PART AUNIT 1:Basic Concepts: Practical sources, Source transformations, Network reduction using Star Deltatransformation, Loop

and node analysis With linearly dependent and independent sources for DCand AC networks, Concepts of super node and super meshUNIT 2:

7 HoursNetwork Topology:

Graph of a network, Concept of tree and co-tree, incidence matrix, tie-set,tie-set

and

cut-set

schedules,

Formulation of

equilibrium

equations

in matrix

form,

Solution ofresistive networks, Principle ofUNIT 3:

duality.

7 HoursNetwork Theorems 1:

Superposition, Reciprocity and Millmans theorems

6 HoursUNIT 4:Network Theorems - II:Thevinins and Nortons theorems; Maximum Power transfer theorem

6 HoursPART BUNIT 5: Resonant Circuits: Series and parallel resonance, frequency-response of series andParallel circuits, Q factor, Bandwidth.6HoursUNIT 6:Transient

behavior

and

initial

conditions:

Behavior

of circuit

elements

under

switchingcondition and their Representation, evaluation of initial and final conditions in RL, RC and RLCcircuits for AC and DC excitations.UNIT 7: Laplace

Transformation &

Applications

: Solution of

networks,

step,

ramp

and

7 Hoursimpulseresponses, waveform

Synthesis7 HoursUNIT 8:Two port network parameters: Definition of z, y, h and transmission parameters, modeling with these parameters, relationship between parameters setsTEXT BOOKS:

6 Hours1. Network Analysis, M. E.2002.

Van Valkenburg, PHI / Pearson Education, 3rd Edition. Reprint2. Networks and systems, Roy Choudhury, 2nd edition, 2006 re-print, New Age InternationalPublications.REFERENCE BOOKS:1. , Engineering Circuit Analysis, Hayt, Kemmerly and DurbinTMH 6th Edition, 20022. Network analysis and Synthesis,International Edition,

Franklin F. Kuo, Wiley3. Analysis of Linear Systems, David K. Cheng, Narosa Publishing House, 11th reprint, 20024. Circuits,

Bruce Carlson, Thomson Learning, 2000. Reprint 2002Question Paper Pattern: Student should answer FIVE full questions out of 8 questions to be set each carrying 20 marks, selecting at least TWO questions from each part.Coverage in the Texts:UNIT 1:

Text 2: 1.6, 2.3, 2.4 (Also refer R1:2.4, 4.1 to 4.6; 5.3, 5.6; 10.9 This book gives conceptsof super node and super mesh)UNIT 2: Text 2: 3.1 to 3.11UNIT 3 and

UNIT 4: Text 2 7.1 to 7.7UNIT 5: Text 2 UNIT 6: Text 1 UNIT 7: Text 1 UNIT 8: Text 1

8.1 to 8.3Chapter 5;7.4 to 7.7; 8.1 to 8.511.1 to 11.INDEX SHEETSL.NOTOPICPAGE NO.

1University syllabus2-3

UNIT 1: Basic Concepts5-14

01Assignment Questions15-17

UNIT - 2: Network Topology18-37

01Assignment Questions38-39

UNIT - 3: Network Theorems 140-47

01Assignment Questions48-49

UNIT - 4: Network Theorems - II50-58

01Assignment Questions59-60

UNIT - 5: Resonant Circuits61-69

01Assignment Questions70-71

UNIT - 6: Transient behavior and initial conditions72-83

01Assignment Questions84-85

UNIT 7: Laplace Transformation & Applications86-95

01Assignment Questions96-97

UNIT 8: Two port network parameters98-110

01Assignment Questions111-112

Unit: 1 : Basic

Concepts

Hrs: 07Syllabus of unit 1:Practical

sources,

Source

transformations,

Network

reduction

using

Star

Deltatransformation,

Loop

and

node analysis

With linearly

dependent

and

independent

sourcesfor

DC and

AC networks Concepts of super node and super mesh.Recommended readings:1. Network Analysis,

M. E.

Van Valkenburg, PHI / Pearson Education2. Networks

and

systems,

Roy

Choudhury,

2 edition,

New Age InternationalPublications .3. Network theory , Ganesh Rao.4. Network analysis , Roy Choudry.BASIC LAWS:1. OHMS LAW

V=IZ

A IAB+

Z B VAB -IAB-Current from A to B

VAB=Voltage of A w.r.t B2. KCL i1i1+i4+i5=i2+i3 i2 i=0

algebraic sumi3 or

iin= iiout

( Iin=-Iout)i4i53. KVL

V2 v=0I2 - +

algebraic sum- Z2 + vrise= vdrop

V1 Z1 - E2

(Vrise= -Vdrop)+ I1E1 + Z3 V3- + -Z4 I3I4 V4Reference DirectionE1-E2=V1-V2+V3-V4=I1Z1-I2Z2+I3Z3-I4Z4CONNECTIONSSERIESPARELLEL

+ V - + V2 - + Vn - I Z1 Z2 Zn+ V -nZ Z K Z1 Z 2 Z 3 Z n1Voltage DivisionVi=(Zi/Z)V+ IV Y1 Y2 Yn- I1 I2 InnY YK Y1 Y2 Y3 Yn1Current DivisionII=(Yi/Y)I

I=V/Z=V1/Z1=V2/Z2=--------------

V=I/Y=I1/Y1=I2/Y2=-------------Problems1.Calculate the voltages V12,V23,V34and VC=15 -100V

in the network shown in Fig,

if Va=17.32+j10 Vb=30

80 0 Vwith Calculator in complex and degree modeV12 = -Vc + Vb 3= (0-15 -100

+30 80

) = 45

800 V * +V23 = Va-Vb+Vc = Va V12= 17.32+10i- 45 800 = 35.61 -74.520

- Va

+ Vc -V34 =

Vb - Va

= 30 80

- 17.32-10i = 23

121.780

1 2 4+ Vb -2.How is current of 10A shared by 3 impedances Z1=2-J5 Z3 = 3 + J4 all connected in parallelAns:

Z2 = 6.708

26.56 andZ = Y-1 = ((2-5i)-1 + (6.708 26.5)-1 +(3+4i)-1 = 3.06

9.550V=1Z =

30.6 9.550 I1 = V/Z1=(30.6

9.550 )

: (2-5i) =5.68

77.750I2 = V Z2

= (30.6 9.550 )

: (6.708 26.56)

= 4.56 -170 I3 = V Z2

= (30.6 9.550 )

: (3+4i) = 6.12 -43.603. In the circuit determine what voltage must be applied across AB in order that a currentof 10 A may flow in the capacitorA

I1 5I2 7

6 8C8

10 BVAC=

(7-8i)(10) = 106.3

-48.80I1 = VAC5+6i

= 13.61

-990 I = I1+I2

= 10 00

+13.61

-990

= 15.576

-59.60V =V1+V2

= 106.3 -48. + (15.576 -59.6) (8+10i)=289 -220Practical

sources:Network

is a system with interconnected electrical elements. Network and circuit are the same.The only difference being a circuit shall contain at least one closed path.Electrical ElementsSources

Passive ElementsIndependentDependantRLC

SourcesSources(Energy(Energy storing(Energy

MNConsumingelement in astoring

Element)magnetic field)element in an

Electric field)

.

Voltage Source(ideal)

Current Source(ideal)

kix

gvx

+kvx

+vix- -B E + A A I

B (a)

(b)

(c)

(d)(a)Current controlled current source (b) Voltage controlled current source (c) Voltage controlled voltage source (d) Current controlled voltage sourceM(Value of source

N(Source quantity is determined by a voltageQuantity is not affected

or current existing at

somein anyway by activities in the reminder of the circuit.)

other Location in the circuit)These appear in the equivalent models for many electronic devices like transistors, OPAMPS and integrated circuits.Practice Voltage source

Mesh (loop)

Loop

Practical current source

Reference nodeTYPES OF NETWORKSLinear and Nonlinear Networks:A network is linear if the principle of superposition holds i.e if e1(t), r1 (t) and e2(t),r2 (t) are excitation+ r2(t).

and response pairs then if excitation is e1 (t) + e2 (t) then the response is r1 (t)The network not satisfying this condition is nonlinearEx:- Linear Resistors, Inductors, Capacitors.Nonlinear Semiconductors devices like transistors, saturated iron core inductor, capacitance of a p-n function.Passive and active Networks:A Linear network is passive if (i) the energy delivered to the network is nonnegativefor any excitation. (ii)excitation is applied.

no voltages and currents appear between any two terminals before anyExample:-

R,L and C.Active network:- Networks containing devices having internal energy Generators, amplifiers and oscillators.Unilateral & Bilateral:The circuit, in which voltage current relationship remains unaltered with the reversalof polarities of the source, is said to be bilateral.Ex:- R, L & CIf V-I relationships are different with the reversal of polarities of the source, thecircuit is said to be unilateral.

Ex:- semiconductor diodes. elements.

Lumped & Distributed:Elements of a circuit, which are separated physically, are known as lumped

Ex:- L & C.elements.

Elements, which are not separable for analytical purposes, are known as distributedEx:- transmission lines having R, L, C all along their length.In the former care Kirchhoffs laws hold good but in the latter case Maxwells laws are required for rigorous solution.Reciprocal:A network is said to be reciprocal if when the locations of excitation and responseare interchanged, the relationship between them remains the same.Source Transformation:In network analysis it may be required to transform a practical voltage source into its equivalentpractical current source and vice versa .

