electrical power system analysis 4. the impedance model and network calculations

Dr Houssem Rafik El Hana Bouchekara 1

Power System Analysis The Impedance Model And Network

Calculations

Dr : Houssem Rafik El- Hana

BOUCHEKARA 2011/2012 1432/1433

KINGDOM OF SAUDI ARABIA Ministry Of High Education

Umm Al-Qura University College of Engineering & Islamic Architecture

Department Of Electrical Engineering


1 THE IMPEDANCE MODEL AND NETWORK CALCULATIONS ............................................. 3

1.1 THE BUS ADMITTANCE AND IMPEDANCE MATRICES ............................................................... 3

1.2 THEVENIN'S THEOREM AND .................................................................................... 6

1.3 MODIFICATION OF AN EXISTING ........................................................................... 12

1.4 DIRECT DETERMINATION OF ................................................................................. 19

1.5 CALCULATION OF ELEMENTS FROM ............................................................. 23


1 THE IMPEDANCE MODEL AND NETWORK CALCULATIONS

The bus admittance matrix o f a large-scale interconnected power system is typically

very sparse with mainly zero elements. In previous chapter we saw how is constructed

branch by branch from primitive admittances. It is conceptually simple to invert to find

the bus impedance matrix but such direct inversion is rarely employed when the

systems to be analyzed are large scale. In practice, is rarely explicitly required, and so

the triangular factors of are used to generated elements of only as they are

needed since this is often the most computationally efficient method. By setting

computational considerations aside, however, and regarding as being already

constructed and explicitly available, the rower system analyst can derive a great deal of in

sight. This is the approach taken in this chapter.

The bus impedance matrix can be directly constructed element by element using

simple algorithms to incorporate one element at a time in to the system representation. The

work entailed in constructing is much greater than that required to construct but

the information content of the bus impedance matrix is far greater than that of We

shall see, for example, that each diagonal element of reflects important characteristics

of the entire system in the form of the Thevenin impedance at the corresponding bus. Unlike

the bus impedance matrix of an interconnected system is never. sparse and contains

zeros only when the system is regarded as being subdivided into independent parts by open

circuits. In Chap.12, for instance, such open circuits arise in the zero-sequence network of

the system.

The bus admittance matrix is widely used for power-flow analysis. On the other

hand, the bus impedance matrix is equally well favored for power system fault analysis.

Accordingly, both and have important roles in the analysis of the power system

network. In this chapter we study how to construct directly and how to explore some

of the conceptual insights which it offers into the characteristics of the power transmission

network.

1.1 THE BUS ADMITTANCE AND IMPEDANCE MATRICES

In Example 7.6 we inverted the bus admittance matrix and called the resultant

the bus impedance matrix . By definition

(1)

and for a network of three independent nodes the standard form is

(2)

Since is symmetrical around the principal diagonal, must also be

symmetrical. The bus admittance matrix need not be determined in order to obtain and

in another section of this chapter we see how may be formulated directly.


The impedance elements of on the principal diagonal are called driving-point

impedances of the buses, and the off-diagonal elements are called the transfer impedances

of the buses. The bus impedance matrix is important and very useful in making fault

calculations, as we shall see later. In order to understand the physical significance of the

various impedances in the matrix, we compare them with the bus admittances. We can

easily do so by looking at the equations at a particular bus. For instance, starting with the

node equations expressed as

(3)

we have at bus of the three independent nodes

(4)

If and are reduced to zero by shorting buses and to the reference node,

and voltage is applied at bus so that current enters at bus , the self-admittance at

bus is

(5)

Thus, the self-admittance of a particular bus could be measured by shorting all other

buses to the reference node and then finding the ratio of the current injected at the bus to

the voltage applied at that bus. Figure 1 illustrates the method for a three-bus reactive

network. The result is obviously equivalent to adding all the admittances directly connected

to the bus, which is the procedure up to now when mutually coupled branches are absent.

Figure 1 also serves to illustrate the off-diagonal admittance terms of At bus the

equation obtained by expanding equation (3) is

(6)

from which we see that

(7)

Thus, the mutual admittance term is measured by shorting all buses except bus

to the reference node and by applying a voltage at bus , as shown in Figure 1. Then,

is the ratio of the negative of the current leaving the network in the short circuit at node

to the voltage . The negative of the current leaving the network at node is used

since is defined as the current entering the network. The resultant admittance is the

negative of the admittance directly connected between buses and , as we would

expect since mutually coupled branches are absent.


