CHAPTER 3

NETWORK ADMITTANCE AND IMPEDANCE MATRICES

As we have seen in Chapter 1 that a power system network can be converted into an equivalent impedance diagram. This diagram forms the basis of power flow (or load flow) studies and short circuit analysis. In this chapter we shall discuss the formation of bus admittance matrix (also known as Ybus matrix) and bus impedance matrix (also known as Zbus matrix). These two matrices are related by

1−= busbus YZ (3.1) We shall discuss the formation of the Ybus matrix first. This will be followed by the discussion of the formation of the Zbus matrix.

3.1 FORMATION OF BUS ADMITTANCE MATRIX

Consider the voltage source VS with a source (series) impedance of ZS as shown in Fig. 3.1 (a). Using Norton’s theorem this circuit can be replaced by a current source IS with a parallel admittance of YS as shown in Fig. 3.1 (b). The relations between the original system and the Norton equivalent are

SS

S

SS Z

YZV

I 1 and == (3.2)

We shall use this Norton’s theorem for the formulation of the Ybus matrix.

Fig. 3.1 (a) Voltage source with a source impedance and (b) its Norton equivalent.

For the time being we shall assume the short line approximation for the formulation of the bus admittance matrix. We shall thereafter relax this assumption and use the π-representation of the network for power flow studies. Consider the 4-bus power system shown in Fig. 3.2. This contains two generators G1 and G2 that are connected through transformers T1 and T2 to buses 1 and 2. Let us denote the synchronous reactances of G1 and G2 by XG1 and XG2 respectively and the leakage reactances of T1 and T2 by XT1 and XT2 respectively. Let Zij, i = 1, …, 4 and j = 1, …, 4 denote the line impedance between buses i and j.

Then the system impedance diagram is as shown in Fig. 3.3 where Z11 = j(XG1 + XT1) and Z22 = j(XG2 + XT2). In this figure the nodes with the node voltages of V1 to V4 indicate the

1.48

buses 1 to 4 respectively. Bus 0 indicates the reference node that is usually the neutral of the Y-connected system. The impedance diagram is converted into an equivalent admittance diagram shown in Fig. 3.4. In this diagram Yij = 1/Zij, i = 1, …, 4 and j = 1, …, 4. The voltage sources EG1 and EG2 are converted into the equivalent current sources I1 and I2 respectively using the Norton’s theorem discussed before.

Fig. 3.2 Single-line diagram of a simple power network.

Fig. 3.3 Impedance diagram of the power network of Fig. 3.2.

Fig. 3.4 Equivalent admittance diagram of the impedance of Fig. 3.3.

1.49

We would like to determine the voltage-current relationships of the network shown in Fig. 3.4. It is to be noted that this relation can be written in terms of the node (bus) voltages V1 to V4 and injected currents I1 and I2 as follows

=

4

3

2

1

2

1

00

VVVV

YII

bus (3.3)

or

=

002

1

4

3

2

1

II

Z

VVVV

bus (3.4)

It can be easily seen that we get (3.1) from (3.3) and (3.4).

Consider node (bus) 1 that is connected to the nodes 2 and 3. Then applying KCL at this node we get

( ) ( )( ) 3132121131211

311321121111

VYVYVYYYVVYVVYVYI

−−++=−+−+=

(3.5)

In a similar way application of KCL at nodes 2, 3 and 4 results in the following equations

( ) ( ) ( )( ) 424323224231222112

4224322312122222

VYVYVYYYYVYVVYVVYVVYVYI

−−++++−=−+−+−+=

(3.6)

( ) ( ) ( )

( ) 4343342313223113

4334232313130VYVYYYVYVY

VVYVVYVVY−+++−−=−+−+−=

(3.7)

( ) ( )

( ) 43424334224

343424240VYYVYVY

VVYVVY++−−=−+−=

(3.8)