These are done as explained belowZS a aES ZL IS ZP ZLfig 1

b

b fig 2Consider a voltage source and a current source as shown in Figure 1 and

2. For the sameload ZL across the terminals a & b in both the circuits, the currents

areIL= ES

in fig 1andIL = IS .Z P in fig 2

Z s+ZLZ p + ZL

For equivalence= ES

= IS . Z P

ZS+ZL Z P+ZLTherefore ES = IS Z P

and ZS = Z PTherefore

IS = ES

= ES

Z P ZSTransformation from a practical voltage source to a practical current source eliminates a node.Transformation from a practical current source to a current source eliminates a mesh.A practical current source is in parallel with an impedance Zp is equivalent to a voltage sourceEs=Is Zp in series with

Zp.A practical voltage source Es in series with a impedance ZsEs/Zs in parallel with Zs.

is equivalent to a current sourceSOURCE SHIFTING:Source shifting is occasionally used to simplify a network. This situation arises because of the factthan an

ideal voltage source cannot be replaced by a current source. Like wise ideal current sourcecannot be replaced by a voltage source. But such a source transformation is still possible if the following techniques are fallowed.c cZ3 Z3+ x a Z1 x Z2 b Ea Z1 +

Z2 b- - ++ x E E xE_-O O(a) E shift operationIZ1 Z1 Z1I Z2 Z4Z3

Z4I Z2Z3

Z2 Z4 IZ3II(b) I shift operationSources with equivalent terminal characteristics+V1 -+V2 -

+ + V1+V2 V1 -

+- V2=V1

+- v1=v2

(i)Series voltage sources

(ii) Parallel voltage sources(ideal)i1i1i1 i2 i1+i2 i1=i2(iii) Parallel current sources

(iv)Series current sources(ideal)Z+ +V z V I I- Z -(v)Voltage source with parallel Z

(vi)Current source with series Z-

+ V+ +V - I V I I-(vii) V and I

in Parallel

(viii) V and I in SeriesDelta-star transformation:A set of star connected (Y o:r T) immittances can be replaced by an equivalent set of mesh (' or ) connected immittances or vice versa. Such a transformation is often necessary to simplify passive networks, thus avoiding the need for any mesh or nodal analysis.For equivalence, the immittance measured between any two terminals under specified conditions must be the same in either case. DELTA to Y transformation:Consider three -connected impedances ZAB, ZBC and ZCA across terminals A, B and C. It is required to replace these by an equivalent set ZA, ZB and ZC connected in star.A ZABB

ZACC ZBC

A ZAC ZCB ZBIn ', impedance measured between A and B with C open isZAB (ZBC + ZCA) ZAB+ ZBC + ZCA

With C open, in Y, impedance measured between A and B is

ZA+ZB.For equivalence ZA+ZB = ZAB (ZBC + ZCA) ZAB+ ZBC + ZCA

-----------(1)Similarly for impedance measured between B and C with A openZB + ZC =

ZBC (ZCA + ZAB) ZAB+ ZBC + ZCA

-----------------------------(2)For impedance measured between C and A with B open ZCA (ZAB + ZBC)ZC + ZA =

ZAB+ ZBC + ZCA

--------------------------------(3)Adding (1),

(2) and (3)2 (ZA+ ZB + ZC) = 2 (ZAB ZBC + ZBCZCA + ZCAZAB) ZAB+ ZBC + ZCA

(ZAB ZBC + ZBCZCA + ZCAZAB)

ZA =

ZAB+ ZBC + ZCA

- (ZB + ZC)Substituting for ZB + ZC from (2)ZA = ZCA ZAB = ZAB+ ZB + ZCA

Z CA Z AB Z ABSimilarly by symmetry

ZB =ZC =

ZAB ZBC ZAB ZBC ZCA ZABIf ZAB = ZBC = ZCA = Z

then

ZA =

ZB = ZC = ZY = Z .3Y to transformation:Consider three Y connected admittance Ya, Yb and Yc across the terminals A, B and C.

It isrequired to replace them by a set of

equivalent admittances Yab, Ybc and Yca.Admittance measured between A and B with B & C shortedIn Y

YA (YB + YC) YA+ YB + YCIn

YAB + YCAA YABB

YACYBC

A YAC

C YCYBB YA (YB + YC)For equivalence YAB + YCA = YA+ YB + YC

-------------------------(1)Admittance between B and C with C & A shortedYBC + YAB = YB (YC + YA) YA+ YB + YC

------------------------------------(2)Admittance between C and A with A & B shortedYCA + YBC = YC (YA + YB) YA+ YB + YC

---------------------------------(3)Adding (1), (2) and (3) YAB + YBC + YCA = YAYB + YBYC + YCYA

YA+ YB + YCYAB = YA YB - (YBC + YCA) YASubstituting from (3)

YA YB

=

YA YB : YA+ YB + YC

YBC =

YB YCYA+ YB + YC

: YCA =

YA+ YB + YCIn terms of impedances,

ZA ZB + ZBZC + ZCZAZAB = YA + YB + YC =YA YB ZCSimilarly ZBC = ZA ZB + ZBZC + ZCZA ZAZCA = ZA ZB + ZBZC + ZCZA

ZBIf ZA = ZB = ZC = ZY

then

ZAB =

ZBC = ZCA = Z = 3ZY .Assignment questions:1) Distinguish the following with suitable examples. i) Linear and non-linear elements.ii) Unilateral and trilateral elements.iii) Independent and dependent sources.2) Write the mesh equation for the circuit shown in Fig. 1 and determine mesh currents usingmesh analysis.i .3) Establish star delta relationship suitably4) Using source transformation, find the power delivered by the'50 Vin the circuit shown in Fig.

voltage source

5) Find the power delivered by the 5A current source in the circuit shown in Fig.4 by using the nodal method.

6) For the network shown in fig. 2, determine the voltage V using source shift and / or source transformation techniques only. Then verify by node equations7) Use mesh current method to determine the current in the capacitor of 6: of the bridge networkshown in fig. 5.

8) Use node equations to determine what value of E will cause Vx to be zero for the network shown in fig. 6.9) Obtain the delta connected equipment of the network shown in fig. 1.

10) c) Find the voltage across the capacitor of 20 : reactance of the network shown in fig. 2, by reducing the network to contain one source only, by source transformation techniques.11) c) Use 3 mesh equations for the network shown in fig. 5 to determine R and C. such that the current in 3+j4: is zero . Take Z = 50 rad/sec.

Unit: 2

Network

Topology: Hrs: 07Syllabus of unit :Graph

of a

Network,

Concept

of tree

and

Co-tree, incidence

matrix,

tie-set,

tie-set

andCut-set resistive

schedules, networks,

Formulation

Of equilibrium

equations in

matrix

form,

Solution ofPrinciple of

dualityRecommended readings:1. Network Analysis,

M. E.

Van Valkenburg, PHI / Pearson Education2. Networks

and

systems,

Roy

Choudhury,

2 edition,

New

Age

InternationalPublications .3. Network theory , Ganesh Rao.4. Network analysis , Roy Choudry.Net work TopologyDefinition:The term circuit topology refers to the science of placement of elements and is a study of the geometric configurations.Circuit topology is the study of geometric properties of a circuit useful behavior

for describing the circuitTerms used in Topology:The following terms are often used in network topologyGraph:In the given network if all the branches are

represented by line segments then the resulting figureis called the graph of a network (or linear graph). The internal impedance of an ideal voltage source is zero and hence it is replaced by a short circuit and that of an ideal current source is infinity and hence it is represented by an open circuit in the graph.Example:

Network

GraphNodeIt is a point in the network at which two or more circuit elements are joined.2, 3 and 4 are nodes.

In the graph shown 1,Branch (or Twig):It is a path directly joining two nodes.

There may be several parallel paths between two nodes.Oriented GraphIf directions of currents are marked in all the branchesof a graph then it is called an oriented (or directed) graph .Connected graphA network graph is connected if there is a path between any two nodes .In our

furtherdiscussion,let us

assume that the graph is connected. Since, if it is not connected each disjoint partmay be analysed separately as a connected graph.1 24 3Unconnected graphIf there is no path between any two nodes,then the graph is called an unconnected graph.1 2 4

1 2 43 53 5 3 5Planar graphA planar graph is a graph drawn on a two dimensional plane so that no two branches intersect at apoint which is not a node.B C A BA ED D CNon planar graphA graph on a two dimensional plane such that two or morebranches intersect

at a point other than node on a graph.Tree of a graphTree is a set of branches with all nodes not forming any loop or closed path.(*)

Contains all the nodes of the given network or all the nodes of the graph(*) No closed path(*) Number of branches in a tree =

n-1 ,

where n=number of nodesA 2 B

C A B C2 45 6 5 D D Two possible treesGraphCo- treeA Co- tree is a set of branches which are removed so as to form a tree or in other words, a co- treeis a set of branches which when added to the tree gives the complete graph. Each branch so removed is called a link.Number of links = l = b (n-1)

Where

b = Total number of branchesn = Number of nodesIncidence MatrixIncidence matrix is a matrix representation to show which branches are connected to which nodesand what is their orientation in a given graph(*) The rows of the matrix represent the nodes and the columns graph.(*) The elements of the incidence matrix will be +1, -1 or zero

represents the branches of the(*) If a branch is connected to a node and its orientation is away from the node the corresponding element is marked +1(*) If a branch is connected to a node and its orientation is towards the node then the corresponding element is marked 1(*) If a branch is not connected to a given node then the corresponding element is marked zero.Incidence MatrixComplete Incidence matrix

Reduced incidence matrix(i)Complete incidence matrix:An incidence matrix in which the summation of elements in any column in zero is called a complete incidence matrix.(ii)Reduced incidence matrix:The reduced incidence matrix is obtained from a complete incidence matrix by eliminating arow.

