2.graph theory.pdf

1

1. GRAPH THEORY

1.1 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 the 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 components 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 a junction point or inter-connection point of two or 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 representation 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 graph.

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Tree: The tree of a network connects all the nodes of the network but contains no closed

path.

Twig: 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 Gs is a

node of G and every branch of Gs is also a branch of G.

Rank: The rank of a 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 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.

Example: The following fig 1.1 shows all the above things clearly

Fig.(1.1)

3

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 of 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.

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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

The elements aij of the complete incidence matrix [A1] are found as follows

aij = 1, if ith

element is incident and oriented away from the jth

node.

aij = -1, if ith element is incident and oriented towards the j

th node.

aij = 0, if ith

element is not incident to the jth

node.

Consider the graph as shown in fig.(1.3). The element node incidence matrix 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).

5

The matrix A can be partitioned into two sub-matrices (i) At of dimension (n-1)x(n-1)

corresponding to the branches which are twigs and (ii) Al 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)

Where AT 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 kij of the branch path incidence matrix [K] are found as follows

kij = 1, if the ith

branch is in the path from the jth

bus to reference and is

oriented in the same direction.

kij = -1, if the ith

branch is in the path from the jth bus to reference and is

oriented in the opposite direction.

kij = 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 At relate the branches to paths and

branches to buses respectively. So there is one to one correspondence between paths and

buses.

6

At KT=U

KT= 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 ith

element is incident to and oriented in the opposite

direction as the jth basic cut set.

bij =0, if the ith

element is not incident to the jth

basic cut set.

The basic cut set incidence matrix for the example of fig. (1.5) is

The sub matrix Bl can be obtained from the bus impedance matrix A. The incidence of links

to buses is given by the sub matrix Al and the incidence of elements to buses is given by the

sub matrix Ab. Since there is a one to one correspondence of the elements and basic cut-sets,

Bl Ab gives the incidence of links to buses.

i.e Bl At= Al

Bl = Al At-1

= Al KT ( Since At

-1 = K

T) --- (1.3)

1.4.5 Augmented cut-set matrix (B1)

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

7

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 of 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 ith element is incident to and oriented in the same direction as

the jth basic loop.

Cij = -1, if the ith

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

CT V = 0 --- (1.4)

1.4.7 Augmented loop incidence matrix (C1)

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.

8

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 tree. It

shows the incidence of branches to paths.

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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

Determine the incidences matrices A, K, B, B1, C, C

1 for the figure as shown below

and verify the following

i) Cb = -BT

l ii) C1 (B

1)

T = U iii) At K

T = U iv) Bl = Al K

T

Solution:

The oriented graph for the given network is

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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

11

Verification:-

i) Cb = -BT

l from the matrices Cb and Bl values, we have

Cb = -BT

l

12

1.4.8 Interrelationships between the matrices A, B and C of the network graph

The matrix A can be partitioned into two sub-matrices

i) Ab 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

b

l

AA

A --- (1.5)

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

BT C = C B

T = 0 --- (1.6)

Since B = lB

U, and C =

U

Cb , it follows

[ U/ BT

l] U

Cb = 0

[CbU+BlTU]=0 [Cb+Bl

T]U=0

Since U 0 Cb+BlT = 0

[BT

l] [ Cb] = 0

CbT = - Bl

Cb = - BT

l --- (1.7)

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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

b

l

AA

A --- (1.8)

It can be shown that Ab is non-singular. Furthermore, the fundamental cut set matrix for a

given tree is given by

B = A Ab-1

= b

l

A

A Ab

-1

= 1

bl AA

U --- (1.9)

Since B = lB

U --- (1.10)

We have 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

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By choosing node 1 as reference node, we can write reduced incidence matrix as

Therefore Ab =

011

001

100

, Al = 0

1

1

1

0

0

1)01(1

011

001

100

bA

Adj Ab =

T

010

011

100

=

001

110

010

001

1-1-0

01-0

Ab Adj1

b

bA

A

001

1-1-0

01-0

0

1

1

1

0

01

bl AA = 1

1

1

1

0

1

Bl = Al Ab-1

= 1

1

1

1

0

1

15

l

b

B

UB =

110

111

100

010

001

Since Cb= -BlT

= -

T

1

1

1

1

0

1 =

11

11

01

C = U

Cb =

10

01

11

11

01

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 - ey

Vxy – Zxy ixy + exy = 0

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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 = current 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

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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.


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 basic loops.


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

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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 busY and bus impedance matrix busZ 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 AT, we obtain

AT i + A

T j = A

T [y] V --- (1.20)

According to KCL, the algebraic sum of currents meeting at any node is equal to zero.

i.e AT i = 0 --- (1.21)

similarly algebraic sum of AT 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 [IBUS*]

T VBUS and the sum of powers in the primitive network is

[j*]

T V. the power in the primitive and interconnected networks must be equal, i.e the

transformation of the variables must power invariant.

