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
Z bus = Ybus
(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
IS =
VS
1
and YS =
ZS
ZS
(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 πrepresentation 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
V1 
 I1 
 
I 
 2  = Ybus V2 
V3 
0
 
 
0
V4 
(3.3)
V1 
 I1 
V 
 
 2  = Z bus  I 2 
V3 
0
 
 
0
V4 
(3.4)
or
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
I1 = Y11V1 + Y12 (V1 − V2 ) + Y13 (V1 − V3 )
= (Y11 + Y12 + Y13 )V1 − Y12V2 − Y13V3
(3.5)
In a similar way application of KCL at nodes 2, 3 and 4 results in the following equations
I 2 = Y22V2 + Y12 (V 2 − V1 ) + Y23 (V2 − V3 ) + Y24 (V 2 − V4 )
= −Y12V1 + (Y22 + Y12 + Y23 + Y24 )V2 − Y23V3 − Y24V 4
0 = Y13 (V3 − V1 ) + Y23 (V3 − V2 ) + Y34 (V3 − V4 )
= −Y13V1 − Y23V2 + (Y13 + Y23 + Y34 )V3 − Y34V4
0 = Y24 (V4 − V2 ) + Y34 (V4 − V3 )
= −Y24V2 − Y34V3 + (Y24 + Y34 )V4
(3.6)
(3.7)
(3.8)
Combining (3.5) to (3.8) we get
0  V1 
− Y12
− Y13
 I1  Y11 + Y12 + Y13
I  
Y22 + Y12 + Y23 + Y24
− Y12
− Y23
− Y24  V2 
 2 = 
0 
Y13 + Y23 + Y34
− Y13
− Y23
− Y34  V3 
 
  
0
Y24 + Y34  V4 
− Y24
− Y34
0 
Comparing (3.9) with (3.3) we can write
(3.9)
1.50
Ybus
0 
− Y12
− Y13
Y11 + Y12 + Y13

Y22 + Y12 + Y23 + Y24
− Y12
− Y23
− Y24 

=

Y13 + Y23 + Y34
− Y13
− Y23
− Y34 


0
Y24 + Y34 
− Y24
− Y34

(3.10)
In general the format of the Ybus matrix for an n-bus power system is as follows
Ybus
 Y1
− Y
=  12
 M

− Y1n
− Y12
Y2
M
− Y2 n
− Y13 L − Y1n 
− Y23 L − Y2 n 
M
O
M 

− Y3n L Ynn 
(3.11)
where
n
Yk = ∑ Ykj
(3.12)
j =1
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
Ybus
5
4
0 
− 13
 5 − 13.5 2.5
2 
per unit
= j
 4
− 9 2.5 
2.5


2
2.5 − 4.5
 0
Consequently the bus impedance matrix is given by
Z bus
 0.1531
 0.0969
= j
 0.1264

0.1133
0.0969 0.1264 0.1133
0.1531 0.1236 0.1367 
per unit
0.1236 0.2565 0.1974 

0.1367 0.1974 0.3926 
It can be seen that like the Ybus matrix the Zbus matrix is also symmetric.
1.51
Let us now assume that the voltages EG1 and EG2 are given by
EG1 = 1∠30° pu and EG 2 = 1∠0° pu
The current sources I1 and I2 are then given by
I1 =
1∠30°
EG1
=
= 4∠ − 60° pu
Z11 0.25∠90°
I2 =
1∠0°
EG 2
= 4∠ − 90° pu
=
Z 22 0.25∠90°
We then get the node voltages from (3.4) as
V1   j 0.1531
V   0.0969
 2 = 
V3   j 0.1264
  
V4   j 0.1133
j 0.0969
j 0.1531
j 0.1236
j 0.1367
j 0.1264
j 0.1236
j 0.2565
j 0.1974
j0.1133 4∠ − 60°
j0.1367  4∠ − 90°

0
j 0.1974  


0
j 0.3926 

Solving the above equation we get the node voltages as
V1  0.9677∠18.45°
V  0.9677∠11.55°
 2 = 
 per unit
V3  0.9659∠15.18°
  