Hence the summation of elements in any column is not zero.Example 1:

Consider the following network and the oriented graph as shownNetwork

Oriented graph(*)

There are four nodes

A, B, C and D and six branches 1, 2, 3, 4, 5 and 6.

Directions of currents are arbitrarily chosen.(*) The incidence matrix is formed by taking nodes as rows and branches as columnsNodesBranches

123456

A-1+1+1000

B0-10-1+10

C00-1+10+1

D+1000-1-1

P =-111000

0-10-110

00-1101

1000-1-1

In the above example the fourth row is negative of sum of the first three rows. Hence the fourth canbe eliminated as we know that it can be obtained by negative sum of first three rows. this we get the reduced incidence matrix.

As a result ofPR =Properties of a complete incidence matrix(*)

Sum of the entries of each column is zero(*)

Rank of the matrix is (n-1), where

n is the no of nodes(*)

Determinant of a loop of complete incidence matrix is always zeroExample

2 : The incidence matrix of a graph is as shown.

Draw the corresponding graph.Solution : The sum

of each column of the

1 0-1 1

0 0 1 -1 1 0 0 0given matrix is zero.

Hence it

0 -1 0 -1 0 1is a compete incidence matrix.

0 0 -1 1 -1 0Number of nodes

= n = 4

(say A. B. C

and D)Number of branches

= b = 6

(say 1, 2, 3,

4, 5 and 6)Prepare a tabular column as shown.NodesBranches

123456

A10001-1

B-111000

C0-10-101

D00-11-10

From the tabular column, the entries have to be interpreted as follows:From the first column the entries for A and B are ones

. Hence branch 1

is connected betweennodes A and B .

Since for node

A entry is

+1 and for node B it is

-1,

the current leaves

node Aand enters node B and so on.From these interpretations the required graph is drawn as shown.6

A B C1 25 3 4DExample 3:

The incidence matrix of a graph is as shown.

Obtain the corresponding graph 110 0 000

0-11 1 000

000 -1 110

0 0 0

0 0 -1 1Solution:-

Given incidence matrix

is a reduced incidence matrix as the sum of each column is notzero. Hence it is first converted in to

a complete incidence matrix by adding the deleted row.

Theelements of each column of the new row is filled using the fact that sum of each column of a complete incidence matrix is zero.In the given matrix in first, third, fifth and the seventh

Column the sum is made zero by adding 1in the new row and the corresponding node is E. also the graph for the matrix are as shown.

The complete incidence matrix so obtained

andNodesBranches

1234567

A1100000

B0-111000

C000-1110

D00000-11

E-10-10-10-1

Graph:6 7 22ETie set Analysis:In order to form a tree from a network several branches need to be removed so that the closed loopsopen up. All such removed branches are called links and they form a Co-tree.

Alternatively when alink is replaced in a tree, it forms a closed loop along with

few of the tree branches.

A current canflow around this closed loop.

The direction of the loop current is assumed to be the same as that ofthe current in the link. set.

The tree branches and the link that form a loop is said to constitute a tie DefinitionA tie set is a set of

branches contained in a loop such that the loop has at least one link and theremainder are twigs (tree branches)6 6z

4B54B5

ACAC

x23 1 3D D Graphtree (In thick lines) Co-tree (In dotted lines)We see that by replacing the links 1, 4 and 5 three loops are formed and hence three loop currentsx, y and z flow as shown.

The relationships obtained between loop currents, tree branches andlinks can be scheduled as followsTie set scheduleTie set

Tree branches

Link

Loop current12,35x

21,42y

34,56z

Tie set matrix (Bf)The Tie set schedule shown above can be arranged in the form of a matrix where in the loop currents constitute the rows and branches of the network constitute the columns Entries inside thematrix are filled by the following

procedure :Let an element of the tie set matrix be denoted by mikThen

mik =

1 Branch

K is in loop i and their current directions are same-1 If branch K is in

loop i and their current directions are opposite.If the branch K is not in loop iBy following this procedure we get the Tie set matrix which is shown below:LoopcurrentssBranches

123456

x0+1+10+10

y+1+10+100

z000+1-1-1

OrBf =

0 1 1 0 1 01 1 0 1 0 00 0 0 1

-1 -1Analysis of the net work based on Tie set scheduleFrom the tie set schedule we make the following observations.(i) Column wise addition for each column gives the relation between branch and loop currentsThat is

i1 =i2 =

yx + yi3 =i4 =

xy+ z

i5 =

x - zi6 = - zPutting the above equations in matrix form, we geti1 0 1 0 i2i3 1 0 0 i5i6

1 1 041 0 -1 z0 0 -1

0 1 1 yIn compact form IB

= B fT ILWhere

IB =

Branch current matrixB fT IL

= Transpose of the tie- set matrix= Loop current column matrix(ii)Row wise addition for each row gives the KVL equations for each fundamental loopRow - 1 : Row - 2 :

V1 + V2 + V3 = 0V1 + V2 + V4 = 0Row - 3 :

V4 - V5 - V6 = 0V10 1 11 1 0

0 1 0 V21 0 0 V30 0 0

1 -1 -1 V4 = 0 V5V6In compact form

B f VB

= 0 - - -

(1)Where

VB =

Branch

voltage column matrix

andB f =

Tie

- set matrixAExample: For the network shown in figure, write aTie-set schedule and then find all the branch 50Currents and voltages VC

5 510 B10 5DSolution:1

The graph and one possible tree is shown: 5A25B4 6C 3 D

1 5x y 24 6 z3Loop

Branch NumbersTie set Matrix: Bf =10 01-1 0

10 01-1

01 -101

Branch Impedence matrix

ZB =

5 0 0 0 0 00100 0 00

005 0 00

000 10 00

000 0 50

000 0 05

ZL =1 0 0 1 -1 05 0 0 0 0 01 0 0

0 1 0 0 1 -10 10 0 0 0 00 1 0

0 0 1 -1 0 10 0 5 0 0 00 0 1

0 0 0 10 0 01 0 -1

0 0 0 0 5 0-1 1 0

0 0 0 0 0 50 -1 1

ZL20 -5 -10= -5 20 -5

-10 -5 20Loop Equations :

ZL IL

= - Bf Vs

-5020 -5 -10 x

1 0 0 1

- 1 0 0-5 20 -5-10 -5 20

y = - z

0 1 00 0 1

0 1 -1 0-1 0 1 00020x-5y-10z =50-5x+20y-5z = 0-10x-5y+20z =0Solving the equations, we get

x = 4.17 Amps y = 1.17 AmpsAnd

z = 2.5

AmpsCut set AnalysisA cut set of a graph is a set of branches whose removal , cuts the connected graph in to two

partssuchA

that the replacement of any one branch of the cut set renders the two parts connected.AExample

1 54 B 3 D 4 B D2 6 3C CDirected graph

Two separate graphs created by

the cut set

(1, 2,

5, 6)Fundamental cut set is a cut set that contains only one tree branch and the others are linksFormation of Fundamental cut set(*) (*) (*)

Select a treeSelect a tree branchDivide the graph in to two sets of nodes by drawing a dotted line through the selected treebranch and appropriate links while avoiding interruption with any other tree branches.Example 1 : A

For the given graph write the cut set

schedule1 5 A4 B 3 D 1 52 4 B 3 D6 2 6C The fundamental cut set of theSelected tree is shown in figure CNote

that

FCS - 1

yields

node A and the set of nodes ( B,

C, D)The Orientation of the fundamental cut set is usually assumed to be the same as the orientation ofthe tree branch in it,

Which is shown by an arrow.

By following the same procedure the FCS- 2and FCS -3

are formed as shown below:A A1 51 54 B 3 D B 32 D2 4 66C CFCS -2

FCS -3It should be noted that for each tree branch there will be a fundamental cut set.

For a graphhaving n number of nodes the number of twigs is (n-1).Therefore there will be (n-1) (n-1) fundamental cut-sets.Once the fundamental cut sets are identified and their orientations are fixed, it is possible to write a schedule, known as cut set schedule which gives the relation between tree branch voltages and all other branch voltages of the graph.Let the element of a cut set schedule be denoted by

Qik

then,Qik = 1

If branch K is in cut set I and the directionof the current in the branch

K is same as cut set direction.-1 If

branch

k is in cut set I and

the direction of the current in

k isOpposite to the cut set direction.Cut set

Schedule

O If

branch is k is not in cut set iTree branch voltageBranch Voltages

123456

e1100-1-10

e2010101

e300101-1

The elements of the cut set schedule may be written in the form of a matrix known as the cut setmatrix.Qf =Analysis of a network using cut set schedule(*)

Column wise addition of the cut set schedule gives the relation between

tree branchvoltage and the branch voltages for the above

cut the scheduleV1 = e1V2 = e2V3 = e3V4 = - e1 + e2V5 = - e1+ e3In matrix formV6 =e2 -e3

V1100

V2010e1

V3001e2

V4=-110e3

V5-101

V601-1

In compact formVB = QT VT

. (1)

Where

VB =Branch voltage MatrixQTf =

Transpose of cut set matrixand

VT = Tree branch voltage matrix.(*)

Row wise addition given KCL at each nodeI1 - I4 -I5=0

I2 + I4 +I6=0

I3 + I5 +I6=0

In matrix

form100-1- 1 0I1

01010 1I2

00101 -1I3=0

I4

I5 I6

In compact form

Qf IB =

0 (2) Where Qf

= cut set matrixIB =

Branch current matrix(*)

Let us consider a network having b branches.

