# 2.1 Introduction. 2.2 Terms and definitions

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1 .1 Introduction An important step in the procedure for solving any circuit problem consists first in selecting a number of independent branch currents as (known as loop currents or mesh currents) variables, and then to express all branch currents as functions of the chosen set of branch currents. Alternately a number of independent node pair voltages may be selected as variables and then express all existing node pair voltages in terms of these selected variables. For simple networks involving a few elements, there is no difficulty in selecting the independent branch currents or the independent node-pair voltages. The set of linearly independent equations can be written by inspection. However for large scale networks particularly modern electronic circuits such as integrated circuits and microcircuits with a larger number of interconnected branches, it is almost impossible to write a set of linearly independent equations by inspection or by mere intuition. The problem becomes quite difficult and complex. A systematic and step by step method is therefore required to deal with such networks. Network topology (graph theory approach) is used for this purpose. By this method, a set of linearly independent loop or node equations can be written in a form that is suitable for a computer solution.. Terms and definitions The description of networks in terms of their geometry is referred to as network topology. The adequacy of a set of equations for analyzing a network is more easily determined topologically than algebraically. Graph (or linear graph): A network graph is a network in which all nodes and loops are retained but its branches are represented by lines. The voltage sources are replaced by short circuits and current sources are replaced by open circuits. (Sources without internal impedances or admittances can also be treated in the same way because they can be shifted to other branches by E-shift and/or I-shift operations.) Branch: A line segment replacing one or more network elements that are connected in series or parallel.

3 Network Topology 11 Figure.1 Relation between nodes, links, and branches Let B Total number of branches in the graph or network N total nodes L link branches Then N 1 branches are required to construct a tree because the first branch chosen connects two nodes and each additional branch includes one more node. Therefore number of independent node pair voltages N 1 number of tree branches. Then L B (N 1) B N +1 Number of independent loops B N +1. Isomorphic graphs Two graphs are said to be ismorphic if they have the same incidence matrix, though they look different. It means that they have the same numbers of nodes and the same numbers of branches. There is one to one correspondence between the nodes and one to one correspondence between the branches. Fig.. shows such graphs. Figure.

4 11 Network Theory. Matrix representation of a graph For a given oriented graph, there are several representative matrices. They are extremely important in the analytical studies of a graph, particularly in the computer aided analysis and synthesis of large scale networks...1 Incidence Matrix A n It is also known as augmented incidence matrix. The element node incidence matrix A indicates in a connected graph, the incidence of elements to nodes. It is an N B matrix with elements of A n (a kj ) a kj 1, when the branch b j is incident to and oriented away from the k th node. 1, when the branch b j is incident to and oriented towards the k th node., when the branch b j is not incident to the k th node. As each branch of the graph is incident to exactly two nodes, n a kj for j 1,,, B. k That is, each column of A n has exactly two non zero elements, one being +1 and the other 1. Sum of elements of any column is zero. The columns of A n are lineraly dependent. The rank of the matrix is less than N. Significance of the incidence matrix lies in the fact that it translates all the geometrical features in the graph into an algebraic expression. Using the incidence matrix, we can write KCL as A n i B, where i B branch current vector. But these equations are not linearly independent. The rank of the matrix A is N 1. This property of A n is used to define another matrix called reduced incidence matrix or bus incidence matrix. For the oriented graph shown in Fig..(a), the incidence matrix is as follows: Nodes branches 1 a b A n 1 1 c d 1 1 Note that sum of all elements in each column is zero. Figure.(a)

5 Network Topology 11.. Reduced incidence matrix Any node of a connected graph can be selected as a reference node. Then the voltages of the other nodes (referred to as buses) can be measured with respect to the assigned reference. The matrix obtained from A n by deleting the row corresponding to the reference node is the elementbus incident matrix A and is called bus incidence matrix with dimension (N 1) B. A is rectangular and therefore singular. In A n, the sum of all elements in each column is zero. This leads to an important conclusion that if one row is not known in A, it can be found so that sum of elements of each column must be zero. From A,wehaveAi B, which represents a set of linearly independent equations and there are N 1 independent node equations. For the graph shown in Fig.(a), with d selected as the reference node, the reduced incidence matrix is Nodes branches A a b c Note that the sum of elements of each column in A need not be zero. j 1 j Note that if branch current vector, j B j j j Then Ai B representing a set of independent node equations. Another important property of A is that determinant AA T gives the number of possible trees of the network. If A [A t : A i where A t and A i are sub-matrices of A such that A t contains only twigs, then det A t is either+1or 1. To verify the property that det AA T gives the number of all possible trees, consider the reduced incidence matrix A of the example considered. That is, Then, Det AA T T T 8