V4  0.9662∠13.56°
∆∆∆
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
I A   K
 I  =  LT
 x 
L  VA 
M  Vx 
(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)
I A = KVA + LVx
(3.14)
−1 T
0 = I x = L VA + MVx ⇒ Vx = − M L VA
T
Substituting (3.15) in (3.14) we get
(3.15)
1.52
(
)
I A = K − LM −1LT VA
(3.16)
Therefore we obtain the following reduced bus admittance matrix
reduced
Ybus
= K − LM −1LT
(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
j5 
− j13
 j4
' L=
K =

 j 5 − j13.5
 j 2.5
0
 − j 9 j 2.5 
and M = 


j 2
 j 2.5 − j 4.5
We then have
j 6.8978 
− j10.8978
reduced
= K − LM −1LT = 
Ybus
 per unit
 j 6.8978 − j10.8978
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
Ax = b
(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
L a1n   x1   b1 
L a2 n   x2   b2 
O
M  M  =  M 
  

L an −1, n   xn −1  bn −1 
L ann   xn   0 
a12
 a11
 a
a22
 21
 M
M

an −1,1 an −1, 2
 an1
an 2
(3.19)
We can then eliminate the kth row and kth column to obtain a reduced (n − 1) number of
equations of the form
 a11
 a
 21
 M

an −1,1
L a1, n −1 
L a2, n −1 
O
M 

L an −1, n −1 
a12
a22
M
an −1, 2
new
 x1   b1 
x  b 
 2 = 2 
 M   M 

  
 xn −1  bn −1 
(3.20)
The elimination is performed using the following elementary operations
akjnew = akj −
akn anj
(3.21)
ann
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
2× 2
Y24Y41
Y Y
= j 5, Y22new = Y22 − 24 42 = − j13.5 + j
= − j12.6111
4.5
Y44
Y44
Y Y
2 × 2.5
Y Y
= Y23 − 24 43 = j 2.5 + j
= j3.6111, Y4new = Y44 − 44 44 = 0
Y44
4.5
Y44
Y21new = Y21 −
Y23new
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
new
bus
Y
j5
j4 
− j13

=  j 5 − j12.6111 j 3.6111 
 j 4
j 3.6111 − j 7.6111
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 πequivalent.
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
Ybus
Y13
0
− Y12

Y11 + Y12 + Y13 + Ych1


Y22 + Y12 + Y23 + Y24 + Ych 2
− Y12
− Y23
− Y24

=


Y13 + Y23 + Y34 + Ych 3
− Y13
− Y23
− Y34


0
Y24 + Y34 + Ych 4 
− Y24
− Y34

(3.22)
where
Ych12 + Ych13
2
Ych12 + Ych 23 + Ych 24
Ych 2 =
2
Y +Y +Y
Ych3 = ch13 ch 23 ch 34
2
Ych 24 + Ych34
Ych 4 =
2
Ych1 =
Fig. 3.6 Admittance diagram of the power system Fig. 3.2 with line charging capacitors.
(3.23)
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
 I1  Y11
 I  Y
 2  =  21
 I 3  Y31
  
 I 4  Y41
Y12 Y13
Y22 Y23
Y32 Y33
Y42 Y43
Y14  V1 
Y24  V2 
Y34  V3 
 
Y44  V4 
(3.24)
We can then write
Y11 =
I1
V1 V
(3.25)
2 =V3 =V 4 = 0
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
Y12 =
I1
V2 V =V
1
(3.26)
3 =V 4
=0
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
V1   Z11
V   Z
 2  =  21
V3   Z 31
  
V4   Z 41
Z12
Z 22
Z 32
Z 42
Z13
Z 23
Z 33
Z 43
Z14   I1 
Z 24   I 2 
Z 34   I 3 
 
Z 44   I 4 
(3.27)
The driving point impedance at bus-1 is then defined as
Z11 =
V1
I1
(3.28)
I 2 = I3 = I 4 =0
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
Z12 =
V1
I2
(3.29)
I1 = I 3 = I 4 = 0
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
V1 
 I1 
M=Z M
orig  
 
Vn 
 I n 
(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
 V1  
M  Z
orig
 =
Vn  
  