Figure 1: Circuit for measuring . and .

We have made this detailed examination of the bus admittances in order to

differentiate them clearly from the impedances of the bus impedance matrix. Conceptually,

we solve equation (3)by premultiplying both sides of the equation by to yield

(8)

and we must remember when dealing with that and are column vectors of

the bus voltages and the currents entering the buses from current sources, respectively.

Expanding equation (8) for a network of three independent nodes yields

(9)

(10)

(11)

From equation (10) we see that the driving-point impedance is determined by

open-circuiting the current sources at buses and and by injecting the source current

at bus . Then,

(12)

Figure 2: Circuit for measuring and .


Figure 2 shows the circuit described. Since is defined by opening the current

sources connected to the other buses whereas is found with the other buses shorted, we

must not expect any reciprocal relation between these two quantities. The circuit of Figure 2

also enables us to measure some transfer impedances, for we see from equation (9) that

with current sources and open-circuited

(13)

and from equation(11)

(14)

Thus, we can measure the transfer impedances and by injecting current at

bus and by finding the ratios of and to with the sources open at all buses except

bus . We note that a mutual admittance is measured with all but one bus short-circuited

and that a transfer impedance is measured with all sources open-circuited except one.

Equation (9) tells us that if we inject current into bus with current sources at

buses and open, the only impedance through which flows is . Under the same

conditions, equations (10) and (11) show that is causing voltages at buses and

expressed by

(15)

It is important to realize the implications of the preceding discussion, for is

sometimes used in power-flow studies and is extremely valuable in fault calculations.

1.2 THEVENIN'S THEOREM AND

The bus impedance matrix provides important information regarding the power

system network, which we can use to advantage in network calculations. In this section we

examine the relationship between the elements of and the Thevenin impedance

presented by the network at each of its buses. To establish notation, let us denote the bus

voltages corresponding to the initial values of the bus currents I by . The

voltages to

are the effective open-circuit voltages, which can be measured by

voltmeter between the buses of the network and the reference node. When the bus

currents are changed from their initial values to new values , the new bus voltages

are given by the superposition equation

(16)

where represents the changes in the bus voltages from their original values. (a)

shows a large-scale system in schematic form with a representative bus Ⓚ extracted along

with the reference node of the system. Initially, we consider the circuit not to be energized


so that the bus currents and voltages are zero. Then, into bus Ⓚ a current of amp

(or per unit for in per unit) is injected in to the system from a current source

connected to the reference node. The resulting voltage changes at the buses of

Figure 3: Original network with bus ⓚ and reference node extracted. Voltage at bus ⓝ is caused by

current entering the network. Thevenin equivalent circuit at node ⓚ·

the network, indicated by the incremental quantities to , are given by

(17)

with the only nonzero entry in the current vector equal to in row . Row-by-

column multiplication in equation (17) yields the incremental bus voltages


Ⓚ

Ⓚ

(18)

which are numerically equal to the entries in column of multiplied by the

current . Adding these voltage changes to the original voltages at the buses according to

equation(16) yields at bus Ⓚ

(19)

The circuit corresponding to this equation is shown in Figure 3 (b) from which it is

evident that the Thevenin impedance at a representative bus of the system is given by

(20)

where is the diagonal entry in row and column of With set equal to 2,

this is essentially the same result obtained in equation(12) for the driving-point impedance

at bus of Figure 2.

In a similar manner, we can determine the Thevenin impedance between any two

buses ⓙ and Ⓚ of the network. As shown in Fig. 8.4(a), the otherwise dead network is

energized by the current injections .at bus ⓙ and at bus Ⓚ. Denoting the changes in

the bus voltages resulting from the combination of these two current injections by to

we obtain

ⓙ

(21)

in which the right-hand vector is numerically equal to the product of and column

added to the product of and column of the system . Adding these voltage

changes to the original bus voltages according to equation (16), we obtain at buses ⓙ and

(22)

(23)

Adding and subtracting , in equation (22), and likewise, , in equation

(23), give


(24)

(25)

Since is symmetrical, equals and the circuit corresponding to these two

equations is shown in Fig. 8.4(b), which represents the Thevenin equivalent circuit of the

system between buses ⓙ and . Inspection of Fig. 8.4(b) shows that the open-circuit

voltage from bus to bus ⓙ is

, and the

Figure 4: Original network with current source at bus ⓙ and at bus ; Thevenin equivalent

circuit: short-circuit connection; impedance between buses ⓙ and .

impedance encountered by the short-circuit current from bus to bus ⓙ in

Figure 4 is evidently the Thevenin impedance.

