1

St. Vincent Pallotti College of Engineering & Technology

Department of Electrical Engineering

Computer Application in Power System

By Dr. Nitin K. Dhote

2

Unit Syllabus Details

1

Formation of network matrices by singular

transformation

Connected graph, orientred graph, tree and links

Basic cut sets and basic loops

Incidence matrices

Primitive Network

2Algorithm for formation of network matrices

Algorithm for formation of Z bus and Ybus without mutual

coupling

3

Three phase networkThree phase balanced network element with balanced and

unbalanced excitation

Incidence and network matrices for 3 phase element

Algorithm for formation of three phase bus impedance matrics

UNIT1- Q1 & Q2 UNIT2-Q3&Q4 UNIT3&5-Q5&Q6UNIT4&6-Q7&Q8 ; SOLVE 4Q , EACH QUESTION CARRY 20 MKS

3

4

Load Flow StudiesG.S. method and N.Raphson method, fast decoupled

technique

Representation of tap changing transforemer

Phase shifting transformer

Elementaryload flow equations

5

Short circuit studiesshort circuit calculations using bus impedance matrix for

balanced and un balanced fault

Algorithm for short circuit studies

6

Transient stability studies

Modelling of synchronous machine

Numerical solution of swing equation by modified Euler,s

method

Numerical solution of swing equation by Runge Kutta

method

algorithm for transient stability study

4

SR. NO. TITLE OF BOOKS / EDITIONAUTHOR /

PUBLICATION

NO. OF

COPIES

AVAILABLE

IN LIBRARY

1Computer methods in

power system analysisStagg and Ele Abaid,

TMHNil

2Electric Energy system

theory and introductionOile J. Elgard,TMH 24

3Elements of power

system analysisW.D.stevenson,TMH 13

4Computer analysis of

power systemR.N.Dhar Nil

5

STUDY OF INTERCONNECTIONS OF POWER SYSTEM COMPONENTS

6

1.Graph:

2. Elements:

3. Nodes:

4. Connected Graph:

7

1.Graph:

2. Elements:

3. Nodes:

4. Connected Graph:

Ϩ Ϩ

Ϩ

G1

G2

G3

1 23

4

1

2

3

45

7

6

8

CONNECTED GRAPH

Total elements--7

Total nodes--5

Ground node-- Ref.node

Total buses --4

1 2 3 4

0

1 2 3

456

7

Ground Node

Ϩ Ϩ

Ϩ

G1

G2

G3

1 23

4

1

2

3

45

7

6

9

Each element is assigned a direction .

1 2 3 4

0

1 2 3

456

7

Ground Node

10

Branches :

0

4321

1

2

3

4

Tree 1

0

4321

1

3

4

6

Tree 2

0

4321

1

2

3

5

Tree 3

11

12

Links :

Rules:

1 2 3 4

0

1 2 3

456

7

Ground Node

13

0 1 2 3 4

Ele =1 1 -1

2 1 -1

3 1 -1

4 -1 1

5 1 -1

6 1 -1

7 1 -1

Branches

Links

Nodes

1 2 3 4

0

1 2 3

456

7

Ground Node

14

Rules:a = 1; If element is directed away from the node.a = -1; If element is directed towards the node.a = 0; If element is not connected to the node.

1 2 3 4

Ele =1 -1

2 -1

3 -1

4 -1 1

5 1 -1

6 1 -1

7 1 -1

Branches

Links

Buses

15

Ele =1

2

3

4

5

6

7

Branches

Links

1 2 3 4

-1

-1

-1

-1 1

1 -1

1 -1

1 -1

Ref. Buses

16

Basic Cut set:

Rules to form Basic Cut set: a) Circular or closed loop cut.b) Should cut only one branch , may cut one or more

links.c) Should not cut any element twice.d) Direction is given by branch.

( Centrally inward or outward)

17

Ground Node

1 2 3 4

0

1 2 3

456

7

AB

C

D

A : Basic Cutset ( Branch 1)B : Basic Cutset ( Branch 2)C : Basic Cutset ( Branch 3)D : Basic Cutset ( Branch 4)

18

b = 1; If element is directed in same direction of basic cutset.

b = -1; If element is directed in opposite direction of basic cutset.b = 0; If element is not connected to the basic cutset.