V p  0 L
0   I1 
 I1 



M
M  M 
= Z new  
 In 
0  In 
 
 
0 Z p   I p 
 I p 
Fig. 3.7 A new bus is added to the reference bus.
(3.31)
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
Vk = Z k1I1 + Z k 2 I 2 + L + Z kk (I k + I p ) + L + Z kn I n
(3.32)
In a similar way the current Ip will also alter the voltages of all the other buses as
Vi = Z i1I1 + Z i 2 I 2 + L + Z ik (I k + I p ) + L + Z in I n , i ≠ k
(3.33)
Furthermore the voltage of the bus p is given by
V p = Vk + Z b I p
= Z k1I1 + Z k 2 I 2 + L + Z kk I k + L + Z kn I n + (Z kk + Z b )I p
(3.34)
Therefore the new voltage current relations are
 V1  
M 
 =
Vn  
  
V p   Z k1
Z orig
L
Z kn
Z k 1   I1 
 I1 
M
M

M  
= Z new  
 In 
Z kn   I n 
 
 
Z kk + Z b   I p 
 I p 
(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
Vi = Z i1I1 + Z i 2 I 2 + L + Z ij (I j + I b ) + Z ik (I k − I b ) + L + Z in I n
= Z i1I1 + Z i 2 I 2 + L + Z ij I j + Z ik I k + L + Z in I n + (Z ij − Z ik )I b
(3.36)
Similarly we have
V j = Z j1I1 + Z j 2 I 2 + L + Z jj I j + Z jk I k + L + Z jn I n + (Z jj − Z jk )I b
(3.37)
Vk = Z k1I1 + Z k 2 I 2 + L + Z kj I j + Z kk I k + L + Z kn I n + (Z kj − Z kk )I b
(3.38)
and
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
Vk − V j = Z b I b ⇒ 0 = Z b I b − Vk + V j
(3.39)
Substituting (3.37) and (3.38) in (3.39) we get
0 = Z b I b + (Z j1 − Z k1 )I1 + L + (Z jj − Z kj )I j + (Z jk − Z kk )I k + L
+ (Z jn − Z kn )I n + (Z jj − 2 Z jk + Z kk )I b
(3.40)
We can then write the voltage current relations as
V1  
M 
 =
Vn  
  
 0   Z j1 − Z k1
where
Z orig
L
Z jn − Z kn
Z j 1 − Z k 1   I1 
 I1 



 
M
 M  = Z  M 
new
 In 
Z jn − Z kn   I n 
 
 
Z bb   I p 
 I p 
(3.41)
1.59
Z bb = Z b + Z jj − 2Z jk + Z kk
(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
Z bus ,1 = j 0.25
(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
 j 0.25
Z bus , 2 = 
 0
0 
j 0.25
(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
Z
in
bus , 3
0
j 0.25 
 j 0.25

j 0.25 − j 0.25
= 0
 j 0.25 − j 0.25
j 0.7 
Eliminating the last row and last column using the Kron’s reduction of (3.31) we get
 j 0.1607
Z bus ,3 = 
 j 0.0893
j 0.0893
j 0.1607
(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.
Z bus , 4
 j 0.1607
=  j 0.0893
 j 0.1607
j 0.0893
j 0.1607
j 0.0893
j 0.1607 
j 0.0893
j 0.4107
(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
in
Z bus
,5
 j 0.1607
 j 0.0893
=
 j 0.1607

− j 0.0714
− j 0.0714
j 0.0714 
j 0.0893 j 0.4107 − j 0.3214

j 0.0714 − j 0.3214
j 0.7928 
j 0.0893
j 0.1607
j 0.1607
j 0.0893
Fig. 3.13 Network of step-5.
Using the Kron’s reduction we get the following matrix
Z bus ,5
 j 0.1543
=  j 0.0957
 j 0.1318
j 0.0957
j 0.1543
j 0.1182
j 0.1318
j 0.1182
j 0.2804
(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
Z bus , 6
 j 0.1543
 j 0.0957
=
 j 0.1318

 j 0.0957
j 0.0957
j 0.1543
j 0.1318
j 0.1182
j 0.1182
j 0.1543
j 0.2804
j 0.1182
j 0.0957
j 0.1543
j 0.1182

j 0.6543
(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
in
Z bus
,7
 j 0.1543 j 0.0957
 j 0.0957 j 0.1543