IK VKEach of the branches has a representation as

shown inReferring to the figure Ik =

Yk Vk + Isk

ISK

Since the network has b branches, one such equation could be written for every branch

of thenetwork ie

I1 =

Y1 V1 + Is1I2 =

Y2 V2 + Is2------------------------------

Ib =

Yb Vb + IsbPutting the above set of equations in a matrix form we getIB =

YB VB + Is

(3)IB = Branch current matrix of order bx1YB =VB =

Branch Voltage

column matrix of order bx1

andIB =

Source current matrix of order bx1.Substituting

(3)

in (2)

we get

Qf YB VB

+ IS = 0But from (1) we have VB =

Qf YB VBQf T VT

+ Qf IS = 0Hence equation (4) becomesQf YB QTf VT + YC VT +

Qf IS = 0QfIS = 0or YC VT = - Qf IS

.

(5)Where YC = Qf YB Qf T

is called cut

- set

admittance matrixAction Plan for cut set analysis.(*) Form the cut- set

matrix Qf(*) Construct the branch admittance matrix YB(*) Obtain the cut set admittance matrix using the equation YC

= Qf YB Qf T(*) Form the KCL or equilibrium equations using the relation

YC VT = - Qf ISThe elements of the source current matrix are positive if the directions of the branch current and the source connect attached to that branch are same otherwise negative.(*) (*)

The branch voltages are found using the matrix equation VB = Qf T VTFinally the branch currents are found using the matrix equation IB = YB VB

+ ISExample 2A51D4

: For the directed graph obtain the cut set matrix6E B28 3 7CSolution :

The tree (marked by thick lines)and the link (marked by doffed lines)are as shown.The fundamental cut sets are formed at nodes A B

C and D keeping

E as reference nodeFcs- 1--( 1,5,6)AFCS-1

Fcs- 2--(2,6,7 )1

FcsFcs- 3- 4----( 3,(4,7,5,8)8)D FCS44 E 2BFCS2

FCS3 C

Hence the cut set schedule is as follows:Tree branch voltageBranch

12345678

e11000-1100

e201000-110

e3001000-11

e40001100-1

Hence the required cut set matrixQf =1000-1100

01000-110

001000-11

0001100-1

Example

3: Find the branch voltages using the concept of cut-sets11 11 11V 1Solution :

The voltage source is transformed in to an equivalent current source. It should be notedthat all the circuit Passive elements must be admittances and the net work should contain only current sources.The graph for the network is shown. (Shown by dotted lines) are shown

A possible tree (shown with thick lines) and co tree1mhoFCS 1 =

3, 1, 5

1mho

1 mhoFCS 2 =

4, 2, 5

1mho

1mho

1mhoFCS 3 =

6, 1, 21ACut set schedule5 FCS23 4A 8FCS1 61 2FCS3Qf =YB =-1 0 101-1

-1 0 1 0 -101 0 0 0 00100

0 1 0 0 00010

YC=0 1 0 1 100 0 1 0 00-110

1 -1 0 0 010 0 0 1 00001

0 0 0 0 10

0 0 0 0 01

3-1-1

=-13-1

-1-13

Equilibrium Equations

; YC VT = - Qf IS3 -1-1e3-1 0 1 0 -1 0-1

-1 3-1e4=- 0 1 0 1 1 00

-1 -13e61 -1 0 0 0 10

0

0

0

3 e3 e4 e6 = - 1-e3 + 3 e4 e6 = 0- e3 e4 + 3e6 = 1Solving we get

e3=-0.25 volt e4=0e6=0.25 voltDUALITY CONCEPTTwo electrical networks are duals if the mesh equations that characterize one have the same mathematical form as the nodal equations of the other.Example 1Consider an

R-L-C

series

network

excited

by avoltage

so source

V as

shown in

the

figure.

Theequation generatingthe circuit behavior is Ri+Ldi +1 idt

=V ..(1)Figure 1

dt CNow

consider

the

parallel

G-C-L

networkfed by a Current Source i is shown in the figure. The equation generating theCircuit behavior is GV+ CdV +1 vdt = i..(2)dt L

Figure 2Comparing the equations (1) and (2),we get the similarity between the networks of fig(1) and fig(2).The solution of equation (1) will be identical to the solution of equation (2) when thefollowing

exchanges are madeR G, L C, CL and V iHence networks of figure (1) and (2) are dual to each other.Table of dual Quantities1.Voltage Source2.Loop currents3.Iductances4.Resistances5.Capacitances6. On KVL basis7.Close of switch

Current source Node voltages Capacitances Conductances InductancesOn KCL basisOpening of switchNote:Only planar networks have duals.Procedure for drawing dual networkThe duals of planar networks could be obtained by a graphical technique known as the dot method. The dot method has the following procedure.Put a dot in each independent loop of the network. These dots correspond to independent nodes in the dual network.Put a dot outside the network. This dot corresponds to the reference node in the dual network. Connect all internal dots in the neighboring loops by dotted lines cutting the common branches.These

branches

that

are

cut

by dashed

lines

will

form

the

branches

connecting

thecorresponding independent nodes in the dual network.Join all internal dots to the external dot by dashed lines cutting all external branches. Duals ofthese node.

branches will form the

branches connecting the

independent nodes and the

referenceExample 1:

Draw the exact dual of the electrical circuit shown in the figure.Solution: Mark two independent nodes 1 and 2 and a reference node 0as shown in the figure.3ohm

4F

Join

node

1 and

2 by

a dotted

line

passing2sin6t1

4ohm6H 2

through the inductance of 6H.this element will appear as capacitor of 6 F between node 1 and2 in the dual.Join

node

1 and

reference

node

through adotted line passing the voltage source of 2sin6t volts. This will appear as a current source of2sin6t amperes between node1 and referenceO node.Join node 1 and reference node through a dotted line passing through 3 ohms resistor. This element appears as 3mho conductance between node1 and reference node in the dual.Join node 2 and reference node through a dotted line passing through the capacitor

of 4 Farads.This element will appear as 4 Henry inductor

between node 2 and reference node in the dualJoin node 2 and reference node through a dotted line passing through the resistor of 4 ohms. This element will appear as 4 mho conductance between node 2 and reference node.The Dual network drawn using these procedural steps is shown.

36

Assignment questions:1) Explain incidence matrix of a network of a network graph? Give suitable example.2) Define the following with suitable examples i) Planar and non-planar graphii) Twigs and links3) For the network shown in Fig. 8 write the graph of the network and obtain the tie-set schedule considering J1, J2, J5 as tree branches. Calculate all branches current.

4) Explain briefly trees, cotrees, and loops in a graph of network with suitable example5) Explain with examples the principal of duality6) Draw the oriented graph of the network shown in Fig. 10 selects a tree, write the set schedule and obtain equilibrium equations

7) Under what conditions do you consider topology for network analysis? For the graph shown inFig. 3, for a co- tree (4,5,2,8), write tie set and cut set matrices.

(10)

8) For the network shown in Fig. 4, draw its dual. Write in intergo differential formi) Mesh equations for the given networkV(t) = 10 sin 40t.

ii) Node equations for the dual.9) What are dual networks? What is their significance ? Draw the dual of the circuitshown

in fig. 8 .

10) For the network shown in Fig. 5, perform source shifts, draw a graph, select tree with branches 1,2 and 3 and obtain tie set and matrices.

Unit: 3 Network

Theorems

1: Hrs: 06Syllabus of unit :Superposition,

Reciprocit y

and

Millmans theoremsRecommended readings:1. Network Analysis,

M. E.

Van Valkenburg, PHI / Pearson Education2. Networks

and

systems,

Roy

Choudhury,

2 edition,

New

Age

InternationalPublications .3. Network theory ,

Ganesh Rao.4. Network analysis , Roy Choudry.NETWORK THEOREMSMesh current or node voltage methods are general methods which are applicable to any network. A number of simultaneous equations are to be set up. Solving these equations, the response in all the branches of the network may be attained. But in many cases, we require the response in one branch or in a small part of the network. In such cases, we can use network theorems, which are the aidesto simplify

the

analysis. To

reduce

the

amount of

work

Involved

by considerable

amount, ascompared to mesh or nodal analysis. Let us discuss some of them.SUPERPOSITION THEOREM:The response of a linear network with a number of excitations applied simultaneously isequal to the sum of the responses of

the network when each excitation is applied individuallyreplacing all other excitations by their internal impedances.Here the excitation means an independent source. Initial voltage across a capacitor and the initial current in an inductor are also treated as independent sources.This theorem is applicable only to linear responses and therefore power is not subject to superposition.During replacing of sources, dependent sources are not to be replaced. Replacing an ideal voltage source is by short circuit and replacing an ideal current source is by open circuit.In any linear network containing a number of sources, the response (current in or voltage across an element) may be calculated by superposing all the individual responses caused by eachindependent

source

acting

alone,

with

all

other

independent

voltage

sources

replaced

by shortcircuits

and

all

other

independent

current

sources

replaced

by open

circuits.