Combining (3.5) to (3.8) we get

+−−−++−−−−+++−

−−++

=

4

3

2

1

34243424

343423132313

24232423122212

1312131211

2

1

0

0

00

VVVV

YYYYYYYYYYYYYYYYY

YYYYYII

(3.9)

Comparing (3.9) with (3.3) we can write

1.50

+−−−++−−−−+++−

−−++

=

34243424

343423132313

24232423122212

1312131211

0

0

YYYYYYYYYYYYYYYYY

YYYYY

Ybus (3.10)

In general the format of the Ybus matrix for an n-bus power system is as follows

−−−

−−−−−−

=

nnnnn

n

n

bus

YYYY

YYYYYYYY

Y

L

MOMMM

L

L

321

223212

113121

(3.11)

where

∑=

=n

jkjk YY

1

(3.12)

It is to be noted that Ybus is a symmetric matrix in which the sum of all the elements of the kth column is Ykk.

Example 3.1: Consider the impedance diagram of Fig. 3.2 in which the system parameters are given in per unit by

Z11 = Z22 = j0.25, Z12 = j0.2, Z13 = j0.25, Z23 = Z34 = j0.4 and Z24 = j0.5 The system admittance can then be written in per unit as

Y11 = Y22 = − j4, Y12 = − j5, Y13 = − j4, Y23 = Y34 = − j2.5 and Y24 = − j2 The Ybus is then given from (3.10) as

−−

−−

=

5.45.2205.295.24

25.25.13504513

jYbus per unit

Consequently the bus impedance matrix is given by

=

3926019740136701133.01974025650123601264013670123601531009690

0.1133126400969015310

...

....

.......

jZbus per unit

It can be seen that like the Ybus matrix the Zbus matrix is also symmetric.

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Let us now assume that the voltages EG1 and EG2 are given by

pu 01 andpu 301 21 °∠=°∠= GG EE The current sources I1 and I2 are then given by

pu 0949025.0

01

pu 6049025.0

301

22

22

11

11

°−∠=°∠

°∠==

°−∠=°∠

°∠==

ZEI

ZEI

G

G

We then get the node voltages from (3.4) as

°−∠°−∠

=

00

904604

3926019740136701133.01974025650123601264013670j123601531009690

j0.1133126400969015310

4

3

2

1

.j.j.jj

.j.j.j.j..j.j.

.j.j.j

VVVV

Solving the above equation we get the node voltages as

°∠°∠°∠°∠

=

56.139662.018.159659.0

11.550.967745.180.9677

4

3

2

1

VVVV

per unit

∆∆∆ 3.1.1 Node Elimination by Matrix Partitioning

Sometimes it is desirable to reduce the network by eliminating the nodes in which the current do not enter or leave. Let (3.3) be written as

=

x

AT

x

A

VV

MLLK

II

(3.13)

In the above equation IA is a vector containing the currents that are injected, Ix is a null vector and the Ybus matrix is portioned with the matrices K, L and M. Note that the Ybus matrix contains both L and LT due to its symmetric nature.

We get the following two sets of equations from (3.13)

xAA LVKVI += (3.14)

AT

xxAT

x VLMVMVVLI 10 −−=⇒+== (3.15) Substituting (3.15) in (3.14) we get

1.52

( ) AT

A VLLMKI 1−−= (3.16) Therefore we obtain the following reduced bus admittance matrix

Treducedbus LLMKY 1−−= (3.17)

Example 3.2: Let us consider the system of Example 3.1. Since there is no current

injection in either bus 3 or bus 4, from the Ybus computed we can write

−

−=

=

−

−=

5.45.25.29

and25.2

04'

5.135513

jjjj

Mjj

jL

jjjj

K

We then have

−

−=−= −

8978.108978.68978.68978.101

jjjj

LLMKY Treducedbus per unit

Substituting I1 = 4∠−60° per unit and I2 = 4∠−90° per unit we shall get the same values of V1 and V2 as given in Example 3.1.