6 11 Network Theory Fig..(b) shows all possible trees corresponding to the matrix A. Figure.(b) To verify the property that the determinant of sub matrix A t of A A t ; A i is +1 or 1. For tree [,, Nodes branches 1 From a A b 1 1 A t ; A i c Det A i For another tree [,, Nodes branches 1 a A b 1 1 A t ; A i c Det A i Loop equations and fundamental loop matrix (Tie-set Matrix) From the knowledge of the basic loops (tie-sets), we can obtain loop matrix. In this matrix, the loop orientation is to be the same as the corresponding link direction. In order to construct this matrix, the following procedure is to be followed.

7 Network Topology j Draw the oriented graph of the network. Choose a tree.. Each link forms an independent loop. The direction of this loop is same as that of the corresponding link. Choose each link in turn.. Prepare the tie-set matrix with elements b ij, where b ij 1; when branch b J in loop i and is directed in the same direction as the loop current. 1; when branch b J is in loop i and is directed in the opposite direction as the loop current: ; when branch b J is not in loop i. Tie-set matrix is an i b matrix. Consider the example of Fig..(a). Figure. Selecting (,, ) as tree, the co-tree is (1, ). Fig.. leads to the following tie-set. with V B Loops # branches 1 M v 1 v v v v Then MV B gives the following independent loop equations: v 1 + v v v + v v Looking column wise, we can express branch currents in terms of loop currents. This is done by the following matrix equation. J B M T I L

8 118 j Network Theory The above matrix equation gives J 1 i 1, J i 1 + i, J i, J i 1 i Note that J stands for branch current while i stands for loop current. In this matrix, (i) Each row corresponds to an independent loop. Therefore the columns of the resulting schedule automatically yield a set of equations relating each branch current to the loop currents. (ii) As each column expresses a branch current in terms of loop currents, the rows of the matrix automatically yield the closed paths in which the associated loop currents circulate. Expressions for branch currents in terms of loop currents may be obtained in matrix form as J B M T I L. where M is the tie-set matrix of L B. In the present example, J B J 1 J J J and I L i1 i. Cut-set matrix and node pair potentials Figure.(a) A directed graph Figure.(b) Two separate graphs created by the cut set {1,,, } A cut-set of a graph is a set of branches whose removal, cuts the connected graph into two parts such that the replacement of any one branch of the cut-set renders the two parts connected. For example, two separated graphs are obtained for the graph of Fig..(a) by selecting the cut-set consisting of branches [1,,,. These seperated graphs are as shown in Fig..(b). Just as a systematic method exists for the selection of a set of independent loop current variables, a similar process exists for the selection of a set of independent node pair potential variables.