(26)

This result is readily confirmed by substituting in equations (24)

and (25) and by setting the difference between the resultant equations equal to

zero. As far as external connections to buses ⓙ and are concerned, Figure 4

represents the effect of the original system. From bus ⓙ to the reference node we can trace


the Thevenin impedance and the open-circuit voltage ; from bus

to the reference node we have the Thevenin impedance and

the open-circuit voltage ; and between buses and ⓙ the Thevenin impedance of

equation(26) and the open-circuit voltage

is evident. Finally, when the branch

impedance is connected between buses ⓙ and of Figure 4 , the resulting current

is given by

(27)

We use this equation in Sec. 8.3 to show how to modify when a branch

impedance is added between two buses o f the network.

Example 1

A capacitor having a reactance of 5.0 per unit is connected between the reference

node and bus ④ of the circuit of Examples 7.5 and 7.6. The original emfs and the

corresponding external current injections at buses and ④ are the same as in those

examples. Find the current drawn by the capacitor.

Solution

The Thevenin equivalent circuit at bus ④ has an emf with respect to reference given

by per unit, which is the voltage at bus ④ found in Example

7.6 before the capacitor is connected. The Thevenin impedance at bus is calculated in

Example 7.6 to be per unit, and so Figure 5 follows. Therefore, the

current leap drawn by the capacitor is

Example 2

If an additional current equal to per unit is injected into the

network at bus ④ of Example 7.6, find the resulting voltages at buses and ④.

Solution

The voltage changes at the buses due to the additional injected current can be

calculated by making use of the bus impedance matrix found in Example 7.6. The required

impedances are in column 4 of . The voltage changes due to the added current injection

at bus ④ in per unit are


Figure 5: Circuit for Examples 8.1 and 8.2 showing: Thevenin equivalent circuit; phasor diagram at bus

④.

By superposition the resulting voltages are determined from equation (16) by adding

these changes to the original bus voltages found in Example 7.6. The new bus volt ages in

per unit are

Since the changes in voltages due to the injected current are all at the same angle

shown in Figure 5 and this angle differs little from the angles of the original voltages, an

approximation will often give satisfactory answers. The change in voltage magnitude at a

.bus may be approximated by the product of the magnitude of the per-unit current and the

magnitude of the appropriate driving-point or transfer impedance. These values added to

the original voltage magnitudes approximate the magnitudes of the new voltages very

closely. This approximation is valid here because the network is purely reactive, but it also

provides a good estimate where reactance is considerably larger than resistance, as is usual


in transmission systems. The last two examples illustrate the importance of the bus

impedance matrix and incidentally show how adding a capacitor at a bus causes a rise in bus

voltages. The assumption that the angles of voltage and current sources remain constant

after connecting capacitors at a bus is not entirely valid if we are considering operation of a

power system. We shall consider such system operation in Chap. 9 using a computer power-

flow program.

1.3 MODIFICATION OF AN EXISTING

In Sec. ‎1.2 we see how to use the Thevenin equivalent circuit and the existing

to solve for new bus voltages in the network following a branch addition without having to

develop the new Since is such an important tool in power system analysis we now

examine how an existing may be modified to add new buses or to connect new lines to

established buses. Of course, we could create a new and invert it, but direct methods of

modifying are available and very much simpler than a matrix inversion even for a small

number of buses. Also, when we know how to modify we can see how to build it

directly. We recognize several types of modifications in which a branch having impedance

is added to a network with known The original bus impedance matrix is identified as

, an N N matrix.

In the notation to be used in our analysis existing buses will be identified by numbers

or the letters and . The letter or will designate a new bus· to be added to the

network to convert to an (N + 1) X (N + 1) matrix. At bus the original voltage will be

denoted by the new voltage after modifying will be , and

will

denote the voltage change at that bus. Four cases are considered in this section.

CASE 1. Adding from a new bus ⓟ to the reference node.