19

A B C D

Ele =1 12 13 14 15 -1 1 16 -1 17 -1 1

Branches

Links

Basic Cutsets

1 2 3 4

0

1 2 3

456

7

A B

CD

20

Ground Node

1 2 3 4

0

1 2 3

456

7A

B

CD

E F

G

21

A B C D

Ele

=11

2 13 14 15 -1 1 16 -1 17 -1 1

Branches

Links

Basic CutsetsE F G

1

1

1

Tie Cutsets

g

ef

1 2 3 4

0

12

3

4

56

7

22

23

c = 1; If element is directed in same direction of basic loop.

c = -1; If element is directed in opposite direction of basic loop.

c = 0; If element is not connected to the basic loop.

24

e f g

Ele =1 12 1 -1 13 -1 -14 -15 16 17 1

Branches

Links

0

4321

12

3

45

6

7

f

g

e-- Basic loop ( Link5 )

f-- Basic loop ( Link6 )

g-- Basic loop ( Link7 )

e

Basic Loops

24

250

4321

12

3

45

6

7

f

g

e

a

b

c

d

26

Ele

=1

2

3

4

5

6

7

Branches

Links

0

4321

12

3

45

6

7

f

g

a--Open loop (Branch 1)

b--Open loop (Branch 2)

c--Open loop (Branch3)

d--Open loop (Branch 4)

e

a

b

c

d

e f g

1

1 -1 1

-1 -1

-1

1

1

1

a b c d

1

1

1

1

BASIC LOOPSOPEN LOOPS

27

1 2 3 4

0

12

3

4

Ref. Node

K1

K2

K3K4

Buses: 1,2 ,3,4Ref. Node: 0

BRANCH PATH I : BUS1BRANCH PATH II : BUS2BRANCH PATH I II: BUS3BRANCH PATH I V: BUS4

28

17. Branch Path Incidence Matrix K

Rules:k = 1; If branch is directed in same direction of branch path.

k = -1; If branch is directed in opposite direction of branch path .k= 0; If branch is not connected to the branch path.

29

K1 K2 K3 K4

Br

=1-1

2 -13 -1 -14 -1

Branch Paths

1 2 3 4

0

12

3

4

Ref. Node

K1

K2

K3K4

30

Pb.2 For the figure shown , determine matrices A, A^, B,B^, C, C^ & K. Select node 1 as reference and select elements 2&5 as a links.

1

23

4

1

2

3

4

5

31

2 33 444

-1

-1

-1

1 -1

1 -1

1

3

4

2

5

Br

Link

Ref. 1

1

1

1

1

2 3

4

1

2

4

5

REF.

32

1

2 3

4

1

2

4

5

REF.

A B

C

A B C

1 1

3 1

4 1

2 -1 1

5 -1 1

Br

Link

33

A B C

1 1

3 1

4 1

2 -1 1

5 -1 1

Br

Link

D E

1

1

34

d

1 1

3 -1

4

2 1

5

eloops

Br

Links

e

1

-1

1

1

2 3

4

1

4REF.

2

5

d

e

35

a b c

1

1

1

Br

Links

1

3

4

2

5

d e

1

-1 1

-1

1

1

36

K1 K2 K3

1 -1

3 -1

4 -1

BR

1

2 3

4

1

4REF.

K1 K2

K3

37

Pb.3 For the Figure shown , determine incidence matrices. Select node 1 as reference and elements 1,2,4 and 6 as branches.

1

4 5

3

2

1

2

4

6

3

5

7

REF.

38

1

4 5

3

2

1

2

4

6

3

5

7

ELEMENT NODE INCIDENCE MATRIX :

REF.

REF. 1

-1

1

.

1

1

2

4

6

3

5

7

2 3 4 5

1

-1

1 -1

1 -1

-1

-1 1

-1 1

39

BASIC CUTSETS AND TIE CUTSETS

1

4 5

3

2

1

2

4

6

3

5

7

A

C

B

D

E

F

G

40

C

B1

4 5

3

2

1

24

6

3

5

7A

D

E

F

G

AUGMENTED CUTSET INCIDENCE MATRIX

E F G

1

1

1

A B C D

1 1

2 1

4 1

6 1

3 1 -1

5 1 -1

7 -1 -1 1 -1

41

BASIC LOOPS AND OPEN LOOPS

1

4 5

3

2

1

2

4

6

3

e

5

fa

b cd

7

g

a,b,c,d : Open Loopse,f,g : Basic Loops

42

1

4 5

3

2

1

24

6

3

e

5

fa

b c d

7g

AUGMENTED LOOP INCIDENCE MATRIX :

1

2

4

6

3

5

7

e f g

1

-1 1

1 -1 -1

1 1

1

1

1

a b C d

1

1

1

1

43

BRANCH PATHS

1

4 5

3

2

1

2

4

6

K1

K2

K3

K4

REF.