=  j 0.1318 j 0.1182

 j 0.0957 j 0.1543
 j 0.0360 − j 0.0360
j 0.1318
j 0.1182
j 0.2804
j 0.1182
j 0.1622
j 0.0957
j 0.0360 
j 0.1543 − j 0.0360
j 0.1182 j 0.1622 

j 0.6543
j 0.5360 
j 0.5360
j1.0982 
Eliminating the 5th row and column through Kron’s reduction we get the final Zbus as
Z bus , 7
 j 0.1531
 0.0969
= 
 j 0.1264

 j 0.1133
j 0.0969
j 0.1531
j 0.1264
j 0.1236
j 0.1236
j 0.1367
j 0.2565
j 0.1974
j0.1133
j0.1367 
j 0.1974 

j 0.3926
(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.
Ybus
  Z aa + Z ab
  Z Z
 =  aa ab
1
1  
1
−
+
Z ab Z bb  
Z ab
1
 1
Z + Z
ab
=  aa
1
 −

Z ab
−
1
Z ab
1 
Z ab 

Z ab + Z bb 
Z ab Z bb 
−
The determinant of the above matrix is
Ybus =
Z aa + Z ab + Z bb
Z aa Z ab Z bb
Therefore the Zbus matrix is
−1
Z bus = Ybus
 Z ab + Z bb
1  Z ab Z bb
=

1
Ybus 
 Z ab
1

Z ab 

Z aa + Z ab 
Z aa Z ab 
Solving the last two equations we get
Z bus
Z
=  11
 Z12
 Z aa (Z ab + Z bb )
Z12   Z aa + Z ab + Z bb
=
Z aa Z bb
Z 22  
 Z aa + Z ab + Z bb
Z aa Z bb

Z aa + Z ab + Z bb 

Z bb (Z aa + Z ab ) 
Z aa + Z ab + Z bb 
(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.,
Z th , a =
Z aa (Z ab + Z bb )
= Z11
Z aa + Z ab + Z bb
(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.,
Z th ,b =
Z bb (Z aa + Z ab )
= Z 22
Z aa + Z ab + Z bb
Hence the driving point impedances of the two buses are their Thevenin impedances.
(3.52)
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.,
Z th , ab =
Z ab (Z aa + Z bb )
Z aa + Z ab + Z bb
With the values given in (3.50) we can write
Z11 + Z 22 − 2Z12 =
=
Z aa (Z ab + Z bb ) Z bb (Z aa + Z ab )
Z aa Z bb
−2
+
Z aa + Z ab + Z bb Z aa + Z ab + Z bb
Z aa + Z ab + Z bb
1
[Z aa Z ab + Z ab Zbb ]
Z aa + Z ab + Z bb
Comparing the last two equations we can write
Z th , ab = Z11 + Z 22 − 2Z12
(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
Z th , k =
Vk
= Z kk
Ik
(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
V ° = Z bus I °
(3.55)
Now suppose the currents are changed by ∆I such that the voltages are changed by ∆V. Then
V = V ° + ∆V = Z bus (I ° + ∆I )
(3.56)
Comparing (3.55) and (3.56) we can write
∆V = Z bus ∆I
(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
 0   Z1 j ∆I j + Z1k ∆I k 

 M  
M

  
 ∆I j   Z jj ∆I j + Z jk ∆I k 
∆V = Z bus   = 

∆I k   Z kj ∆I j + Z kk ∆I k 

 M  
M

  
 0   Z nj ∆I j + Z nk ∆I k 
(3.58)
We can therefore write the following two equations form (3.58)
V j = V jo + ∆V j = V jo + Z jj ∆I j + Z jk ∆I k
Vk = Vko + ∆Vk = Vko + Z kj ∆I j + Z kk ∆I k
The above two equations can be rewritten as
V j = V jo + (Z jj − Z jk )∆I j + Z jk (∆I j + ∆I k )
(3.59)
Vk = Vko + Z kj (∆I j + ∆I k ) + (Z kk − Z kj )∆I k
(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
Voc , kj = Vko − V j0
(3.61)
and the short circuit current through these two buses is
I sc , kj = ∆I j = − ∆I k
(3.62)
Also during the short circuit Vk − Vj = 0. Therefore combining (3.59) and (3.60) we get
(
)
Vk − V j = Vko − V jo + (2Z kj − Z jj − Z kk )I sc , kj = 0
(3.63)
Combining (3.61) to (3.63) we find the Thevenin impedance between the buses k and j as
Z th , kj =
Voc , kj
= Z jj + Z kk − 2 Z kj
I sc , kj
(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.