3Figure 3.2 Use of KCL in nodal analysisFigure 3.3 Illustration of nodal analysisFigure 3.5Figure 3.8 Nodal analysis with voltage sourcesFigure 3.13 Assignment of currents and voltages around mesh 1Figure 3.14 Assignment of currents and voltages around mesh 3Figure 3.12 A two-mesh circuitFigure 3.18 Mesh analysis with current sourcesFigure 3.26 The principle of superpositionFigure 3.27 Zeroing voltage and current sourcesFigure 3.28 One-port networkFigure 3.29 Illustration of equivalent-circuit conceptFigure 3.31 Illustration of Thevenin theorumFigure 3.32 Illustration of Norton theoremFigure 3.34 Equivalent resistance seen by the loadFigure 3.35 An alternative method of determining the Thevenin resistanceFigure 3.46Figure 3.47Figure 3.48Figure 3.49 A circuit and its Thevenin equivalentFigure 3.57 Illustration of Norton equivalent circuitFigure 3.58 Computation of Norton currentFigure 3.63 Equivalence of Thevenin and Norton representationsFigure 3.64 Effect of source transformationFigure 3.65 Subcircuits amenable to source transformationFigure 3.71 Measurement of open-circuit voltage and short-circuit currentFigure 3.73 Power transfer between source and loadFigure 3.74 Source loading effectsFigure 3.77 Representation of nonlinear element in a linear circuitFigure 3.78 Load lineFigure 3.79 Graphical solution equations 3.48 and 3.49Figure 3.80 Transformation of nonlinear circuit of Thevenin equivalent© The McGraw-Hill Companies, Inc. 2000McGraw-Hill1PRINCIPLES AND APPLICATIONS OF ELECTRICAL ENGINEERINGTHIRD EDITIONG I O R G I O R I Z Z O N IC H A P T E R3Resistive Network Analysis© The McGraw-Hill Companies, Inc. 2000McGraw-Hill2PRINCIPLES AND APPLICATIONS OF ELECTRICAL ENGINEERINGTHIRD EDITIONG I O R G I O R I Z Z O N IFigure 3.2 Use of KCL in nodal analysisi1R1vavbi3vcvdR3R2i2By KCL : i1 – i2 – i3 = 0. In the node voltage method, we express KCL by va – vbR1–vb – vcR2–vb – vdR3= 0© The McGraw-Hill Companies, Inc. 2000McGraw-Hill3PRINCIPLES AND APPLICATIONS OF ELECTRICAL ENGINEERINGTHIRD EDITIONG I O R G I O R I Z Z O N IFigure 3.3 Illustration of nodal analysisi1vavbi3vc = 0i2R1iSR3R2Node aNode bNode cR1R3R2iSiSVa/R1+(Va-Vb)/R2 =IsVb/R3+(Vb-Va)/R2=0OrVa(1/R1+1/R2)+Vb(-1/R2)=IsVa(-1/R2) +Vb(1/R2+1/R3)=0or, in matrix form03/12/12/12/12/11/1 IsVbVaRRRRRR0322221IsVbVaGGGGGG© The McGraw-Hill Companies, Inc. 2000McGraw-Hill4PRINCIPLES AND APPLICATIONS OF ELECTRICAL ENGINEERINGTHIRD EDITIONG I O R G I O R I Z Z O N IFigure 3.5I2I1R4R1R2R3Node 1I2I1R4R1R2R3Node 2Example 3.1R1=1K, R2=2K, R3=10K,R4=2KI1=10mA, I2=50mA,V1/R1+(V1-V2)/R2+(V1-V2)/R3=I1V2/R4+(V2-V1)/R2+(V2-V1)/R3=-I2Or(1/R1+1/R2+1/R3)V1+ (-1/R2-1/R3)V2=I1(-1/R2-1/R3)V1 + (1/R2+1/R3+1/R4)V2= I2Plugging the numbers1.6 V1- 0.6 V2=10-0.5V1 +1.1 V2=-50By solving the above Eq.V1=-13.57V2=-52.86© The McGraw-Hill Companies, Inc. 2000McGraw-Hill5PRINCIPLES AND APPLICATIONS OF ELECTRICAL ENGINEERINGTHIRD EDITIONG I O R G I O R I Z Z O N IFigure 3.8 Nodal analysis with