The addition of the new bus ⓟ connected to the reference node through without

a connection to any of the buses of the original network cannot alter the original bus

voltages when a current is injected at the new bus. The

Figure 6: Addition of new bus ⓟ connected through impedance to existing bus ⓚ.

Voltage at the new bus is equal to then,


(28)

We note that the column vector of currents multiplied by the new will not alter

the voltages of the original network and will result in the correct voltage at the new bus ⓟ.

CASE 2. Adding from a new bus ⓟ to an existing bus

The addition of a new bus ⓟ connected through to an existing bus with

injected at bus ⓟ will cause the current entering the original network at bus to become

the sum of . injected at bus plus the current coming through , as shown in Figure

6.

The current flowing into the network; at bus will increase the original voltage

by the voltage just like in equation(19); that is,

(29)

and will be larger than then new by the voltage . So,

(30)

and substituting for , we obtain

(31)

We now see that the new row which must be added to In order to find is

Since must be a square matrix around the principal diagonal, we must add a

new column which is the transpose of the new row. The new column accounts for the

increase of all bus voltages due to , as shown in equation(17). The matrix equation is


(32)

Note that the first elements of the new row are the elements of row of

and the first elements of the new column are the elements of column of

CASE 3. Adding from existing bus to the reference node

To see how to alter by connecting an impedance from an existing bus to

the reference node, we add a new bus ⓟ connected through to bus . Then, we short-

circuit bus ⓟ to the reference node by letting equal zero to yield the same matrix

equation as equation (32) except that is zero. So, for the modification we proceed to

create a new row and new column exactly the same as in Case 2, but we then eliminate the

row and column by Kron reduction, which is possible because of the zero in

the column matrix of voltages. We use the method developed in equation(7.50) to find each

element in the new matrix, where

(33)

CASE 4. Adding between two existing buses ⓙ and

(8.33) To add a branch impedance between buses ⓙ and already established

in , we examine Figure 7, which shows these buses extracted, from the original

network. The current flowing from bus to bus ⓙ is similar to that of Figure 4. Hence,

from equation (21) the change in voltage at each

Figure 7: Addition of impedance between existing buses ⓙ and ⓚ.


bus ⓗ caused by the injection at bus ⓙ and – at bus is given by

(34)

which means that the vector of bus voltage changes is found by subtracting

column from column of and by multiplying the result by . Based on the definition

of voltage change, we now write some equations for the bus voltages as follows:

(35)

And using equation (34) gives

(36)

Similarly, at buses ⓙ and

(37)

(38)

we need one more equation since is unknown. This is supplied by equation (27),

which can be rearranged in to the form

(39)

From equation (37) we note that equals the product of row of and the

column of bus currents ; likewise, of equation (38) equals row, of multiplied by

1. Upon substituting the expressions for and

in equation (39), we obtain

(40)

By examining the coefficients of equations (36) through (38) and eq.(40), we can

write the matrix equation

(41)


in which the coefficient of in the last row is denoted by

(42)

The new column is column minus column of with in the row.

The new row is the transpose of the new column. Eliminating the row and

column of the square matrix of equation(41) in the same manner as previously, we see that

each element in the new matrix is

(43)

We need not consider the case of introducing two new buses connected by

because we could always connect one of these new buses through an impedance to an

existing bus or to the reference bus before adding the second new bus.

Removing a branch. single branch of impedance between two nodes can be

removed from the network by adding the negative of between the same terminating

nodes. The reason is of course, that the parallel combination of the existing branch ( ) and

the added branch amounts to an effective, open circuit.

Table 1 summarizes the procedures of Cases 1 to 4.

TABLE.1 Modification of existing


Example 3

Modify the bus impedance matrix of Example 7.6 to account for the connection of a

capacitor having a reactance of 5.0 per unit between bus ④ and the reference node of the

circuit of Fig.7.9. Then, find using the impedances of the new matrix and the current

sources of Example 7.6. Compare this value of with that found in Example 2.


Solution.

We use equation (32) and recognize that is the matrix of Example 7.6,

that subscript , and that per unit to find

The terms in the fifth row and column were obtained by repeating the fourth row

and column of and noting that

Then, eliminating the fifth row and column, we obtain for from equation

(33)

and other elements in a similar manner to give

The column matrix of currents by which the new is multiplied to obtain the new

bus voltages is the same as in Example 7.6. Since both and are zero while and are

nonzero, we obtain

as found in Example 2.