9 Network Topology j 119 It is already known that the cut set is a minimal set of branches of the graph, removal of which divides the graph in to two connected sub-graphs. Then it separates the nodes of the graph in to two groups, each being one of the two sub-graphs. Each branch of the tie-set has one of its terminals incident at a node on one sub-graph. Selecting the orientation of the cut set same as that of the tree branch of the cut set, the cut set matrix is constructed row-wise taking one cut set at a time. Without link currents, the network is inactive. In the same way, without node pair voltages the network is active. This is because when one twig voltage is made active with all other twig voltages are zero, there is a set of branches which becomes active. This set is called cut-set. This set is obtained by cutting the graph by a line which cuts one twig and some links. The algebraic sum of these branch currents is zero. Making one twig voltage active in turn, we get entire set of node equations. This matrix has current values, q ij 1, if branch J is in the cut-set with orientation same as that of tree branch. 1, if branch J is in the cut-set with orientation opposite to that of tree branch., if branch J is not in the cut-set. and dimension is (N 1) B. Row-by-row reading, it gives the KCL at each node and therefore we have QJ B. The procedure to write cut-set matrix is as follows: (i) Draw the oriented graph of a network and choose a tree. (ii) Each tree branch forms an independent cut-set. The direction of this cut-set is same as that of the tree branch. Choose each tree branch in turn to obtain the cut set matrix. Isolate the tree element pairs and energize each bridging tree branch. Assuming the bridging tree branch potential equals the node pair potential, thus regarding it as an independent variable. (iii) Use the columns of the cut-set matrix to yield a set of equations relating the branch potentials in terms of the node pair potentials. This may be obtained in matrix form as V B Q T E N where v and e are used to indicate branch potential and node voltage respectively. In the example shown in Fig. (c), (,, ) are tree branches. Links are shown in dotted lines. If two tree branch voltages in and are made zero, the nodes a and c are at the same potential. Similarly the nodes b and d are at the same potential. The graph is reduced to the form shown in Fig..(d) containing only the cut-set branches. Then, we have i i 1 i i Similarly by making only e to exist (with e and e zero), the nodes a, b and c are at the same potential, reducing the graph to the form shown in Fig..(e). Thus, i + i + i

10 1 j Network Theory Figure.(c) Figure.(d) This corresponds to cut set as shown in Fig.. (f). Figure.(e) Figure.(f) For the remaining cut-set, e and e are made zero as in Fig..(g). e is active and hence, the nodes a, d, and c are at the same potential. Thus i 1 + i + i The corresponding cut-set is shown in Fig.(h). Figure.(g) Figure.(h) Therefore, the cut set schedule is Node pair branches voltages # 1 e 1 v e v e v Q

11 Network Topology 11 Note that QJ B gives the following equilibrium equations: J 1 J J + J J + J + J J 1 + J + J Looking column-wise, we can express branch voltages in terms of node pair voltages as v 1 e 1 + e 1 1 v e 1 + e + e v e Or 1 e 1 v e 1 e V B v e 1 1 e v e 1 + e 1 1 That is V B Q T E N. Netwrok Equilibrium Equations Loop equations: A branch of a netwrok can, in general, be represented as shown in Fig.., Figure. where E B is the voltage source of the branch. I B is the current source of the branch Z B is the impedance of the branch J B is the current in the branch The voltage-current relation is then given by V B (J B + I B ) Z B E B For a general network with many branches, the matrix equation is V B Z B (J B +I B ) E B (.1) where V B, J B, I B, and E B are B 1 vectors and Z B is the branch impedance matrix of B B.

12 1 Network Theory Each row of the tie-set matrix corresponds to a loop and involves all the branches of the loop. As per KV L, the sum of the corresponding branch voltages may be equated to zero. That is MV B (.) where M is the tie-set matrix. In the same matrix, each column represents a branch current in terms of loop currents. Transposed M is used to give the relation between branch currents and loop currents. J B M T I L (.) This equation is called loop transformation equation. Substituting equation (.1) in (.), we get Substituting equation (.) in (.), we get MZ B {J B + I B } ME B (.) or MZ B M T I L + MZ B I B ME B MZ B M T I L ME B MZ B I B With MZ B M T Z L,wehaveZ L I L E L If there are no current sources in the network, then Z L I L ME B E L (.) where Z L is the loop impedance matrix and E L is the resultant loop voltage source vector. Node equations: Next, each row of the cut-set matrix corresponds to a particular node pair voltage and indicates different branches connected to a particular node. KCL can be applied to the node and the algebraic sum of the branch currents at that node is zero. QJ B (.) Each column of cut-set matrix relates a branch voltage to node pair voltages. Hence, V B Q T E N (.) This equation is known as node transformation equation. branch is Current voltage relation for a J B Y B (V B +E B ) I B For a netwrok with many branches the above equation may be written in matrix form as J B Y B V B + Y B E B I B (.8) where Y B is branch admittance matrix of B B.