Initial

Capacitorvoltages and initial inductor currents, if any, are to be treated as independent sources.To prove this theorem consider the network shown in fig.IaISIS ES

Ia1

We consider only one-voltage sources and only one current sources for simplicity. It is required to calculate Ia with Is acting alone the circuit becomesIS Z1 Z3Ia1 =

Z1 + Z2 + Z3 Z4 Z3 + Z4Z3 + Z4= IS Z1 Z3

(Z1 + Z2 + Z3) Z4 + (Z1 + Z2) Z3

------------------------------------(1)Ia2ESwith ES acting alone -EIa1 =

Z4 + (Z1 + Z2) Z3Z1 + Z2 + Z3= -ES (Z1 + Z2 + Z3) (Z1 + Z2 + Z3) Z4 + (Z1 + Z2) Z3

----------------------------------------(2)Next converting the current source to voltage source, the loop equationsIS Z1 I1 I2 ESI2 =

Z1+Z2+Z3 IS Z1 -Z3 -ES

Z1+Z2+Z3 -Z3-Z3 Z3+Z4= ISZ1Z3 - ES (Z 1+Z2+Z3) (Z1+Z2+Z3) Z4 + (Z1+Z2) Z3

--------------------------------- (3)From equation (1), (2) and (3)Hence proof

Ia1 + Ia2 = I2 = IaReciprocity Theorem :In an initially relaxed linear network containing one independent

source only.

The ratioof the

response

to the

excitation

is invariant

to an

interchange

of the

position

of theexcitation and the response. i.e if a single voltage source Ex in branch X produces a current response Iy

the branch Y,then the removal of the voltage source from branch x and its insertion in branch Y will produce the current response Iy in branch X.Similarly if the single current source Ix

between nodes X and X produces the voltageresponse Vy between nodes Y and Y then the removal of the current source from X and X and its insertion between Y and Y will produce the voltage response Vy between the nodes X and X.Between the excitation and the response, one is voltage and other is current.

It should benoted that after the source and response are interchanged, other parts of the network will not remain the same.

the current and the voltages inProof :Z1 Z2I1+AE Z3_-Z4Consider a network as shown in which the excitation is E and the response isI in Z4. The reading of the ammeter isEI1 = .Z1 + Z3 ( Z2 + Z4)

Z3Z2+Z3+Z4Z2+Z3+Z4E Z3I1 =

Z1 ( Z2+Z3+Z4) + Z3(Z2 + Z4)

(1)Next interchange the source and ammeter.EZ1 Z2Z4AZ3 +I2 E_Now the reading of the Ammeter is :E Z3I2 = .( Z2 + Z4)

+ Z1 Z3 Z1+Z3Z1 + Z3E Z3I2 =

Z1 ( Z2+Z3+Z4) + Z3(Z2 + Z4)

(2)From (1) & (2)I1 = I2It can be similarly be shown for a network with current sources by writing node equations.Transfer Impedance :The transfer impedance between any two pairs of terminals of a linear passive network is the ratio of the voltage applied at one pair of terminals to the resulting current at the other pair of terminals .With this definition the reciprocity theorem can be stated as :Only one value of transfer impedance is associated with two pairs of terminals of a linear passive network .I2Z1 a c+ E1

-b d

I2Z1 a c+ I1_Z2Z2b dw.r.t figs shown E1 =

E2 = ZTI2 I1If E1 = E2

then

I1 =

I2.Millmans Theorem:Certain simple combinations of potential and current source equivalents are of use because theyoffer

simplification

in solutions of

more

extensive

networks

in which

combinations

occur.Millmans

Theorem

says

that

if a

number

of voltage

sources

with

internal

impedances

areconnected in parallel across two terminals, then the entire combination can be replaced by a single voltage source in series with single impedance.The single voltage is the ratioSum of the product of individual voltage sources and their series admittancesSum of all series admittancesand the single series impedance is the reciprocal of sum of all series admittances.E1 Z1E2 Z2E3 Z3En ZnLet E1, E2.En be the voltage sources and Z1, Z2Zn are their respectiveimpedances.

All

these

are

connected

between A & B

with

Y=1/Z,

according to

MillmansTheorem, the single voltage source that replaces all these between A & B isnEAB = EK YK K=1

n YK K=1

AndThe single impedance is

Z = 1

n YK K=1

Proof: Transform each voltage into its equivalent current source.E1/Z1Z1B E2/Z2 A Z2

Then the circuit is as in Fig.En/ZnZnWith Y=1/Z the circuit is simplified as

E1 Y1+E2 Y2 +..EnYn=

EKYKB AY1+ Y2 +..Yn=

YKWhich is a single current source in series with a single admittance?Retransforming this into the equivalent voltage sourceEY Y

Z= 1/YA B- +The theorem can be stated as If a number of current sources with their parallel admittances are connected in series between terminals A and B, then they can be replaced by a single current source in parallel with a single admittance. The single current source is the ratioSum of products of individual current sources and their impedancesSum of all shunt impedancesAnd the single shunt admittance is the reciprocal of the sum of all shunt impedances.Let

I1, I2,

..In be

the

n number of

current

sources

and

Y1,Y2..Yn be

theirrespective shunt admittances connected in series between A & B.

Then according to MillmansTheorem they can be replaced by single current I AB in parallel with a single admittance Y AB whereIAB=And

IKZKYAB= 1 ZK

ZKI1 I2 InA BY1 Y2 YnTransform each current source into its equivalent voltage source to get the circuit as in fig A B- + - + - +I1/Y1 Y1 I2/Y2 Y2 In/Yn Yn6I Z= I1Z1+I2Z2

Z +Z2+..

I AB

k k6ZkRetransforming to equivalent current source

YAB16ZkAssignment questions:1) Find the condition for maximum power transfer in the following network type AC

Source,complex source impedance and complex load impedance but only load resistance varying.2) In the

circuit

shown in fig .10 find

the load connected at AB

for which the powertransferred will

be maximum. Also find maximum power

3) In the circuit shown in Fig. 11. Find Vx and prove reciprocity theorem.4) ) Determine the current through 2: resister of the network shown in Fig. Principle

using superposition

5) State Millmans theorem, using the same calculate current through the load in the circuit shown in Fig.

7) Calculate the current I shown in fig. 19 using Millmans theorem

8) state and explain i) Reciprocity theorem ii) Millmanns theorem.Using superposition theorem, obtain the response I for the network shown in Fig.12.9) Use millmans theorem to determine the voltage Vs of the network shown in Fig.6Given thatER = 230 00 V, EY = 230 -1200V, and EB = 230 1200V

Unit: 4 Network Theorems - II : Hrs: 06Syllabus of unit :Thevinins and Nortons theorems; Maximum Power transfer theoremRecommended readings:1. Network Analysis,

M. E.

Van Valkenburg, PHI / Pearson Education2. NetworksPublications .

and

Systems,

Roy

Choudhury,

2 editions,

New

Age

International3. Network theory , Ganesh Rao.4. Network analysis , Roy Choudry.Thevinins and Nortons Theorems:If we are interested in the solution of the current or voltage of a small part of the network, it is convenient from the computational point of view to simplify the network, except thatsmall part in question, by a simple equivalent. Nortons theorem.

This is achieved by Thevinins Theorem orThevinins Theorem :If two linear networks one M with passive elements and sources and the other N with passive elements only and there is no magnetic coupling between M and N, are connected together at terminals A and B, then with respect to terminals A and B, the network M can be replaced by an equivalent network comprising a single voltage source in series with asingle

impedance.

The

single

Voltage

source

is the

Open

circuit

Voltage

across

theterminals A and B and single series impedance is the impedance of the network M as viewed from A and B with independent voltage sources short circuited and independent current sources open circuited. Dependent sources if any are to be retained.Arrange the networks M and N such that N is the part of the network where response is required.MA.N

.

B

To prove this theorem, consider

the circuit shown in Fig.Z1 Z2+ +E1 E2- -ZL

Z4 + IS_Suppose

the

required

response

is the

current

IL in

ZL.

Connected

between A

and B.According to Thevinins theorem the following steps are involved to calculate ILStep 1:Remove

ZL and

measure

the

open

circuit

voltage

across

AB.

This

is also

called asThevinins voltage and is denoted as VTHZ1 Z2+ -E1 E2- A +B.

Zs IS_VTH

= VAB

= E 1

E1 I S Z SZ1+Z 2 + Z S

Z1 + E2 (E1

+ E2) (Z1+Z 2 + Z S) ( E1 I S Z S ) Z1 VTH

= VAB

= Z1+Z 2 + Z SStep 2:To obtain the single impedance as viewed from A and B,

replace the network in Fig.replacing

the

sources.

This

single

impedance

is called

Thevinins

Impedance

and isdenoted by ZTHZ1 Z2A+ZS_BZ TH =

Z1 (Z 2 + Z S)Z1+Z 2 + Z SStep 3 :Write the thevinins network and re introduce ZLZTHVTHThen the current in ZL isVTHIL =

ZTH + ZL( E1=

+ E2) ( Z1+Z 2 + Z S ) ( E1 I SZ S ) Z1Z1+Z 2 + Z SZ1(Z 2 + Z S)Z1+Z 2 + Z S

+ ZL( E1

+ E2) ( Z1+Z 2 + Z S ) ( E1 I SZ S ) Z1=

Z1(Z 2 + Z S)

+ Z2 (Z1+Z 2 + Z S)To verify the correctness of this, write loop equations for the network to find the current in ZL( E1

+ E2) Z1( E1 - IS Zs) Z1+Z 2 + Z SZ1+Z LZ1

Z1Z1+Z 2 + Z S( E1

+ E2) ( Z1+Z 2 + Z S ) ( E1 I SZ S ) Z1=

(Z 1 + Z L)

(Z1+Z 2 + Z S) Z1 2( E1

+ E2) ( Z1+Z 2 + Z S ) ( E1 I SZ S ) Z1=

Z1(Z 2 + Z S)

+ Z2 (Z1+Z 2 + Z S)Nortons Theorem :-Z1 Z2+ +E1 E2 Z5 IS- 1 - 2ZL_The

Thevinins

equivalent

consists of a

voltage

source

and a

series

impedance

. If

thecircuit is transformed to its equivalent current source, we get Nortons equivalent. Nortons theorem is the dual of the Thevinins theorem.