It is of interest to note that may be calculated directly from equation (27) by

setting node ⓙ equal to the reference node. We then obtain for and

since is already calculated in Example 1.


1.4 DIRECT DETERMINATION OF

We could determine by first finding and then inverting it, but this is not

convenient for large-scale systems as we have seen. Fortunately, formulation of using a

direct building algorithm is a straightforward process on the computer.

At the outset we have a list of the branch impedances showing the buses to which

they are connected. We start by writing the equation for one bus connected through a

branch impedance to the reference as

(44)

and this can be considered as an equation involving three matrices, each of which

has one row and one column. Now we might add a new bus connected to the first bus or to

the reference node. For instance, if the second bus is connected to the reference node

through we have the matrix equation

(45)

and we proceed to modify the evolving matrix by adding other buses and

branches following the procedures described in Sec.8.3. The combination of these

procedures constitutes the building algorithm. Usually, the buses of a network must be

renumbered internally by the computer algorithm to agree with the order in which they are

to be added to as it is built up.

Example 4

Determine for the network shown in Figure 8, where the impedances labeled 1

through 6 are shown in per unit. Preserve all buses.

Solution .

The branches are added in the order of their labels and numbered subscripts on

will indicate intermediate steps of the solution. We start by establishing bus with its

impedance to the reference nod e and write

We then have the bus impedance matrix

To establish bus with its impedance to bus we follow equation(32) to write


The term above is the sum of and . The elements in the new

row and column are the repetition of the elements of row 1 and column 1 of the matrix

being modified.

Figure 8: Network. Branch impedances are in per unit and branch numbers are in parentheses.

Bus with the impedance connecting it to bus is established by writing

Since the new bus is being connected to bus , the term above is the sum

of of the matrix being modified and the impedance of the branch being connected to

bus from bus , The other elements of the new row and column are the repetition of

row 2 and column 2 of the matrix being modified since the new bus is being connected to

bus .

If we now decide to add the impedance from bus to the reference

node, we follow equation (32) to connect a new bus ⓟ through and obtain the

impedance matrix


where above is the sum of . The other elements in the new row and

column are the repetition of row 3 and column 3 of the matrix being modified since bus is

being connected to the reference node through

We now eliminate row ⓟ and column ⓟ by Kron reduction. Some of the elements

of the new matrix from equation (33) are

When all the elements are determined, we have

We now decide to add the impedance from bus to establish bus ④

using equation (32), and we obtain

The off-diagonal elements of the new row and column are the repetition of row 3

and column 3 of the matrix being modified because the new bus ④ is being connected to

bus . The new diagonal element is the sum of of the previous matrix and .

Finally, we add the impedance between buses and ④. If we let and in

equation (41) equal 2 and 4, respectively, we obtain the elements for row 5 and column 5.


and from equation(42)

So, employing previously found, we write the 5 x 5 matrix

and from equation(43) we find by Kron reduction

which is the bus impedance matrix to be determined. All calculations have been

rounded off to five decimal places.

Since we shall again refer to these results, we note here that the reactance diagram

of Figure 8 is derived from Fig.7.10 by omitting the sources and one of the mutually coupled

branches. Also, the buses of Fig.7.10 have been renumbered in Figure 8 because the

building algorithm must begin with a bus connected to the reference node, as previously

remarked.

The building procedures are simple for a computer which first must determine

the types of modification involved as each branch impedance is added. However, the

operations must follow a sequence such that we avoid connecting an impedance between

two new buses.

As a matter of interest, we can check the impedance values of by the network

calculations of Sec.8.1.

Example 5

Find of the circuit or Example 4 by determining the impedance measured

between bus and the reference node when currents injected at buses , , and ④ are

zero.

Solution:


The equation corresponding to equation(12) is

Were cognize two parallel paths between buses and of the circuit of Figure 8

with the resulting impedance of

This impedance in series with combines in parallel with to yield

which is identical with the value found in Example 4. Although the network

reduction method of Example 5 may appear to be simpler by comparison with other

methods of forming such is not the case because a different network reduction is

required to evaluate each element of the matrix. In Example 5 the network reduction to find

, for instance, is more difficult than that for finding , The computer could make a

network reduction by node elimination but would have to repeat the process for each node.