13 Network Topology 1 The matrix nodal equations may be obtained from equations.,., and.8. Substituting equation (.8) in (.) Substituting equation (.) in (.9) QY B V B + QY B E B QI B (.9) QY B Q T E N Q (I B Y B E B ) In the absence of voltage sources, the equation becomes Y N E N QI B I N where Y N is the node admittance matrix and I N is the node current vector. Worked Examples EXAMPLE.1 Refer the circuit shown in Fig..(a). Draw the graph, one tree and its co-tree. Figure.(a) SOLUTION We find that there are four nodes (N )and seven branches (B ). The graph is then drawn and appears as shown in Fig.. (b). It may be noted that node d represented in the graph (Fig..(b)) represents both the nodes d and e of Fig..(a). Fig..(c) shows one tree of graph shown in Fig..(b). The tree is made up of branches, and. The co-tree for the tree of Fig..(c) is shown in Fig..(d). The co-tree has L B N links. Figure.(b)Graph Figure.(c) Tree

14 1 j Network Theory Figure.(d) Co-tree EXAMPLE. Refer the network shown in Fig..8(a). Obtain the corresponding incidence matrix. Figure.8(a) Figure.8(b) SOLUTION The network shown in Fig..8(a) has five nodes and eight branches. The corresponding graph appears as shown in Fig..8(b). The incidence matrix is formed by following the rule: The entry of the incidence matrix, a k 1, if the current of branch k leaves the node i 1, if the current of branch enters node i, if the branch k is not connected with node i. Incidence matrix: Nodes Branch numbers a b c d e f g h

15 Network Topology j 1 Some times, we represent an incidence matrix as follows: A where subscript indicates that there are five nodes in the graph. It may be noted that from a given incidence matrix, the corresponding graph can be drawn uniquely. EXAMPLE. For the incidence matrix shown below, draw the graph. SOLUTION a b c d Figure.9 Observing the matrix, it can be seen that it is a reduced incidence matrix. Branches 1,, and are to be connected to the reference node. Branch appears between the nodes a and b, between b and c, between c and d and 8 between a and c. With this infromation, the oriented graph is drawn as shown in Fig..9. Orientation is +1 for an arrow leaving a node d 1 for an arrow entering a node. EXAMPLE. Draw the graph of a netwrok of whose the incidence matrix is as shown below p q r s

16 1 j Network Theory SOLUTION Sum of the elements in columns, 9 are not zero. Therefore the given matrix is a reduced matrix. Taking as reference node, the oriented graph is as shown in Fig..1 after making the nodes in an order. Figure.1 EXAMPLE. For the graph shown in Fig..11(a), write the incidence matrix. Express branch voltage in terms of node voltages and then write a loop matrix and express branch currents in terms of loop currents. SOLUTION With the orientation shown in Fig..11(a), the incidence matrix is prepaed as shown below. a b c d e Figure.11(a) Branch voltages in terms of node voltages are v 1 e a e d v e b e d etc. Figure.11(b) For loop (tie-set) matrix, L B N With twigs (1,,, ), we have chords (links) (,,, 8) and the corresponding tree is as shown in Fig.11 (b). Introducing the chords one at a time, the tie-set matrix is prepared as shown below.

17 Network Topology 1 i 1 J i J i J i J The branch currents in terms of loop currents are J 1 i 1 i, J i 1 i etc. EXAMPLE. For the network shown in Fig..1(a), determine the number of all possible trees. For a tree consisting of (1,, ) (i) draw tie set matrix (ii) draw cut-set matrix. Figure.1(a) SOLUTION If the intention is to draw a tree only for the purpose of tie-set and cut-set matrices, the ideal current source is open circuited and ideal voltage source is short circuited. The oriented graph is drawn for which d is the reference. Refer Fig..1(b), A a b c

18 18 Network Theory Therefore, possible number of trees 1. Fig..1(b) shows the corresponding graph, tree, co-tree, and loops 1,,. Co-tree (links) Figure.1(b) (i) Tie-set matrix for twigs (1,, ) is branches Loop Currents 1 i 1 J i J 1 1 i J (ii) Cut-set matrix is With e 1 acting (e e ) With e acting (e 1 e ) With e acting (e 1 e ) Figure.1(c)