ThusIf two linear networks, one M with passive elements and sources and the other N with passive elements only and with no magnetic coupling between M and N, are connected together at terminals A and B, Then with respect to terminals A and B, the network M canbe replaced by a single current source in parallel with a single impedance.

The singlecurrent

source

is the

short

circuit

current

in AB

and

the

single

impedance

is theimpedance of the network M as viewed from A and B with independent sources being replaced by their internal impedancesThe proof of the Nortons theorem is simpleConsider the same network that is considered for the Thevinins Theorem and for the same response.Step 1:

Short the terminals A and B and measure the short circuit current in AB, this is Nortonscurrent source.Z1 Z2+ + ZsE1 - E2 - IN=Isc=E1+E2 + E2+ISZS

Z1 Z2+ZS=(E1 + E2)(Z2 + ZS )+(E2 +IS ZS )Z1Z1 (Z2+ZS )Step 2: This is the same as in the case of thevnins theoremStep 3: write the Nortons equivalent and reintroduce ZLA ZnB Then the current in ZL isIL=IN. Zn

Zn+ ZL(E1+E2)(Z2 +Zs)+(E2+IsZs)Z1

. Z1

(Z2 +Zs)= Z1(Z2 +Zs) Z1 +Z2+Zs

Z1(Z2 +Zs )

+ ZLZ1 +Z2 +Zs= (E1+E2)(Z2 +Zs)+(E2+IsZs)Z1

Z1 (Z2 +Zs) +ZL(Z1

+Z2+Zs )= (E1+E2) ( Z1 +Z2 +Zs) - (E1 -IsZs)Z1Z1 (Z2 +Zs) + ZL (Z1+Z2+Zs)Verification is to be done as in Thevinins TheoremDetermination of Thevinins or Nortons equivalent when dependent

sources are presentSince

IL= VTH

= IN .ZTH

Z TH +ZL Z TH +ZLZTH can also be determined as

ZTH

=VTHIN

= o.c voltage across ABs.c current in ABWhen

network

contains

both

dependent

and

independent

Sources. It

is convenient to determine ZTH by finding both the open circuit voltage and short circuit currentIf the network contains only dependent sources both VTH and IN are zero in the absence of independent sources. Then apply a constant voltage source (or resultant source) and the ratio of voltage to current gives the ZTH . However there cannot be an independent source ie, VTH or I N in the equivalent network.Maximum Transfer Theorem:-When a linear network containing sources and passive elements is connected at terminals A and B to a passive linear network, maximum power is transferred to the passive network when its impedance becomes the complex conjugate of the Thevinins impedance of the source containing network as viewed from the terminals A and B.Fig represents a network with sources replaced by its Thevinins equivalent of source of ETH volts and impedance Zs, connected to a passive network of impedance z at terminals A & B. With Zs =Rs+JXs and z=R+JX, The proof of the theorem is as followsCurrent in the circuit isI = ETH (Rs+R)2+(Xs+X)2

(1)Zs a+ETh Zpower delivered to the load is= E2Th

bP=I2R.R

(2)( Rs+R)2 +(Xs+X)2As P = (R,X) and since P is maximum when dP=0We have dP=

P .dR + P .dX

(3)R XPower is maximum when PR

=0 and PX

=0 simultaneously PR

= (Rs+R)2+(Xs+X)2 R{2(Rs+R)} =0D2ie, (Rs+R)2+(Xs+X)2 2R{2(Rs+R)}

=0

(4)P = =R{2(Xs+X)} = 0 X D2ie 2R(Xs+X)=0From (5) we have X= -XsSubstituting in (4) (Rs+R)2 =2R(Rs+R), ie , R=Rs

(5) (6)ie, Rs+R= 2RAlternatively as P == E2R (Rs+R)2 +(Xs+X)2= E2Z Cos(Rs+ZCos )2+(Xs+ZSin)2= E2Z CosZs2+Z2+2ZZsCos(-s)

(7)ie P=f(Z,)for Pmax

dP =

P .dZZ

+ P

.d =0P = 0 = {Zs 2+Z 2+2 Z Zs Cos( -s)} Cos -Z Cos {2Z+2Zs Cos( -s)}Zie Zs 2+Z2=2 Z2+2Z Zs Cos( -s).

Or | Z |=| Zs | (8)then withP = 0 = {Zs 2+Z 2+2 Z Zs Cos(-s)}Z(-Sin)-ZCos {ZS2+Z 2 2Z Zs Sin(-s)}(Zs2+Z2) Sin =2Z Zs {Cos Sin (-s) - Sin Cos (-s)}= - 2Z Zs SinsSubstituting (8) in (9)2 Zs Sin = -2 Zs 2 Sins = -s

(9)Z = Zs

-sEfficiency of Power Transfer:With Rs=RL and Xs= - XL Substituting in (1) P Lmax =E2THR = E2 H

(2R) 2 4Rand the power supplied is Ps = E2TH 2R = E2TH(2R)2 2RThen tra = PLPs

= E2TH 4R E2TH 2R

= 1 = 50%2This means to transmit maximum power to the load 50% power generated is the loss. Such a low efficiency cannot be permitted in power systems involving large blocks of power where RL is very large compared to Rs. Therefore constant voltage power systems are not designed to operate on the basis of maximum power transfer.However in communication systems the power to be handled is small as these systems are low current circuits. Thus impedance matching is considerable factor in communication networks.However between R & X if either R or X is restricted and between Z and if either |Z| or isrestricted the conditions for Max P is stated as followsCase (i) :- R of

Z is varied keeping X constant with R only Variable, conditions for max powertransfer is

(Rs+R)2+(Xs+X)2 2R(Rs+R)=0Rs2+ R2+ 2RsR+(Xs+X)2-2RsR-2R2=0R2= Rs2+(Xs+X)2R= Rs 2 (Xs X)2Case (ii):- If Z contains only R ie, x=0 then from the eqn derived aboveR=|Zs|.

Rs 2 Xs2Case (iii):- If |Z| is varied keeping T constant then from (8) Case (iv):- If |Z| is constant but T is variedThen from eqn (9) (Z2+Zs2) Sin T =-2Z Zs SinTs

|Z|=|Zs|SinT = -2ZZs

Sin Ts(Z2+Zs2)Then power transfer to load may be calculated by substituting for R and X for specified condition. For exampleFor case(ii) Pmax is given byPmax =

E2R(Rs+R)2+(Xs+X)2== E2Zs = (Rs+Zs)2+Xs2

E2Zs Rs2+2RsZs+Zs2+Xs2= E2

2(Zs+Rs)

(ie Rs2+Xs2= Zs2)Assignment questions:1) State and explain superposition theorem and Nortons theorem2) Obtain the Thevenins equivalent of network shown in Fig.X and Y.

between terminals

3) Obtain the Thevenins and Nortons equivalent circuits across terminals A and B for the circuit shown in Fig. 4.4) ) In the circuit of Fig. 7 obtain I by Thevenins theorem

5) state and prove Thevenins theorem6) Find Thevenins equivalent circuit across AB using Millmans theorem and find the current through the load (5+j5) : shown in Fig. 8.

7) Calculate Thevenins equivalent circuit across AB for the network shown in fig. 9.

Unit:

5: Resonant

Circuits: Hrs: 06Syllabus of unit :Series

and

parallel

resonance,

frequency response of series and Parallel circuits, Q factor,Bandwidth.Recommended readings:1. Network Analysis,

M. E.

Van Valkenburg, PHI / Pearson Education2. Networks

and

Systems,

Roy

Choudhury,

2 edition,

New

Age

InternationalPublications .3. Network theory , Ganesh Rao.4. Network analysis , Roy

Choudry.Resonant CircuitsResonance is an important phenomenon which may occur in circuits containing both inductors capacitors.

andIn a two terminal electrical network containing at least one inductor and one capacitor, we define resonance as the condition, which exists when the input impedance of the network is purelyresistive.

In other words a network is in resonance when the voltage and current at the network input terminals are in phase.Resonance condition is achieved either by keeping inductor and capacitor same and varyingfrequency or by keeping the frequency same and varying inductor and capacitor.

Study ofresonance is very useful in the area of communication.

The ability of a radio receiver to select thecorrect frequency transmitted by a broad casting station and to eliminate frequencies from other stations is based on the principle of resonance.The resonance circuits can be classified in to two categoriesSeries Resonance Circuits. Parallel Resonance Circuits.1.Series Resonance Circuit

R L CA series resonance circuit is one in which a coil anda capacitance are connected in series across an alternating Ivoltage ofV

varying frequency

as shown in figure.The response I

of the circuit

is dependent on the impedance of the circuit,Where

Z= R +jXL - jXCand

I= V / z

at any value of frequencyWe have XL

=2fL

XL varies as fand

XC = 1 XC varies inversely as f2fCaIn other words, by varying the frequency it is possible to reach a point where

XL = XC . Inthat case Z = R

and hence circuit will be under resonance.