1.5 CALCULATION OF ELEMENTS FROM

When the full numerical form of is not explicitly required for an application, we

can readily calculate elements of as needed if the upper-and lower-triangular factors of

are available. To see how this can be done, consider postmultiplying by a vector

with only one nonzero element in row and all other elements equal to zero.

When is an matrix, we have

(46)

Thus, postmultiplying by the vector shown extracts the th column, which we

have called the vector that is


Since the product of and equals the unit matrix, we have

(47)

If the lower-triangular matrix and the upper-triangular matrix of are

available, we can write equation(47) in the form

(48)

It is now apparent that the elements in the column vector

can be found from

equation(48) by forward elimination and back substitution, as explained in Sec.7.8. If only

some of the elements of

are required, the calculations can be reduced accordingly. For

example, suppose that we wish to generate and of for a four-bus system. Using

convenient notation for the elements of and , we have

(49)

We can solve this equation for

in two steps as follows:

(50)

Where:

(51)

By forward substitution equation(50) immediately yields


and by back substitution of these intermediate results in equation (51) we find the

required elements of column 3 of ,

If all elements of

are required, we can continue the calculations,

The computational effort in generating the required elements can be reduced by

judiciously choosing the bus numbers.

In later chapters we shall find it necessary to evaluate terms like ( )

involving differences between columns ⓜ and ⓝ of If the elements of are not

available explicitly, we can calculate the required differences by solving a system of

equations such as

(52)

Where

is the vector formed by subtract

ting column from, column of , and in row and in row

of the vector shown.

In large-scale system calculations considerable computational efficiency can be

realized by solving equations in the triangularized form of equation (52) while the full

need not be developed. Such computational considerations underlie many of the formal

developments based on in this text.

Example 6

The five-bus system shown in Fig.8.9 has per-unit impedances as marked. The

symmetrical bus admittance matrix for the system is given by

and it is found that the triangular factors of are


Use the triangular factors to calculate , the

Thevenin impedance looking into the system between buses ④ and ⑤ of Fig.8.9.

Solution

Since is symmetrical, the reader should check that the row elements of equal

the column elements of divided by their corresponding diagonal elements. With is

representing the numerical values of , forward solution of the

system of equations

Figure 9: Reactance diagram for Example 8.6, all values are per-unit impedances.

yields the intermediate values


Backsubstituting in the system of equations

where represent the numerical values of , we find from the last two rows that

The desired Thevenin impedance is therefore calculated as follows :

Inspection of Figure 9 verifies this result.


Problem 1

Construct the bus impedance matrix for the network given by the following figure.

Figure 10: Impedance diagram.

Solution:


Problem 2:


Problem 3

Problem 4

Problem 5

A transmission line exists between buses 1 and 2 with per unit impedance 0.4.

Another line of impedance 0.2 p.u. is connected in parallel with it making it a doubl-circuit

line with mutual impedance of 0.1 p.u. Obtain by building algorithm method the impedance

of the two-circuit system.

Solution


Problem 6

The double circuit line in the problem E 4.1 is further extended by the addition of a

transmission line from bus (1). The new line by virtue of its proximity to the existing lines has

a mutual impedance of 0.05 p.u. and a self – impedance of 0.3 p.u. obtain the bus

impedance matrix by using the building algorithm.


Solution


Problem 7

The system E4.2 is further extended by adding another transmission line to bus 3 w

itil 001£ in pedance of 0 .3 p.u .0 bta.in tile ZBUS

Solution:


Problem 8

The system in E 4.3 is further extended and the radial system is converted into a ring

system joining bus (2) to bus (4) for reliability of supply. Obtain the ZBUS. The self

impedance of element 5 is 0.1 p.u

Solution:


Problem 9:

Compute the bus impedance matrix for the system shown in figure by adding

element by element. Take bus (2) as reference bus

Solution:


Problem 10:

Using the building algorithm construct zBUS for the system shown below. Choose 4

as reference BUS.

Solution:


Problem 11:

Given the network shown in Fig. E.4.1S.


Solution:


Problem 12:

E 4.8 Consider the system in Fig. E.4.17. Obtain ZBUS by using building algorithm.


Solution:


Problem 13

Form the impedance matrix of the electric network shown in the following figure by

using the branch addition method.

Solution

According to the node ordering, we can make the sequence table of branch adding

as follows.

electrical power system analysis 4. the impedance model and network calculations

Documents

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power system network

impedance model

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impedance matrices

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