44

BRANCH PATH INCIDENCE MATRIX K

K1

1

4 5

3

2

1

24

6K2

K3

K4

K1 K2 K3 K4

1 1

2 -1 -1 -1

4 1 1

6 -1

45

Pb.4 For a particular power system, element node incidence matrix is given by

1 2 3 4

1 1 -1

2 1 -1

3 -1 1

4 1 -1

5 1 -1

Draw oriented graph, select 4 & 5 as link and node 1 as ref. node. Determine matrices B^, c^ and K.

46

ORIENTED GRAPH

1

32

4

1

2

3

4

5

BASIC CUTSETS

1

32

4

1

2

3

4

5A

B

C

47

BASIC CUTSET INCIDENCE MATRIX

1

32

4

1

2

3

4

5A

B

C

A B C

1 1

2 1

3 1

4 -1 1 -1

5 -1 1

48

AUGMENTED CUTSET INCIDENCE MATRIX

1

32

4

1

2

3

4

5A

B

C

A B C

1 1

2 1

3 1

4 -1 1 -1

5 -1 1

D

E

D E

1

1

49

1

32

4

1

2

3

4

5

d

e

1

2

3

4

5

d e

1 1

-1 -1

1

1

1

ab c

a b c

1

1

1

50

32

4

1

2

3

1

K1

K2

K3

REF.

K1 K2 K3

1 -1

2 -1 -1

3 1

51

[Ab]br X b X [KT]b X br = [Ubr]

[Al]l X b X [KT]b X br = [Bl]lXbr

52

Pb.5 4 bus sample system is shown in fig. Consider ground as ref. and elements 4 &5 as links. Determine A^, B^,C^ and K and also verify the relation sUbr = Ab.K T ; Al.K T = Bl and Cb = -Bl T

G

G

1 2 3

4

1

2 3

45

6

53

ORIENTED GRAPH

ELEMENT NODE INCIDENCE MATRIX

0 1 2 3 4

1 1 -1

2 1 -1

3 1 -1

6 1 -1

4 1 -1

5 1 -1

1 2 3

4

0REF.

1

2 3

45

6

54

BUS INCIDENCE MATRIX

1 2 3 4

1 -1

2 1 -1

3 1 -1

6 -1

4 1 -1

5 1 -1

Ab

Al

55

AUGMENTED CUTSET INCIDENCE MATRIX

1 2 3

4

0REF.

1

2 3

45

6 A

B

C

D

A B C D E F

1 1

2 1

3 1

6 1

4 -1 -1 -1 1 1

5 -1 1 1

E

F

56

AUGMENTED LOOP INCIDENCE MATRIX

1 2 3

4

0REF.

1

2 3

45

6

f

a

b c

d

a b c d e f

1 1 1 1

2 1 1

3 1 1

6 1 -1 -1

4 1

5 1

e

57

BRANCH PATH INCIDENCE MATRIX K

1 2 3

4

0

1

2 3

6 K4

K1 K2 K3 K4

1 -1 -1 -1

2 -1 -1

3 -1

6 -1

58

-1

1 -1

1 -1

-1

-1

-1 -1

-1 -1 -1

-1

1

1

1

1

1 -1

1 -1

-1

-1 -1

-1 -1 -1

-1

-1 -1 -1 1

-1 1

1 1

1

1

-1 -1

60

Pb.6 For the power system shown , select element 1,2,6,7 as links and ground node as reference. Determine matrices A^, B^,C^ and K . Verify the relations 1. Ab. K T= Ubr 2. Al.K T =Bl and 3. Cb= -Bl T