Inspecting the reduced Ybus matrix we can state that the admittance between buses 1 and 2 is −j6.8978. Therefore the self admittance (the admittance that is connected in shunt) of the buses 1 and 2 is −j4 per unit (= −j10.8978 + j6.8978). The reduced admittance diagram obtained by eliminating nodes 3 and 4 is shown in Fig. 3.5. It is to be noted that the impedance between buses 1 and 2 is the Thevenin impedance between these two buses. The value of this impedance is 1/(−j6.8978) = j0.145 per unit.

∆∆∆

Fig. 3.5 Reduced admittance diagram after the elimination of buses 3 and 4. 3.1.2 Node Elimination by Kron Reduction

Consider an equation of the form

bAx = (3.18) where A is an (n×n) real or complex valued matrix, x and b are vectors in either Rn or Cn. assume that the b vector has a zero element in the nth row such that (3.18) is given as

1.53

=

−−−−−

01

2

1

1

2

1

21

,12,11,1

22221

11211

n

n

n

nnnn

nnnn

n

n

b

bb

xx

xx

aaaaaa

aaaaaa

MM

L

L

MOMM

L

L

(3.19)

We can then eliminate the kth row and kth column to obtain a reduced (n − 1) number of equations of the form

=

−−−−−−

−

−

1

2

1

1

2

1

1,12,11,1

1,22221

1,11211

nn

new

nnnn

n

n

b

bb

x

xx

aaa

aaaaaa

MM

L

MOMM

L

L

(3.20)

The elimination is performed using the following elementary operations

nn

njknkj

newkj a

aaaa −= (3.21)

Example 3.3: Let us consider the same system of Example 3.1. We would like to

eliminate the last two rows and columns. Let us first eliminate the last row and last column. Some of the values are given below

6111.125.4225.13,5

44

42242222

44

41242121 jjj

YYYYYj

YYYYY newnew −=

×+−=−==−=

0,6111.35.4

5.225.244

4444444

44

43242323 =−==

×+=−=

YYYYYjjj

YYYYY newnew

In a similar way we can calculate the other elements. Finally eliminating the last row and last column, as all these elements are zero, we get the new Ybus matrix as

−−

−=

6111.76111.346111.36111.1254513

jjjjjj

jjjY new

bus

Further reducing the last row and the last column of the above matrix using (3.21), we obtain the reduced Ybus matrix given in Example 3.2.

∆∆∆ 3.1.3 Inclusion of Line Charging Capacitors

So far we have assumed that the transmission lines are modeled with lumped series impedances without the shunt capacitances. However in practice, the Ybus matrix contains the shunt admittances for load flow analysis in which the transmission lines are represented by its

1.54

π-equivalent. Note that whether the line is assumed to be of medium length or long length is irrelevant as we have seen in Chapter 2 how both of them can be represented in a π-equivalent.

Consider now the power system of Fig. 3.2. Let us assume that all the lines are represented in an equivalent-π with the shunt admittance between the line i and j being denoted by Ychij. Then the equivalent admittance at the two end of this line will be Ychij/2. For example the shunt capacitance at the two ends of the line joining buses 1 and 3 will be Ych13/2. We can then modify the admittance diagram Fig. 3.4 as shown in Fig. 3.6. The Ybus matrix of (3.10) is then modified as

++−−−+++−−−−++++−

−+++

=

434243424

3433423132313

242322423122212

13121131211

0

0

ch

ch

ch

ch

bus

YYYYYYYYYYYYYYYYYYYY

YYYYYY

Y

(3.22) where

2

2

2

2

34244

3423133

2423122

13121

chchch

chchchch

chchchch

chchch

YYY

YYYY

YYYY

YYY

+=

++=

++=

+=

(3.23)

Fig. 3.6 Admittance diagram of the power system Fig. 3.2 with line charging capacitors.