voltage sourcesR2R1vSR4vavciSR3+_vbVa=Vs(Vs-Vb)/R1-vb/R2-(Vb-Vc)/R3=0 (Vb-Vc)/R3+Is-Vc/R4=0Or(1/R1+1/R2+1/R3)Vb+(-1/R3)Vc=Vs/R1(-1/R3)Vb+ (1/R3+1/R4)Vc=IsOr in Matrix formIsRVsVcVbRRRRRRR 1/4/13/13/13/13/12/11/1 IsRVsVcVbGGGGGGG1/4333321© The McGraw-Hill Companies, Inc. 2000McGraw-Hill6PRINCIPLES AND APPLICATIONS OF ELECTRICAL ENGINEERINGTHIRD EDITIONG I O R G I O R I Z Z O N IFigure 3.13 Assignment of currents and voltages around mesh 1R3R4vSR1R2+_ i1i2v2v1+ –+–Mesh 1: KVL requires thatvS – v1 – v2 = 0, where v1 = i1R1,v2 = (i1 – i2) R1.© The McGraw-Hill Companies, Inc. 2000McGraw-Hill7PRINCIPLES AND APPLICATIONS OF ELECTRICAL ENGINEERINGTHIRD EDITIONG I O R G I O R I Z Z O N IFigure 3.14 Assignment of currents and voltages around mesh 3R3R4vSR1R2+_i1i2v2v3+ –+–v4+–Mesh 2: KVL requires thatv2 + v3 + v4 = 0wherev2 = ( i2 – i1) R2,v3 = i2R3,v4 = i2R4© The McGraw-Hill Companies, Inc. 2000McGraw-Hill8PRINCIPLES AND APPLICATIONS OF ELECTRICAL ENGINEERINGTHIRD EDITIONG I O R G I O R I Z Z O N IFigure 3.12 A two-mesh circuitR3R4vSR1R2+_i1i2I1R1+(I1-I2)R2=Vs(I2-I1)R2 + I2R3 + I2R4=0Or0214322221VsIIRRRRRRRThe advantage of Mesh Current Method is that it uses resistances in the equations, rather than conductances.But Node Voltage Method is physically more sensible.© The McGraw-Hill Companies, Inc. 2000McGraw-Hill9PRINCIPLES AND APPLICATIONS OF ELECTRICAL ENGINEERINGTHIRD EDITIONG I O R G I O R I Z Z O N IFigure 3.18 Mesh analysis with current sources2 4 10 V5 2 Ai1vxi2+_+–5I1 +Vx =10-Vx+2I2+4I2=0I1-I2=2Adding Eqs. 1 and 2 will delete Vx5I1 +6 I2 =10I1-I2=2I1=2 AI2=0P3.1-3.20© The McGraw-Hill Companies, Inc. 2000McGraw-Hill10PRINCIPLES AND APPLICATIONS OF ELECTRICAL ENGINEERINGTHIRD EDITIONG I O R G I O R I Z Z O N IFigure 3.26 The principle of superpositionRvB2+_+_vB1i=R+_vB1iB1The net current throughR is the sum of the in-dividual source currents:i = iB1 + iB2.RvB2+_iB2+© The McGraw-Hill Companies, Inc. 2000McGraw-Hill11PRINCIPLES AND APPLICATIONS OF ELECTRICAL ENGINEERINGTHIRD EDITIONG I O R G I O R I Z Z O N IFigure 3.27 Zeroing voltage and current sourcesiSR1+_vSA circuitiSR1R2The same circuit with vS = 0iSR1+_vSR2R2A circuitR1R2The same circuit with iS = 0+_vS1. In order to set a voltage source equal to zero, we replace it with a short circuit.2. In order to set a current source equal to zero, we replace it with an open circuit.© The McGraw-Hill Companies, Inc. 2000McGraw-Hill12PRINCIPLES AND APPLICATIONS OF ELECTRICAL ENGINEERINGTHIRD EDITIONG I O R G I O R I Z Z O N IFigure 3.28 One-port networkLinearnetworkiv+–i© The McGraw-Hill Companies, Inc. 2000McGraw-Hill13PRINCIPLES AND APPLICATIONS OF ELECTRICAL ENGINEERINGTHIRD EDITIONG I O R G I O R I Z Z O N IFigure 3.29 Illustration of equivalent-circuit conceptR3+_vSR2iv+–R1LoadSource© The McGraw-Hill Companies, Inc. 2000McGraw-Hill14PRINCIPLES AND APPLICATIONS OF ELECTRICAL ENGINEERINGTHIRD EDITIONG I O R G I O R I Z Z O N IFigure 3.31 Illustration of Thevenin