G

G

L L

G

1 321

2

34

5 6

7

61

ORIENTED GRAPH

1

0

323 4

51

2

6

7

REF

62

1

0

323 4

512

67

REF

ELEMENT NODE INCIDENCE MATRIX

0 1 2 3

3 1 -1

4 1 -1

5 -1 1

1 -1 1

2 -1 1

6 -1 1

7 -1 1

63

0 1 2 3

3 1 -1

4 1 -1

5 -1 1

1 -1 1

2 -1 1

6 -1 1

7 -1 1

BUS INCIDENCE MATRIX

64

AUGMENTED CUTSET INCIDENCE MATRIX

1

0

323 4

512

67

REF

AB

C

DE

FG

A B C D E F G

3 1

4 1

5 1

1 1 1 1

2 1 1 1

6 -1 1 1

7 -1 1 1

65

AUGMENTED LOOP INCIDENCE MATRIX

1

0

323 4

512

67

REF

e

d f

g

a b

c

a b c d e f g

3 1 -1 -1

4 1 1 1

5 1 -1 1 -1 -1

1 1

2 1

6 1

7 1

66

BRANCH PATH INCIDENCE MATRIX

1

0

23 4

5

3

REF

K1

K2 K3

K1 K2 K3

3 1

4 -1

5 1 1 1

67

PRIMITIVE ELEMENT : PRIMITIVE MATRIX :

A) IMPEDANCE FORM :

EqZpq

P QEp

epq

ipq+Vpq-

+- +

+-

-

68

Vpq = Zpq ipq – epqVpq +epq = Zpq .ipq

V nx1 + e nx1 = Z nxn . i nx1

69

YpqP Qipq

jpq

+ Vpq -

Ipq +jpq = Vpq . Ypq

i nx1 + j nx1 = Y nxn . V nx1 ,

70

A) BUS REFERENCE FRAMEEBUS = [ZBUS] . IBUS

IBUS = [YBUS]. EBUS

B) BRANCH REFERENCE FRAMEEBR = [ZBR].IBR

IBR = [YBR]. EBR

C) LOOP REFERENCE FRAMEELOOP = [ZLOOP]. ILOOP

ILOOP = [YLOOP].ELOOP

PERFORMANCE EQUATION OF PRIMITIVE ELEMENT IN ADMITTANCE FORM:i + j = [y]. V

PREMULTIPLYING EQN.1 BY AT

AT . i + AT . j = AT . [y] . V

APPLYING KCL, AT . i =0

I BUS = AT . j USING EQNS. (2) AND (3)

I BUS = AT [Y] VTRANSFORMATION FROM PRIMITIVE TO BUS REF. IS POWER INVARIANT

S = IBUS* T . EBUS = j *T .V

FROM EQ.(3) , IBUS = AT. j TAKING CONJUGATE TRANSPOSE

IBUS*T = j*T . A*

71

72

IBUS*T = j*T . A*

A* = A IBUS

*T = j*T . AS = IBUS

* T . EBUS = j *T .V (5)FROM EQ.(5) AND EQ.(6),

j*T . A. EBUS = j *T .VV = A. EBUS

I BUS = AT [Y] VFROM EQS.(4) AND (7),

IBUS = AT[Y]A.EBUS

PERFORMANCE EQUATION IN BUS REF. FRAME IN ADMITTANCE FORM

IBUS = [YBUS] . EBUS

FROM EQNS. (8) AND (9)YBUS = AT . Y. A

PERFORMANCE EQUATION OF PRIMITIVE ELEMENT IN ADMITTANCE FORM:i + j = [y]. V

PREMULTIPLYING EQN.1 BY BT

BT . i + BT . J = BT . [y] . V

APPLYING KCL, BT . i =0

I BR = BT . j USING EQNS. (2) AND (3)

I BR = BT [Y] VTRANSFORMATION FROM PRIMITIVE TO BRANCH REF. IS POWER INVARIANT

S = IBR* T . EBR = j *T .V

FROM EQ.(3) , IBR = BT. j TAKING CONJUGATE TRANSPOSE

IBR*T = j*T . B*

73

74

IBR*T = j*T . B*

* = B IBR

*T = j*T . BS = IBR

* T . EBR = j *T .V (5)FROM EQ.(5) AND EQ.(6),

j*T . B. EBR = j *T .VV = B. EBR

I BR = BT [Y] VFROM EQS.(4) AND (7),

IBR = BT[Y]B.EBR

PERFORMANCE EQUATION IN BRANCH REF. FRAME IN ADMITTANCE FORM

IBR = [YBR] . EBR

FROM EQNS. (8) AND (9)YBR = BT . Y. B

75

V + e = [Z].i

C T .V +C T . e = C T . [Z].iC -- INCIDENCE OF ELEMENTS TO BASIC LOOPS C T .V -- ALGEBRIC SUM OF VOLTAGES OF THE ELEMENTS AROUND EACH BASIC LOOP