1.55

3.2 ELEMENTS OF THE BUS IMPEDANCE AND ADMITTANCE MATRICES

Equation (3.1) indicates that the bus impedance and admittance matrices are inverses of each other. Also since Ybus is a symmetric matrix, Zbus is also a symmetric matrix. Consider a 4-bus system for which the voltage-current relations are given in terms of the Ybus matrix as

=

4

3

2

1

44434241

34333231

24232221

14131211

4

3

2

1

VVVV

YYYYYYYYYYYYYYYY

IIII

(3.24)

We can then write

01

111

432 ===

=VVVV

IY (3.25)

This implies that Y11 is the admittance measured at bus-1 when buses 2, 3 and 4 are short circuited. The admittance Y11 is defined as the self admittance at bus-1. In a similar way the self admittances of buses 2, 3 and 4 can also be defined that are the diagonal elements of the Ybus matrix. The off diagonal elements are denoted as the mutual admittances. For example the mutual admittance between buses 1 and 2 is defined as

02

112

431 ===

=VVVV

IY (3.26)

The mutual admittance Y12 is obtained as the ratio of the current injected in bus-1 to the voltage of bus-2 when buses 1, 3 and 4 are short circuited. This is obtained by applying a voltage at bus-2 while shorting the other three buses.

The voltage-current relation can be written in terms of the Zbus matrix as

=

4

3

2

1

44434241

34333231

24232221

14131211

4

3

2

1

IIII

ZZZZZZZZZZZZZZZZ

VVVV

(3.27)

The driving point impedance at bus-1 is then defined as

01

111

432 ===

=IIII

VZ (3.28)

i.e., the driving point impedance is obtained by injecting a current at bus-1 while keeping buses 2, 3 and 4 open-circuited. Comparing (3.26) and (3.28) we can conclude that Z11 is not the reciprocal of Y11. The transfer impedance between buses 1 and 2 can be obtained by injecting a current at bus-2 while open-circuiting buses 1, 3 and 4 as

1.56

02

112

431 ===

=IIII

VZ (3.29)

It can also be seen that Z12 is not the reciprocal of Y12.

3.3 MODIFICATION OF BUS IMPEDANCE MATRIX

Equation (3.1) gives the relation between the bus impedance and admittance matrices. However it may be possible that the topology of the power system changes by the inclusion of a new bus or line. In that case it is not necessary to recompute the Ybus matrix again for the formation of Zbus matrix. We shall discuss four possible cases by which an existing bus impedance matrix can be modified.

Let us assume that an n-bus power system exists in which the voltage-current relations are given in terms of the bus impedance matrix as

=

n

orig

n I

IZ

V

VMM11

(3.30)

The aim is to modify the matrix Zorig when a new bus or line is connected to the power system. 3.3.1 Adding a New Bus to the Reference Bus

It is assumed that a new bus p (p > n) is added to the reference bus through an impedance Zp. The schematic diagram for this case is shown in Fig. 3.7. Since this bus is only connected to the reference bus, the voltage-current relations the new system are

=

=

p

nnew

p

n

p

orig

p

n

II

I

Z

II

I

Z

Z

VV

VMM

L

MM111

000

0

(3.31)

Fig. 3.7 A new bus is added to the reference bus.

1.57

3.3.2 Adding a New Bus to an Existing Bus through an Impedance

This is the case when a bus, which has not been a part of the original network, is added to an existing bus through a transmission line with an impedance of Zb. Let us assume that p (p > n) is the new bus that is connected to bus k (k < n) through Zb. Then the schematic diagram of the circuit is as shown in Fig. 3.8. Note from this figure that the current Ip flowing from bus p will alter the voltage of the bus k. We shall then have

( ) nknpkkkkkk IZIIZIZIZV ++++++= LL2211 (3.32) In a similar way the current Ip will also alter the voltages of all the other buses as

( ) kiIZIIZIZIZV ninpkikiii ≠++++++= ,2211 LL (3.33) Furthermore the voltage of the bus p is given by

( ) pbkknknkkkkk

pbkp

IZZIZIZIZIZ

IZVV

+++++++=

+=

LL2211

(3.34)

Therefore the new voltage current relations are

=

+

=

p

nnew

p

n

bkkknk

kn

orig

k

p

n

II

I

Z

II

I

ZZZZZ

ZZ

VV

VMM

L

MM11

1

11

(3.35)

It can be noticed that the new Zbus matrix is also symmetric.