19 Network Topology j 19 branches Node - pair voltage 1 e 1 v 1 e v e v a 1 1 b c d EXAMPLE. For the network shown in Fig..1(a), write a tie-set schedule and then find all the branch currents and voltages. Figure.1(a) SOLUTION Fig..1(b) shows the graph for the netwrok shown in Fig..1(a). Also, a possible tree and co-tree are shown in Fig. 1(c). Co-tree is in dotted lines. Figure.1(b) Figure.1(c)

20 1 j Network Theory First, the tie-set schedule is formed and then the tie-set matrix is obtained. Tie-set schedule: Loop Branch numbers currents 1 x y z Tie-set matrix is M The branch impedance matrix is Z B 1 1 The loop impedance matrix is ME B E B Z L MZ B M T

21 Network Topology 11 The loop equations are obtained using the equation, Z L I L ME B 1 x y 1 z Solving by matrix method, we get x.1 A, y 1.1 A, z. A The branch currents are computed using the equations: I B M T I L I 1 I I I I I Hence, I 1 x.1 A, I y 1. A, I z. A, I x z 1. A, I x + y. A, I y + z.8 A The branch voltages are computed using the equation: V B Z B I B E B x y z Hence, V 1 I V, V 1I 1. V, V I 1. V, V 1I 1. V, V I 1. V, V I.1 V EXAMPLE.8 Refer the network shown in Fig..1(a). Find the tie-set matrix and loop currents. Figure.1(a)

22 1 Network Theory SOLUTION In the circuit, Ω in series with current source is shorted (as it is trival), the graph is as shown in Fig..1(b) with 1 as tree branch and as link. Using the equation, MZ B M T I L ME B MZ B I B with M [1 1,we have Figure.1(b) [ MZ B M T [1 1 ME B Z B I B [ 1 1 ([ 1 [ 1 1 [ 1 V x [ [ 1 V x [ V x / ) 8I 1 1 V x but V x I 1 8I 1 1 I 1 I 1 /A As I 1 V x / I, I /A EXAMPLE.9 For the given cut-set matrix, draw the oriented graph SOLUTION Data from the matrix: B, N t, N, L. Tree branch voltages are e 1 v 1, e v, e v and e v. Therefore all these are connected to reference node. Individual cut-sets are which indicates that is common to a and b, is common to b and c, is common to c and d.

23 Network Topology 1 Therefore the graph is EXAMPLE.1 Refer the network shown in Fig..1(a). Solve for branch currents and branch voltages. S S 1S S Figure.1(a) SOLUTION The oriented graph for the network is shown in Fig..1(b). A possible tree and cotree with fundamental cut-sets are shown in Fig..1(c). () () (1) () (1) () () () Figure.1(b) Directed graph for the network shown in Fig.. Figure.1(c) A possible tree (thick lines) and cotree (dotted lines)

24 1 Network Theory Cut-set schedule: Branch Numbers Tree branch voltages 1 e e Cut-set matrix: Q Branch admittance matrix: Y B [ Cut-set admittance matrix: Y N QY B Q T Equilibrium equations: [ [ 1 [ Y N E N QI B [ [ [ e1 e [ e1 Solving using cramer s rule, we get e [ [ 1 e 1 V, e + V 1

25 Network Topology 1 Branch voltage are found using the matrix equation: V 1 V V V V B Q T E N Hence, V 1 V.91V V + V+.8V V + +.8V V +.8V Branch currents are found using the matrix equation (.8): Verification: Refer Fig..1(a). KCL equations and KVL equations: I 1 I I I J B Y B V B I B 1 I 1 V A I V.8 A I V.18 A I V. A V 1 V V V I 1 I.18 A I I + I A V + V + V 1 V V are statisfied. 1

26 1 j Network Theory EXAMPLE.11 For the oriented graph shown, express loop currents in terms of branch currents for an independent set of columns as those pertinent to the links of a tree: (i) Composed of,,, 8 (ii) Composed of 1,,, Verify whether the two sets of relations for i s in tems of J s are equivalent. Construct a tie-set schedule with the currents in the links,,, 8 as loop currents and find the corresponding set of closed paths. SOLUTION For the first set Loop Branch numbers No:# B ; ; ; 8 and L 1; ; ;. i 1 J 1, i J, i J, & i J. Then for the second set, of the mesh currents indicated for the first set, we have J i ) i J J i 1 i ) i 1 J 1 + J J i i ) i J J 8 J 8 i i i J J J 8 By applying KCL for the oriented graph, i 1 J 1 J + J ; i J J J J J J 8 ; i J J J 8 ; i J Thus the two sets of relations for i s in terms of J s are equivalent. The tie-set schedule with the currents in links,,, 8 as loop currents are shown below.