C T . V = 0C T .e -- ALGEBRIC SUM OF EXTERNAL VOLTAGE SOURCE AROUND EACH LOOP

ELOOP = C T . e

ELOOP = C T .[Z]. i

I LOOP iE LOOPe

S= ILOOP * T . ELOOP = i* T .e

I LOOP* T . C T.e = i* T . e

I LOOP * T .C T = i* T

76

I LOOP * T .C T = i* T

i = C *. I LOOP

SINCE C IS REAL MATRIX, C* = Ci = C . I LOOP

ELOOP = C T .[Z]. i (4)

ELOOP = C T .[Z]. C . I LOOP

E LOOP = Z LOOP . I LOOP

Z LOOP = C T . [Z]. C

77

RHS = K . [ ]. K T

YBUS = A T [Y] A= [Y]

A K T = B , B T = K .A T

RHS = B T [Y] B= YBR

78

RHS = Ab T [ ] AbY BR = B T [Y] B

= Ab T [Y] AbB = A K T

B T = KA T

RHS A T [Y] A Ab K T =Ubr

K T = Ab -1

KT Ab = Ubr

K Ab T = UbrAb T = K -1

Ab T K = Ubr

RHS = A T [Y] A= Y BUS

79

1 2

4

3

0.51

0.52

50.2

80

1 2 3

4

0.52

REF.

1 3 4

1 1

3 1 -1

5 -1

2 1

4 -1 1

0.51

0.25

4

1 3 5 2 4

1 1 1 1

3 -1 -1

4 -1 1

1 3 5 2 4

1 2

3 2.5

5 5

2 2

4 1.25

ELEMENTNO.

1 2 3 4 5

Z 0.5 0.5 0.4 0.8 0.2

Y 2 2 2.5 1.25 5

82

1 3 5 2 4

1 1 1 1

3 -1 -1

4 -1 1

1 3 5 2 4

1 2

3 2.5

5 5

2 2

4 1.25

1 3 4

1 1

3 1 -1

5 -1

2 1

4 -1 1

2 2.5 2

-5 -1.25

-2.5 1.25

1

1 -1

-1

1

-1 1

83

6.5 0 -2.5

0 6.25 -1.25

-2.5 -1.25 3.75

1 2 3

4

3

1

5REF.

K1

K3

K2

K1 K2 K3

1 1 1

3 -1

5 -1

84

1 1

-1

-1

6.5 0 -2.5

0 6.25 -1.25

-2.5 -1.25 3.75

1

-1

1 -1

4 -1.25 1.25

2.5 1.25 -3.75

-6.25 1.25

1

-1

1 -1

5.25 -1.25 1.25

-1.25 3.75 -1.25

1.25 -1.25 6.25

85

2

41

3

1 (0.5)

2 (0.5)

5 (0.4)

3 (1.0)

0.2

86

2

41

3

1 (

0.5

)

2 (0.5)

5 (0.4)

3 (1

.0)

0.2

2 3 4

1 -1

4 -1

2 -1

3 1 -1

5 1 -1

** MUTUAL COUPLED BRANCHES 1 & 4 ARE KEPT ADJACENT AND PLACED AT THE TOP OF THE MATRIX

87

1 4 2 3 5

2 -1 1

3 -1 1 -1

4 -1 -1

1 4 2 3 5

1 0.5 0.2

4 0.2 0.4

2 0.5

3 1

5 0.4

Z1 0

0 Z4

88

Y = Z -1 Y1 0

0 Y4

0.5 0.2

0.2 0.4=

-1

=

2.5 -1.25

-1.25 3.125

=0.5

1

0.4

-1

=

2

1

2.5

Y = -j

2.5 -1.25

-1.25 3.125

2

1

2.5

89

-1 1

-1 1 -1

-1 -1

2.5 -1.25

-1.25 3.125

2

1

2.5

-1

-1

-1

1 -1

1 -1

5 -3.75 0

-3.75 6.625 -1

0 -1 3

90

2

41

3

1

2

4K1

K3

K1 K2 K3

1 -1

4 -1

2 -1

REF.