Fig. 3.8 A new bus is added to an existing bus through an impedance. 3.3.3 Adding an Impedance to the Reference Bus from an Existing Bus

To accomplish this we first assume that an impedance Zb is added from a new bus p to an existing bus k. This can be accomplished using the method discussed in Section 3.3.2. Then to add this bus k to the reference bus through Zb, we set the voltage Vp of the new bus to zero. However now we have an (n + 1) × (n + 1) Zbus matrix instead of an n × n matrix. We can then remove the last row and last column of the new Zbus matrix using the Kron’s reduction given in (3.21).

1.58

3.3.4 Adding an Impedance between two Existing Buses

Let us assume that we add an impedance Zb between two existing buses k and j as shown in Fig. 3.9. Therefore the current injected into the network from the bus k side will be Ik − Ib instead of Ik. Similarly the current injected into the network from the bus j side will be Ij + Ib instead of Ij. Consequently the voltage of the ith bus will be

( ) ( )( ) bikijninkikjijii

ninbkikbjijiii

IZZIZIZIZIZIZ

IZIIZIIZIZIZV

−+++++++=

++−+++++=

LL

LL

2211

2211 (3.36)

Similarly we have

( ) bjkjjnjnkjkjjjjjj IZZIZIZIZIZIZV −+++++++= LL2211 (3.37) and

( ) bkkkjnknkkkjkjkkk IZZIZIZIZIZIZV −+++++++= LL2211 (3.38)

Fig. 3.9 An impedance is added between two existing buses.

We shall now have to eliminate Ib from the above equations. To do that we note from Fig. 3.9 that

jkbbbbjk VVIZIZVV +−=⇒=− 0 (3.39) Substituting (3.37) and (3.38) in (3.39) we get

( ) ( ) ( )( ) ( ) bkkjkjjnknjn

kkkjkjkjjjkjbb

IZZZIZZ

IZZIZZIZZIZ

+−+−+

+−+−++−+=

20 111 LL

(3.40)

We can then write the voltage current relations as

=

−−−

−

=

p

nnew

p

n

bbknjnkj

knjn

orig

kj

n

II

I

Z

II

I

ZZZZZZZ

ZZZ

V

VMM

L

MM

11

11

111

0

(3.41)

where

1.59

kkjkjjbbb ZZZZZ +−+= 2 (3.42) We can now eliminate the last row and last column using the Kron’s reduction given in (3.21). 3.3.5 Direct Determination of Zbus Matrix

We shall now use the methods given in Sections 3.3.1 to 3.3.4 for the direct determination of the Zbus matrix without forming the Ybus matrix first. To accomplish this we shall consider the system of Fig. 3.2 and shall use the system data given in Example 3.1. Note that for the construction of the Zbus matrix we first eliminate all the voltage sources from the system.

Step-1: Start with bus-1. Assume that no other buses or lines exist in the system. We add this bus to the reference bus with the impedance of j0.25 per unit. Then the Zbus matrix is

25.01, jZbus = (3.43)

Step-2: We now add bus-2 to the reference bus using (3.31). The system impedance diagram is shown in Fig. 3.10. We then can modify (3.43) as

=

25.00025.0

2, jj

Zbus (3.44)

Fig. 3.10 Network of step-2.