27 Network Topology 1 Loop Branch numbers No: EXAMPLE.1 In the graph shown in Figure.1(a), the ideal voltage source e 1V. For the remaining branches each has a resistance of 1 Ω with O as the reference. Obtain the node voltage e 1, e and e using network topology. SOLUTION Figure.1(a) With e shift, graph is as shown in Figure.1(b). Branches are numbered with orientation. With T(,, ) the cut set matrix is Q Figure.1(b)

28 18 Network Theory Y B Q T QY B Q T QY B E B According to the equation QY B Q T E N QY B E B,wehave 1 e e 1 e 1 Therefore, e 1 e e

29 Network Topology j 19 EXAMPLE.1 For the circuit shown in Fig..1, for a tree consisting of ab, bc, cd form a tie-set schedule and obtain equilibrium loop equations. Choose branch numbers same as their resistance values. Solve for loop currents. Figure.1 SOLUTION The oriented graph is Figure.1(a) Tree with nodes ab, bc, and cd and links are shown in dotted line. The tie set matrix is B! 1 M i 1 i i

30 1 j Network Theory MZ B M T E B Z B I B M (E B Z B I B ) Then the loop equations, 1 1 i 1 i i 1

31 Network Topology 11 The loop currents are i 1 i i EXAMPLE.1 For the network shown in Fig..18(a), prepare a cut-set schedule and obtain equilibrium equations. Number the branches by their ohmic values. Figure.18(a) Figure.18(b) SOLUTION Numbering the branches same as those of ohmic values, the oriented graph is as shown in Fig..18(b). Choosing,, as tree branches, the tie set schedule is as shown below. Figure.18(c)

32 1 Network Theory Q e a e b e c QY B Q T I B Y B E B Q (I B Y B E B )

33 Network Topology 1 Equations in matrix form therefore is e 1 e e Without matrix method: From the matrix Q, looking row wise, we have by KCL J 1 + J + J J J + J J 1 + J + J 1 Looking column wise, we have v 1 e a e c ; v e a e b ; v e b e c ; v e c ; v e a ; v e b Writing set (a) in terms of branch voltages and then (b) branch voltage in terms of node pair voltages, For (i) For (ii) For (iii) v 1 + v + v e a e c + e a e b + e a 1.e a.e b e c (.1) v v 1 + v + e b e a + e b e c + e b 1.e a + e b.e c 1 (.11) v 1 + v 1 + v 1 e a + e c + e c e b v 1 + v + v + e c e a.e b 19 1 e c (.1) (.1), (.11), and (.1) are the required equations.

34 1 Network Theory EXAMPLE.1 Device a tree for the circuit shown in Fig..19(a), for which all the currents pass through 1 Ω. For this tree write the tie-set matrix to obtain equilibrium equations. A W 1A 1W 1W W 1W 1.W.8W Figure.19(a) SOLUTION Performing I shift the network is redrawn as shown in Fig..19(b). A 1A 1A 1 1 A Figure.19(b) The oriented graph is as shown in Fig.19(c) for which the tree is shown in Fig..19(d) to satisfy the condition. Figure.19(c) Figure.19(d) With link, row a of tie-set is With link, row b of tie-set is With link row c of tie-set is 1 1 1

35 Network Topology 1 So that M MZ B M T [ [ [ [ Matrix equation is i 1 i i 1 1

36 1 j Network Theory EXAMPLE.1 For the circuit shown in Fig..(a), construct a tree in which v 1 and v are tree branch voltages, write a cut-set matrix and obtain equilibrium equations. Solve for v 1. Figure.(a) Figure.(b) SOLUTION Performing E shift and I shift, the circuit is redrawn as shown in Fig..(b). With v 1 and v as tree branch voltages the graph is as shown in Fig..(c), with tree branches with full lines and the links in dotted lines. Q QG B Q T :1 : :1 : : + : : : : : : : : : Figure.(c) I B GE B QI B G B E B :1 : : :