[IBUS*]

T VBUS = [j

*]

T V --- (1.24)

Taking the conjugate transpose of eqn.(1.22), we get

[AT]

*T [j

*]

T = [IBUS

*]

T ---

(1.25)

But A is a real matrix, so A* = A

From the matrix property [AT]

T = A

[IBUS*]

T = A [j

*]

T ---

(1.26)

Substituting eq. (1.26) in (1.24), we get

A [j*]

T VBUS = [j

*]

T V

A VBUS = V --- (1.27)


IBUS = AT [y] A VBUS --- (1.28)

But the performance equation of the network is

IBUS = YBUS VBUS --- (1.29)

19

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 YBR and The branch impedance matrix ZBR 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

Pre multiplying both sides with BT, we get

BTi + B

Tj = B

T [y] V --- (1.32)

BT 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)

Similarly, BT j is a vector in which each element is the algebraic sum of the 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 IBR and given by

BT j = IBR --- (1.34)

Substituting eq. (1.33) and (1.34) in (1.32), we get

IBR = BT [y] V --- (1.35)

Power into the network is given by [IBR*]

T VBR, and this is equal to the sum of powers in the

primitive network, [j*]

T V since the power is invariant

[IBR*]

T VBR = [j

*]

T V --- (1.36)

Taking the unique transpose of eq.(1.34), we get

[IBR*]

T = [(B

T)

*]

T (j

*)

T --- (1.37)

Since B is a real matrix, B* = B

From the matrix property [BT]

T = B

[IBR*]

T = B [j

*]

T --- (1.38)


B [j*]

T VBR = [j

*]

T V --- (1.39)


IBR = BT [y] B VBR --- (1.40)


20

IBR = YBR VBR --- (1.41)

From eq. (1.40) and (1.41), we get

YBR = BT [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 YLOOP and loop impedance matrix ZLOOP 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 CT, we get

CT V + C

T e = C

T [Z] i --- (1.44)

CT V is the algebraic sum of the all branch voltages in the loop and is equal to zero.

CT V = 0 --- (1.45)

Similarly CT e gives the algebraic sum of the source voltage around each basic loop.

VLOOP = CT e --- (1.46)


VLOOP = CT [Z] i --- (1.47)

Power into the network is given by [ILOOP*]

T VLOOP, and this is equal to the sum of powers in

the primitive network, [i*]

T e since the power is invariant

[ILOOP*]

T VLOOP = [i

*]

T e --- (1.48)


[ILOOP*]

T C

T e = [i

*]

T e

[ILOOP*]

T C

T = [i

*]

T --- (1.49)

[i*]

T = [C ILOOP

*]

T

i* = C ILOOP

*

i = C* ILOOP --- (1.50)

Since C is a real matrix, C* = C

i = C ILOOP --- (1.51)


VLOOP = CT [Z] C ILOOP --- (1.52)


21

VLOOP = ZLOOP ILOOP --- (1.53)


ZLOOP = CT [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 =

77767574737271

67666564636261

57565554535251

47464544434241

37363534333231

27262524232221

17161514131211

ZZZZZZZ

ZZZZZZZ

ZZZZZZZ

ZZZZZZZ

ZZZZZZZ

ZZZZZZZ

ZZZZZZZ

22

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] =

77

66

55

44

33

22

11

000000

000000

000000

000000

000000

000000

000000

Z

Z

Z

Z

Z

Z

Z

Zloop = CT [z] C

=

1100011

0010011

0001110

77

66

55

44

33

22

11

000000

000000

000000

000000

000000

000000

000000

Z

Z

Z

Z

Z

Z

Z

100

100

010

001

001

111

110

=

77662211

552211

443322

000

0000

0000

ZZZZ

ZZZ

ZZZ

100

100

010

001

001

111

110

=

77662211221122

112255221122

2222443322

ZZZZZZZ

ZZZZZZ

ZZZZZ

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.

23

Element Reactance in p.u

1-2 (1)

1-6 (2)

2-4 (3)

2-3 (4)

3-4 (5)

4-5 (6)

5-6 (7)

0.04

0.02

0.0.

0.02

0.08

0.06

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 system is

24

Primitive impedance matrix for the given system is

[Z] =

05.0000000

006.000000

0008.00000

00002.0000

000003.000

0000002.00

00000004.0

j

j

j

j

j

j

j

Primitive admittance matrix is

[y] = [Z]-1

The bus admittance matrix for the given system is

= a

YBUS = AT

[y] A

[y] A =

20000000

067.1600000

005.120000

00050000

00003.3300

00000500

00000025

j

j

j

j

j

j

j

11000

01100

00110

00011

00101

10000

00001

=

2020000

067.1667.1600

005.125.120

0005050

003.3303.33

500000

000025

jj

jj

jj

jj

jj

j

j

The bus admittance matrix for the given system is

YBUS = AT

[y] A

25

=

1000010

1100000

0110100

0011000

0001101

2020000

067.1667.1600

005.125.120

0005050

003.3303.33

500000

000025

jj

jj

jj

jj

jj

j

j

YBUS =

7020000

2067.3667.1600

067.365.625.123.33

005.125.6250

003.335033.108

jj

jjj

jjjj

jjj

jjj

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 N – 1

a) N + 1 b) N-1 c) N2-1 d) N

2+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 matrix is called Mutual admittance

a) Self admittances b) Mutual admittance c) both d) none

6. The matrix consisting of the self and mutual admittances of the network 1of the power

system is called Bus admittance matrix

a) loop admittance b) node admittance c) Bus admittance d)none

7. If a graph having N number of nodes, the rank of the graph is n-1

a) N + 1 b) N-1 c) N2-1 d) N

2+1

8. The terminal of an element is called a vertex

a) Vertex b) node c)both 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

26

10. The dimension of the bus incidence matrix is ex (n-1)

a) e × (n-1) b) e × n 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, then 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 c) both d) none

15. If Ybus matrix is symmetrical then the corresponding Zbus matrix is symmetrical

a) symmetrical b) unsymmetrical c) both 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 element node

a) bus 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 b) n2 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 d)none

27

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 the 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 d) both a and c

2.graph theory.pdf

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