Step-3: We now add an impedance of j0.2 per unit between buses 1 and 2 as shown in Fig. 3.11. The interim Zbus matrix is then obtained by applying (3.41) on (3.44) as

−−=

7.025.025.025.025.00

25.0025.0

3,

jjjjj

jjZ in

bus

Eliminating the last row and last column using the Kron’s reduction of (3.31) we get

=

1607.00893.00893.01607.0

3, jjjj

Zbus (3.45)

Step-4: We now add bus-3 to bus-1 through an impedance of j0.25 per unit as shown

in Fig. 3.12. The application of (3.35) on (3.45) will then result in the following matrix

1.60

Fig. 3.11 Network of step-3.

=

4107.00893.01607.00893.01607.00893.01607.00893.01607.0

4,

jjjjjjjjj

Zbus (3.46)

Fig. 3.12 Network of step-4.

Step-5: Connect buses 2 and 3 through an impedance of j0.4 per unit as shown in Fig. 3.13. The interim Zbus matrix is then formed from (3.41) and (3.46) as

−−−

−

=

7928.03214.00714.00714.03214.04107.00893.01607.0

0714.00893.01607.00893.00714.01607.00893.01607.0

5,

jjjjjjjj

jjjjjjjj

Z inbus

Fig. 3.13 Network of step-5. Using the Kron’s reduction we get the following matrix

=

0.28040.11820.13180.11821543.00.09570.13180.09571543.0

5,

jjjjjjjjj

Zbus (3.47)

1.61

Step-6: We now add a new bus-4 to bus-2 through an impedance of j0.5 as shown in Fig. 3.14. Then the application of (3.35) on (3.47) results in the following matrix

=

0.65430.11821543.00.09570.11820.28040.11820.1318

1543.00.11821543.00.09570.09570.13180.09571543.0

6,

jjjjjjjjjjjjjjjj

Zbus (3.48)

Fig. 3.14 Network of step-6.

Step-7: Finally we add buses 3 and 4 through an impedance of j0.4 to obtain the network of Fig. 3.3 minus the voltage sources. The application of (3.41) on (3.48) results in the interim Zbus matrix of

−

−=

09821536001622.00360003600536000.65430.11821543.00.09571622.00.11820.28040.11820.1318036001543.00.11821543.00.0957

036000.09570.13180.09571543.0

7,

.j.jj.j.j.jjjjj

jjjjj.jjjjj

.jjjjj

Z inbus

Eliminating the 5th row and column through Kron’s reduction we get the final Zbus as

=

3926019740136701133.01974025650123601264013670j123601531009690

j0.1133126400969015310

7,

.j.j.jj

.j.j.j.j..j.j.

.j.j.j

Zbus (3.49)

The Zbus matrix given in (3.49) is the as that given in Example 3.1 which is obtained by inverting the Ybus matrix.

3.4 THEVENIN IMPEDANCE AND Zbus MATRIX

To establish relationships between the elements of the Zbus matrix and Thevenin equivalent, let us consider the following example.

Example 3.4: Consider the two bus power system shown in Fig. 3.15. It can be seen that the open-circuit voltages of buses a and b are Va and Vb respectively. From (3.11) we can write the Ybus matrix of the system as

1.62

Fig. 3.15 Two-bus power system of Example 3.4.

+−

−+

=

+−

−+=

bbab

bbab

ab

ababaa

abaa

bbabab

ababaabus

ZZZZ

Z

ZZZZZ

ZZZ

ZZZY 1

1

111

111

The determinant of the above matrix is

bbabaa

bbabaabus ZZZ

ZZZY ++=

Therefore the Zbus matrix is

+

+

== −

abaa

abaa

ab

abbbab

bbab

busbusbus

ZZZZ

Z

ZZZZZ

YYZ 1

111

Solving the last two equations we get

( )

( )

+++

++

+++++

=

=

bbabaa

abaabb

bbabaa

bbaa

bbabaa

bbaa

bbabaa

bbabaa

bus

ZZZZZZ

ZZZZZ

ZZZZZ

ZZZZZZ

ZZZZ

Z2212

1211 (3.50)

Now consider the system of Fig. 3.15. The Thevenin impedance of looking into the

system at bus-a is the parallel combination of Zaa and Zab + Zbb, i.e.,

( )11, Z

ZZZZZZZ

bbabaa

bbabaaath =

+++

= (3.51)

Similarly the Thevenin impedance obtained by looking into the system at bus-b is the parallel combination of Zbb and Zaa + Zab, i.e.,

( )22, Z

ZZZZZZZ

bbabaa

abaabbbth =

+++

= (3.52)

Hence the driving point impedances of the two buses are their Thevenin impedances.