37 Network Topology 1 Therefore the equilibrium equation is [.... [ e1 e [ 1 1 Therefore, [ e1 e [ [ 1 1 [ EXAMPLE.1 (a) Construct a tree for the circuit shown in Fig..1(a) so that i 1 is a link current. Assign a complete set of link currents and find i 1 (t). (b) Construct another tree in which v 1 is a tree branch voltage. Assign a complete set of tree branch voltages and find v 1 (t)v. Take i(t) sin 1 t Amp, v(t) 1 cos 1 (t) SOLUTION (a) The circuit after I shift is as shown in Fig..1(b). The oriented graph is as shown in Fig..1(c). With branches as numbered, and 1 as a link the tree is as shown in Fig..1(d) Links are shown in dotted line (links are 1 and ). The tie-set martix is M [ Figure.1(a) The branch impedence matrix is Z B E B Z B I B 1 j1 1 j j1 1 j1 1 j Figure.1(b) j1

38 18 Network Theory M (E B Z B I B ) MZ B M T [ [ j1 [ 1 j1 1 j + j [ j1 1 j1 1 j1 1+j1 Figure.1(c) Equilibrium equations are [ j1 1 j1 1 j1 1+j1 [ I1 I [ Figure.1(d) + j1 Therefore I 1 + j1 1 j1 1 + j1 j1 1 j1 1 j1 1+j1 1. / 18 Amp Therefore I 1 1. sin ( 1 t 18 ) Amp (b) With 1 as tree branch, the oriented tree is as shown in Fig..1(c). Links are shown with dotted lines, with e 1 and e as node pair voltages the cut-sets are shown. The cut-set matrix is Y e a e b [ j1 1 j. Figure.1(e)

39 Network Topology 19 I B Y B E B Q (I B Y B E B ) QY B Q T [ [ j1 1 j. [ j j1. j1 [. 1 j1 1 j. Equilibrium equations are [ [ [ j. 1 E j1 E Therefore E j1 j j1 1. / 18 volts Therefore E 1 1. sin ( 1 t 18 ) volts Dual Networks Two electrical circuits are duals if the mesh equations that characterize one of them have the same mathematical form as the nodal equations that characterize the other. Let us consider the series R a L a C a network excited by a voltage source v a as shown in Fig..(a), and the parallel G b C b L b network fed by a current source i b as shown in Fig..(b). Figure.(a) Series RLC network Figure.(b) Parallel RLC network

40 1 j Network Theory The equations governing the circuit behaviour are: (i) Z di a L a dt + R ai a + 1 C a (ii) i a dt v a di a ) L a dt + R ai a + 1 v a C Z a di a ) L a dt + R ai a + 1 i a dt v a (.1) C a Z dv b C b dt + G bv b + 1 v b dt i b L b Z dv b ) C b dt + G bv b + 1 v b dt i b L b Z dv b ) C b dt + G bv b + 1 v b dt i b (.1) L b Comparing equations (.1) and (.1), we get the similarity between the networks of Fig..(a) and Fig.(b). The solution of equation (.1) will be identical to the solution of equation (.1) when the following exchanges are made: L a! C b ;R a! G b; C a! L b v a! i b ;i a! v b ; Hence, the series network in Fig..(a) and parallel network in Fig..(b) are duals of each other. The advantage of duality is that there is no need to analyze both types of circuits, since the solution of one automatically gives the solution of the other with a suitable change of symbols for the physical quantities. Table.1 gives the corresponding quantitites for dual electrical networks. Table.1: Table of dual quantities Loop basis Node basis 1. A loop made up of several branches 1. A node joining the same number of branches.. Voltage sources. Current sources. Loop currents. Node voltages. Inductances. Capacitances. Resistances. Conductance. Capacitances. Inductances Only planar 1 networks have duals. The duals of planar networks could be obtained by a graphical technique known as the dot method. The dot method has the following procedure: 1. Put a dot in each independent loop of the network. These dots correspond to independent nodes in the dual network. 1 Planar networks are those that can be laid on a plane without branches crossing one another.