Hence the

series

A.C. circuit is to beunder resonance, when inductive

reactance of the circuit is equal to the capacitive reactance. Thefrequency at which the resonance occurs is called as resonant frequency

( fr)Expression for Resonant Frequency ( At resonance XL = XC

fr )Salient Features of Resonant circuit(*)

At resonance XL = XC(*)

At resonance

Z = R

i.e.

impedance is minimum

and hence I = V

is maximumZ(*) (*)

The current at resonance (Ir) is in phase with the voltageThe circuit power factor is unity(*) Voltage across the capacitor is equal and opposite to the voltage across the inductor.Frequency response of a series resonance circuitFor a

R-L-C

Series

Circuit

the current

I is given by V I =

R + j (XL- XC)At resonance

XL = XC

and hence the current at resonance (Ir)

is given by Ir

= V/RAt off resonance frequencies since the impedance of the circuit increases the currentin the circuit will reduce.

At frequencies f Where f> fr , the impedance is going to be moreinductive.

Similarly at frequencies f

< fr

the circuit impedance is going to be

more capacitive.Thus the resonance curve will be as shown in figure.IIrXc > Xll

> Xcfr fQualify factor (or Q factor):Another feature of a resonant circuit is the Q rise of voltage across the resonating elements. If V is the applied voltage across a series resonance circuit at resonance, Ir = VRAny circuit response, which is frequency dependent, has certain limitations.

The output responseduring limited band of frequencies only will be in the useful range. If the output power is equal toor more than half of the maximum power output that band of frequencies is considered to be theuseful band.

If I r is the maximum current at resonance thenPower at resonance =

Pmax

= I2 r RConsider the frequency response characterstic of a series resonant circuit as shown in figureIr0.707 IrF1 Fr F2

freqIn the figure it is seen that there are two frequencies where the output power is half of the maximum power. These frequencies are called as half power points f1 and f2A frequency f1 which is below fr where power is half of maximum power is called as lowerhalf power frequency (or

lower cut off frequency).

Similarly frequency f2 which is above fr isCalled upper half power frequency (or upper

Cut-off frequency)The band of frequencies between f2 and f1 are said to be useful band of frequencies since during these frequencies of operation the output power in the circuit is more than half of themaximum power.

Thus their band of frequencies is called as Bandwidth.i.e

Band width =B.W

= f2 - f1Selectivity :Selectivity is a useful characteristic of the resonant circuit. Selectivity band width to resonant frequency

is defined

as the ratio ofSelectivity

= f2- f1 frIt can be seen that selectivity is the reciprocal of Quality factor. Hence larger the value of QSmaller will

be the selectivity.The Selectivity of a resonant circuit depends on how sharp the output is contained with inlimited band of frequencies.

The circuit is said to be highly selective if the resonance curve fallsvery sharply at off resonant frequencies.Relation between Resonant frequency and cut-off frequenciesLet fr be the resonant frequency of a series resonant circuit consisting of R,L and C elements .From the Characteristic it isseen that at both half frequencies

Ir0.707Irf2 and

f1 the output current is 0.707 Iowhich means

that the magnitude of theimpedance is same at these points. At lower cut-off frequency f1 f1

fr f2Resonance by varying InductanceResonance in RLC series circuit can also be obtained by varying resonating circuit elements. us consider a circuit where in inductance is varied as shown in figure.

LetR L CIrV, f Hza

0.707Ir

Parallel ResonanceA parallel resonant circuit is one in which a coil and a capacitance are connected in parallel across a variable frequency A.C. Supply. The response of a parallel resonant circuit is somewhat different from that of a series resonant circuit.Impedance at resonanceWe know that at resonance the susceptive part of the admittance is zero. Hence Y0 = R R2 +0 2 L 2But R2

+0 2 L 2

= L/CSo Y0 = RC/L or Zo = L/RCWhere Zo is called the dynamic resistance. When coil resistance R is small, dynamic resistance of the parallel circuit becomes high. Hence the current at resonance is minimum. Hence this type of circuit is called rejector circuit.Frequency response characterisiticsThe frequency response curve of a parallel resonant circuit is asshown in the figure. We find that current is minimum at resonance. The half power points are given by the points at which the current is 2 Ir .From the above characteristic it is clear that the characteristic is exactly opposite to that of series resonant.I2IrIrFrequencyf1 fr f2Quality factor ( Q-factor)The quality factor of a parallel resonant circuit is defined as the current magnificationQ = Current through capacitance at resonance Total Current at resonance= IC0 / I0= V / (1/ Z0C) V / Z0= Z0Z0C = (L / RC) Z0C= Z0L / RHence the expression for the Q- factor for both series and parallel resonant circuit are the sameAlso

Band width= f0 / QII A coil and a Practical Capacitor in parallelConsider a parallel resonant circuit in whichthe resistance of the capacitance is also considered

ILRL XLImpedance of the coil = ZL = RL + j ZL ICYL = 1 / ZL = 1/ RL+j ZL2 2 2= RL j ZL / RL

+Z L

I RC XCImpedance of the Capacitor = ZC = RC j / ZC= RC + j / ZC VRc2 +1/Z2C2Therefore total admittance = Y=YL+YC2 2 2= ( RL j ZL / RL

+Z L) + RC + j / ZCRc2 +1/Z2C2At resonance the susceptance part of the total admittance is zero, which gives1/Z0C Z0L=2 2 2RC2 + 1/Z20C2 RL

+Z 0L2 2 2 2 2 2 21/LC [ RL

+Z 0L] = Z 0[RC

+ 1/Z 0C ]Z20 (R 2

L/C) = R 2/ LC 1/ C2Z20 = 1/LC(R 2

L/C)( RC2 L/C)Z0 = 1/ LC

2 L/C ( RC2 L/C)f0 = 1

( RL2 L/C)2S LC

( R 2

L/C)

At Resonance the admittance is purely conductive given byY0 =

RL RC+2 2 2

2 2 2RL +Z 0L

RC + 1/Z 0CExample 1

: Determine the value of RC in theShown in figure to yield ResonanceSolution: Let then

ZL be the impedance of the inductive branchZL = 10+ j 10YL = 1/ (10 +j 10)

RC 10= 10 j 10 = 10- j 20102 +102

200Let ZC be the impedance of the capacitive branch then

-j2

j10ZC =

RC j 2YC =

1RC j 2

= RC j 2RC2 + 4Total admittance of the circuit = Y = YL + YCFor the circuit under Resonance the Susceptance part is zero( 2/ RC2 + 4) -RC2 = 36

(10 / 200) = 0

25 25 mHRC = 6 ohms

AnswerExample 2: An Impedance coil of 25 ohmsResistance and 25 mH inductanceis connected in parallel with a variable capacitor. For what value of Capacitor will the circuit resonate.If 90 volts,400Hz source is used, what will be the lineCurrent under these conditions

C90 Volts,400HzSolution:

Z0 = 2Sf0 = 2S( 400)

0 2 = 1 - R2LC L 26.316 x 106 = 1 - R2

LC L 21 = 0 2 + R2LC L 2= 6.316 x 106 += 7.316 x 106C= 5.467 PF

252 / (25 x 10-3)2Z0 = L/RC =

(25 x 10-3)/ 25 x 5.467 x 10-6= 182.89 ohmsI0 = V0/ Z0 = 90/ 182.89 = 0.492 ampereAssignment questions:1) Explain parallel resonance ? Derive the condition for parallel resonance when RL connected parallel to RC .2) Show that resonant frequency of series resonance circuit is equal to the geometric mean of two half power frequencies.3) In the circuit given below in Fig. 21, an inductance of 0.1 H having a Q of 5 is in parallelwith a capacitor. Determine the value of capacitance and coil resistance at resonant frequency of500 rad/sac.

4) In the circuit shown in Fig. 22, determine the complete solution for the current when the switch S is closed at t = 0. Applied voltage is V (t) = 400 cos 500+Resistance R = 15:, inductance L = 0.2H and capacitance C = 3 F.

5) In the circuit shown in fig. 23 the switch S is moved from a to b at t = 0. Find values of i,at t = 0+ if R = 1:, L = 1H, c= 0.1

F and V=100 V.Assume steadily state is achieved when K is at a.6) A series resonant circuit includes 1 F capacitor and a resistance of 16:? If the bandWidth is a 500 rad/sec, determine i)

ii) Q and

iii) L7) A two branch antiresonant circuit contains L = 0.4H and C = 40F. Resonance is to beachieved by variation of RL and RC. Calculate the resonance frequency for the followingcases:i) RL = 120:, RC = 80:ii)iii)

RL = 80:, RC = 0RL = RC = 100:Unit: 6

Transient

behavior

and

initial

conditions

Hrs: 07Syllabus of unit :Behavior

of circuit

elements under switching

condition and their Representation, evaluationof initial and final conditions in RL, RC and RLC circuits for AC and DC excitations.Recommended readings:1. Network Analysis,

M. E.

Van Valkenburg, PHI / Pearson Education2. Networks

and

systems,

Roy

Choudhury,

2 edition,

New

Age

InternationalPublications .3. Network theory , Ganesh Rao.4. Network analysis , Roy Choudry.Electrical circuits are connected to supply by closing the switch and disconnected from the supply by opening the switch. This switching operation will change the current and voltage in the device. A purely resistive device will allow instantaneous change in current and voltage.An inductive device will not allow sudden change in current or delay the change in current. A capacitive device will not allow sudden change in voltage or delay the change in voltage. Hence when switching operation is performed in inductive or capacitive device the current and voltage in the device will take a certain time to change from preswitching value to steady valueafter switching. This study of switching condition in network is called transient analysis. The state (or condition) of the current from the instant of switching to attainment of steady state is called transient state or transient. The current and voltage of circuit elements during transient period is called transient response.The transient may also occur due to variation in circuit elements. Transient analysis is an useful tool in electrical engineering for analysis of switching conditions in Circuit breakers, Relays, Generators etc.It is also useful for the analysis of faulty conditions in electrical devices. Transient analysis is also useful for analyzing switching Conditions in analog and digital Electronic devices.R-L Series circuit transient:Consider The R-L series circuit shown in the fig. Switch K is closed at t=0. Referring to the circuit, balance equation using

R LK i(t)V t

R t ldi t

t = 0 VKirchoffs law can be written asTaking Laplace Transform we get

i dtV s s

I s u R L^SI s i 0 `Assuming there is no stored energy in the inductorI(0)=0V s V s AR

R

V s putB

V s R

S R L

B L LI s V s 1 1 Therefore

R S

S R L Taking inverse Laplace we get i t

V R t

1 e L R

The equation clearly indicates transient nature of current, which is also shown in figure.LWhere R

Tune constant of the circuit, which is denoted by Z given in seconds.V Z Hence

i t

1 e t R V VPutting t=z we get

i(z) = 0.632 R

Where R

= steady state current. Hence Time constant for anR-L series current circuit is defined as the time taken by the circuit to reach 63.2% of its final steady value.R-C series circuit Transient

Consider the RC circuit shown.

