 5 independent loops. The number of branches in the network is 11 a) 11
b) 7
c) 5
d) 12
23. The diagonal elements of bus impedance matrix is called driving point impedances a) driving point impedances b) transfer impedances c) both
d)none
24. The off diagonal elements of bus impedance matrix is called Transfer impedances a) driving point impedances b) transfer impedances c) both 26
d)none
25. The matrix consisting of driving point impedances and transfer impedances of the network of the power system is called bus impedance matrix a) bus impedance matrix b) loop impedance matrix c)branch impedance matrix d)none 26. --------- is a minimal set of elements of a connected graph, which divides the entire graph into two parts cut set a) cut set
b) tie set
c) both
d)none
27. ---------- incidence matrix gives t he incidence of element to basic loops of a connected graph basic loop a) basic cut set
b) basic loop
c) basic tie set
d) element node
28. Apply the KCL at every bus for developing the bus----------- matrix admittance a) impedance
b)admittance
c) both
d)none
29. A graph consisting n number of nodes, the rank of tree is n-1 a) n
b) n-1
c) n-2
d) 2n
30. Apply the KVL at every bus for developing the bus----------- matrix impedance a) impedance
b)admittance
c) both
d)none
31. Nodal admittance matrix is a _______ matrix symmetric
s) symmetric
b) un symmetric
c) sparse
27
d) both a and c