41 Network Topology j 11. Put a dot outside the network. This dot corresponds to the reference node in the dual network.. Connect all internal dots in the neighbouring loops by dahsed lines cutting the common branches. These branches that are cut by dashed lines will form the branches connecting the corresponding independent nodes in the dual network. As an example, if a common branch contains R and L in series, then the parallel combination of G and C should be put between the corresponding independent nodes in the dual network.. Join all internal dots to the external dot by dashed lines cutting all external branches. Duals of these branches cut by dashed lines will form the branches connecting the independent nodes and the reference node.. Convention for sources in the dual network: (i) a clockwise current source in a loop corresponds to a voltage source with a positive polarity at the dual independent node. (ii) a voltage rise in the direction of a clockwise loop current corresponds to a current flowing toward the dual independent node. EXAMPLE.18 Draw the dual of the circuit shown in Fig..(a). Write the mesh equations for the given network and node equations for its dual. Verify whether they are dual equations. Figure.(a)

42 1 j Network Theory SOLUTION For the given network, the mesh equations are R 1 i 1 + L 1 D (i 1 i )+ 1 C R i + L Di + R (i i )+ 1 C Z Z (i 1 i ) dt v g i i (i i ) dt The dual network, as per the procedure described in the theory is prepared as shown in Fig..(b) and is drawn as shown in Fig..(c). The node equations for this network are Figure. (b) G 1 V 1 + C 1 D (v 1 v )+ 1 L G v + C Dv + G (v v )+ 1 L Z Z (v 1 v ) dt i g (v v ) dt Figure. (c)

43 Network Topology j 1 EXAMPLE.19 For the bridge network shown in Fig..(a), draw its dual. Write the integro-differential form of the mesh equations for the given network and node equations for its dual. The values for resistors one in ohms, capacitors are in farads and inductors are in Henrys. Figure.(a) SOLUTION The dual for the given network is shown in Fig..(c) using the procedure shown in Fig..(b). The integro-differential form for the network is 1i 1 + D (i 1 i )+ 1 D (i i 1 )+Di + 1 i + 1 Z (i i 1 ) dt + 1 Z Z Z (i 1 i ) dt 1 sin t (i i ) dt (i i ) dt Figure.(b)

44 1 j Network Theory Figure.(c) The node equations for the dual network are 1v 1 + D (v 1 v )+ 1 D (v v 1 )+Dv + 1 v + 1 Z (v v 1 ) dt + 1 Z Z Z (v 1 v ) dt 1 sin t (v v ) dt (v v ) dt EXAMPLE. A network with a controlled source is shown in Fig..(a). Draw the dual for the given network and write equations for both the networks. SOLUTION Figure.(a) The dual for the given network is shown in Fig..(c) using the procedure given in Fig..(b). i x Figure.(b)

45 Network Topology 1 S S Figure.(c) Mesh equations for the given network are i x i 1 i i x +1 (i 1 i ) dt e 1t i i.i x i.1e 1t i x +(i i )+1 1 Di The node equations for the dual network are v x v 1 v v x +1 (v 1 v ) dt e 1t v v.v x v.1e 1t v x +(v v )+1 1 Dv Exercise Problems E.P.1 Consider the bridge circuit of Fig. E.P..1. Using node d as the datum, determine the graph, select a tree, find the cut set equations, and determine v a.. Figure E.P..1 Ans:.8 V

46 1 j Network Theory E.P. Identify the 1 trees in the graph of Fig. E.P... Figure E.P.. E.P. Refer the network shown in Fig. E.P.. The ohmic values also represent the branch numbers. Form a tree with tree branches,, and find the various branch currents using the concept of tie-set and cut-set matrices. Figure E.P.. Ans: I 1 :11 A I :9 A I : A I 8:11 A I : A I : A

47 Network Topology j 1 E.P. Refer the network shown in Fig.E.P... Find the current supplied by the 1 V battery and the power dissipated in the 1-ohm resistor (connected across a b) by using tie-set and cut-set matrices. Figure E.P.. Ans: 1. A,.9 W E.P. Refer the network shown in Fig. E.P... Find the power dissipated by the Ω resistor by constructing tie-set and cut-set matrices. Figure E.P.. Ans:. KW

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