1.63

Let us now consider the Thevenin impedance while looking at the system between the buses a and b. From Fig. 3.15 it is evident that this Thevenin impedance is the parallel combination of Zab and Zaa + Zbb, i.e.,

( )bbabaa

bbaaababth ZZZ

ZZZZ+++

=,

With the values given in (3.50) we can write

( ) ( )

[ ]bbababaabbabaa

bbabaa

bbaa

bbabaa

abaabb

bbabaa

bbabaa

ZZZZZZZ

ZZZZZ

ZZZZZZ

ZZZZZZZZZ

+++

=

++−

+++

++++

=−+

1

22 122211

Comparing the last two equations we can write

122211, 2ZZZZ abth −+= (3.53) ∆∆∆

As we have seen in the above example in the relation V = ZbusI, the node or bus

voltages Vi, i = 1, …, n are the open circuit voltages. Let us assume that the currents injected in buses 1, …, k − 1 and k + 1, …, n are zero when a short circuit occurs at bus k. Then Thevenin impedance at bus k is

kkk

kkth Z

IVZ ==, (3.54)

From (3.51), (3.52) and (3.54) we can surmise that the driving point impedance at each bus is the Thevenin impedance.

Let us now find the Thevenin impedance between two buses j and k of a power system. Let the open circuit voltages be defined by the voltage vector V° and corresponding currents be defined by I° such that

°=° IZV bus (3.55) Now suppose the currents are changed by ∆I such that the voltages are changed by ∆V. Then

( )IIZVVV bus ∆+°=∆+°= (3.56) Comparing (3.55) and (3.56) we can write

IZV bus∆=∆ (3.57)

1.64

Let us now assume that additional currents ∆Ik and ∆Ij are injected at the buses k and j respectively while the currents injected at the other buses remain the same. Then from (3.57) we can write

∆+∆

∆+∆∆+∆

∆+∆

=

∆∆

=∆

knkjnj

kkkjkj

kjkjjj

kkjj

k

jbus

IZIZ

IZIZIZIZ

IZIZ

II

ZV

M

M

M

M

11

0

0

(3.58)

We can therefore write the following two equations form (3.58)

kjkjjjojj

ojj IZIZVVVV ∆+∆+=∆+=

kkkjkj

okk

okk IZIZVVVV ∆+∆+=∆+=

The above two equations can be rewritten as

( ) ( )kjjkjjkjjojj IIZIZZVV ∆+∆+∆−+= (3.59)

( ) ( ) kkjkkkjkj

okk IZZIIZVV ∆−+∆+∆+= (3.60)

Since Zjk = Zkj, the network can be drawn as shown in Fig. 3.16. By inspection we can

see that the open circuit voltage between the buses k and j is

0, j

okkjoc VVV −= (3.61)

and the short circuit current through these two buses is

kjkjsc III ∆−=∆=, (3.62) Also during the short circuit Vk − Vj = 0. Therefore combining (3.59) and (3.60) we get

( ) ( ) 02 , =−−+−=− kjsckkjjkjoj

okjk IZZZVVVV (3.63)

Combining (3.61) to (3.63) we find the Thevenin impedance between the buses k and j as

kjkkjjkjsc

kjockjth ZZZ

IV

Z 2,

,, −+== (3.64)

The above equation agrees with our earlier derivation of the two bus network given in (3.53).

1.65

Fig. 3.16 Thevenin equivalent between buses k and j.