91

1

0.1025

0.42

0.63

4

0.1025

1 32

1

0

32

1

2 3

4

92

1

0

32

1

2 3

4

A

B

C

A B C

1 1

2 1

3 1

4 1 1 1

1 2 3 4

A 1 1

B 1 1

C 1 1

93

ELEMENTNO.

Z Y

1 0.1025 9.756

2 0.4 2.5

3 0.6 1.67

4 0.1025 9.756

1 2 3 4

1 9.756

2 2.5

3 1.67

4 9.756

1 1

1 1

1 1

9.756

2.5

1.67

9.756

1

1

1

1 1 1

94

19.512 9.756 9.756

9.756 12.256 9.756

9.756 9.756 11.426

0.0937 -0.034 -0.05

-0.034 0.267 -0.198

-0.05 -0.198 0.3009

1

0

32

1

2 31 2 3

1 -1

2 1 -1

3 1 -1

BR

BUSES

95

-1 1

-1 1

-1

19.512 9.756 9.756

9.756 12.256 9.756

9.756 9.756 11.426

-1

1 -1

1 -1

12.256 -2.5 0

-2.5 4.17 -1.664

0 -1.67 11.42

96

B

DC

G A

REF.

1

2

3

4

5

6

7

ELEMENT NO.

1 2 3 4 5 6 7

Z 0.04 0.04 0.05 0.03 0.07 0.02 0.1

Y 25 25 20 33.33 14.28 50 10

97

1 2 3 4 5 6 7

1 25

2 25

3 20

4 33.33

5 14.28

6 50

7 10

B

DC

G A

1

2

3

4

5

6

7

98

B

DC

G A

1

2

3

45

6

7

A

B

C

D

99

B

DC

G A

1

2

3

4 5

6

7

A

BC

D

A B C D

1 1

2 1

3 1

4 1

5 1 1 1

6 -1 1

7 -1 -1 -1

100

1 2 3 4 5 6 7

A 1 1 -1

B 1 -1 -1

C 1 1 1

D 1 1 -1

1 1 -1

1 -1 -1

1 1 1

1 1 -1

25

25

20

33.33

14.28

50

10

1

1

1

1

1 1 1

-1 1

-1 -1 -1

101

49.28 10 14.28 24.28

10 85 -50 10

14.28 -50 84.28 14.28

24.28 10 14.28 57.61

B

DC

G A

1

2

3

4

A B C D

1 1

2 -1

3 -1

4 -1 1

102

1 2 3 4

A -1

B -1

C 1 -1

D 1

-1

-1

1 -1

1

49.28 10 14.28 24.28

10 85 -50 10

14.28 -50 84.28 14.28

24.28 10 14.28 57.61

1

-1

-1

-1 1

103

85 -50 0 -10

-50 84.28 0 -14.28

0 0 58.33 -33.33

-10 -14.28 -33.33 57.61

104

1

0

4

32

REF.

1

2 3

6

105

ELEMENT NO.

1 2 3 4 5 6

Z 0.1 0.2 0.22 0.43 0.3 0.12

1 2 3 6 4 5

1 0.1

2 0.2

3 0.22

6 0.12

4 0.43

5 0.3

106

1

0

4

32

REF.

1

2 3

d

e

d e

1 1 1

2 1

3 1

6 -1 -1

4 1

5 1

1 2 3 6 4 5

d 1 -1 1

e 1 1 1 -1 1

6

107

1 -1 1

1 1 1 -1 1

0.1

0.2

0.22

0.12

0.43

0.3

1 1

1

1

-1 -1

1

1

0.65 0.22

0.22 0.94

1.67 -0.391

-0.391 1.1553

108

SELF MUTUAL

ELEMENT NO. Z ELEMENT NO.

Z

1 0.6

2 0.5 1 0.1

3 0.5

4 0.4

5 0.2

2 3 4

1 -1

2 -1

3 1 -1

4 -1

5 1 -1

3

21

4

1

2

3

4

5

REF.

A B C

1 1

2 1

3 1

4 1

5 -1 1 1

1 2 3 4 5

1 0.6 0.1

2 0.1 0.5

3 0.5

4 0.4

5 0.2

Y1 0

0 Y4

112

Y BUS = A T [Y] A

9.22 -0.34 -5

-0.34 4.07 -2

-5 -2 7

1 2 3 4 5

1 1.724 -0.344

2 -0.344 2.07

3 2

4 2.5

5 5

113

Y BR = K [Y BUS] K T

Z LOOP = C T [Z] C