Let the switch be closed at t=0.Writing the balance equation using Kirchoffs voltage law ,v t

iR 1 i dt cTaking Laplace transform, we getLet us assume that there is no stored energy in the circuit.Hence

Q o =0 V s s

I s R

I s CS

I s R

1 CS I s

V s 1 R S 1 RC Taking Laplace inverse we get 1 V t

tRC

i t e R

1The sketch of transient current is shown in figure Where RC

Wthe time constant of the circuit.WPutting

1RC in the current equation we getVi(z) = 0.367 RHence time constant of RC series current can be defined as the time taken by current transient to fall to 36.7% of its initial value.Example 1:In the circuit shown in figure the switch K is moved from position 1 to position 2 at time t = 0.The steady state current having been previously established in R-L circuit. Find the current i(t) after switching.Solution:From the given data the circuit is under steady state when switch K is in position 1 under steady10state condition inductance is a short and hence i(0) = 10

= 1 Amp.

When the circuit is switched to position 2, this 1 Amp current constituted the stored energy in thecoil.Writing the balance equation for position 2 we getTaking Laplace transformation20I s 4>s I s i 0 @ 020I s 4>s I s 1@

20i 4 di 0 dtI s

4

4 s 5

1 s 5taking inverse Laplace we geti t

e 5t

Example 2:A series R-C circuit is shown in figure. The capacitor has an initial charge of 800Coulombs on its plates, at the time the switch is closed. Find the resulting current transient.800 u 10 6Solution: From the data given q(0) = CWriting the balance equation we get1100

10i t i(t)dt4 u 10 6Taking Laplace transformation100S

10I(s) 1 >I(s) Q(0)@4 u 10 6 S 106 I(s)10

100 800 u 10= 6 4S S

4 u 10 6 S100 2005 40S 106 30I(s) 4S SI(s)

1200

1200

I(s)

40S 106 30 S 25000

40 S

106 40 Taking Inverse Laplace we geti(t)

30e 25000 t

Example3:For the circuit shown in figure the relay coil is adjusted to operate at a current of 5 Amps. Switch Kis closed at t = 0 and the relay is found to operate at t = 0.347 seconds. Find the value of inductanceL of the relay.Soln: Writing the balance equation for the relay circuitV(t)

Ri(t) L di dtApplying Laplace transformationV(s)S

RI(S) LS>I(S) i(0)@Since there is no mention of initial current in the coil i(0) =010Hence S

I(s) I(s)LSI(S)^SL 1` 10S10I(s)

10

L

A B

S 1 SL

1 S S L

S 1 S L 10 A S 1 BS L L A=10I(S)

B= -1010 10 S 1LTaking Inverse Laplace we get ti(t)

10 10e LThe relay operates at t = 0.347 seconds when the current value reaches 5A. Hence 0.3475 10 10e L 0.34710e L

10 5 5 0.347 5e LSolving the equation we get

L=0.5HExample 4:In figure the switch K is closed. Find the time when the current in the circuitry reaches to 500 mASoln: When the switch is closed Vc (0) = 0When the switch is closed at t = 0

I1 (t)50 = 10I (t) u 70 1 102 100 u 10

6

i2dtTaking Laplace for both the equationsI1(S)

1050S

0.2 (1)5I (S) 70

1 I2 (s) 102 u c S 570I2 (s)

I2 (S) 10100 u 10 6 s 5I (S)

10

1

1 ' 1 (2)2 70S 104

7s 103

7 S 142.86

Taking inverse Laplace for equation (1) and (2) I1 (t) =0.2 AI2 (t)

1 e 142.86 t7Total current from the battery i(t) =I1 + I2i(t)

0.2 1 e 142.86t7when this current reaches 500 mA500 u 10 3

0.2 1 e 142.86 t7Solving we get t = 5.19 10-3 Seconds.R-L-C Series Transient circuit:Assuming zero initial conditions when switch K is closed the balanced equation is given byV iR L di dt

1 idtCTaking Laplace transformation we getV(s)s

1(s)R LSI(s) I(s) CSI(s)

SL 1 CS

V(s)I(s)

V(s)

L

S(R SL 1 )

S2 R S 1CS L LCThe time response of the circuit depends on the poles or roots of the characteristic equationS2 S R 1 0L LCRoots of the characteristic equation are given by R r

R 4 u 1 S1,S2

L L LC22S ,S

R r

R 11 2 2L

2L LCIntial conditions:The reason for studying initial and final conditions in a network is to evaluate the arbitrary constants that appear in the general solution of the differential equations written for the network.In this chapter we concentrate on finding the change in selected variables in a circuit when aswitch is thrown from open to closed or vice versa position. time of throwing the switch

Please note that t = 0

indicates thet = 0- indicates time immediately before throwing the switch andt =0+ indicates time immediately after throwing the switch.We are very much interested in the change in currents and voltages of energy storage elements (inductor and capacitor) after the switch is thrown since these variables along with the sources will dictate the circuit behaviour for t > 0.Initial conditions in a network depend on the past history of the circuit (before t= 0-) and structure of the network at t = 0+.Past history will show up in the form of capacitor voltages and inductor currents.Initial and

final conditions in elementsThe resistor:The cause

effect relation for the ideal resistor is given by

v = Ri. From this equation we find thatthe current through a resistor

will change instantaneously,

if the voltage changesinstantaneously.Similarly voltage will change instantaneously if current changes instantaneously.The inductor : K LInitial condition VThe switch is closed at t= 0The expression for current through the inductor is

given by i(t) = 1

tv dt0- t

L -= 1/L

vdt-

+ 1/Lt

vdt0-= i(0-) + 1/L 0-

vdtPutting t = 0+0+i (0+)i (0+)

= i (0-)= i (0-)

+ 1/L 0-

vdzThe above equation indicates that the current in an inductor can not change instantaneously.

Henceif i

(0-)

=0,

then i(0+) = 0.

This means that at t = 0+ inductor will act as an open circuit,independent of voltage across the terminals.O.C LIf i (0-)

= I0 (i. e.

if a residual

current is present) then i

(0+)

= I0 , meaning that an inductor at t= 0+

can be thought of as a current source of I0 which is as shownI0 IIFinal

(or steady state)

Lcondition I0The final conditionL di/ dt

equivalent circuit of an inductor is derived from the basic relation ship V =Under steady state condition di

= 0 which means v = o

and hence L acts as a shortat t =

( final or steady state) I0

S.CI0L S.CAt t= The capacitorThe switch is closed at

t = 0 .

The expression

At t=0For voltage across the tv = 1/ C i dt-

capacitor

is given by

i(t) v C0- tv(t) = 1/ C i dt + 1/ C i dt- Putting t= 0+v(0+ ) = v(0-) + 1/C

0-0+ i dtV(0+)

= V (0-)

0-which means that the voltage

across the capacitor can not changeinstantaneously. If t=0+

V(o-) = o

then

V (o+) = o

indicating that the Capacitor acts as a short atC S.C

at t=0+V0= Q0 / C - + V0Final (or steady state) conditionThe final-condition equivalent network is determined from the basic relationship i = C dv/dtUnder

steady state condition

dv / dt = 0 which means at t= the Capacitor actsas a open circuit.C

O.CV0- + V0-

-+ - O.CAssignment questions:1) Establish the procedure for

evaluating initial conditions with suitable examples.2) )

In the circuit shown in Fig. 20 V=10V, R= 10:,

L=1H, C= 10PF, and VC=0 . findi(0+),

(0+) and

(0+) it switch K is closed at t=0.

3) ) Find i(t) for the following network shown in Fig. 21 if if the switch K is opened at t= 0, before that the circuit has attained steady state condition .

4) R = 1:, L = 1H and C = 1/2F are in series with a switch across C 2V is applied to the

circuit. Att = 0 the switch is in closed position. At t = 0 the switch is opened. Find at t = 0+ , the voltage across the switch, its first and second derivatives.5) A coil of R = 1000: and L = 1 H is connected to a d.c. voltage of 100V through a changeover switch. At t = , the switch connects a capacitor of C = 0.1F in series with the coil, excluding thevoltage. Solve for i,

and