3 02 0 0 v1 02 v2 3 1325 0125 1 0 0125 05 0125

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Unformatted text preview: as four nonreference nodes, so we need four node equations. This implies that the size of the conductance matrix G, is 4 by 4. The diagonal terms of G, in siemens, are G11 = 1 1 + = 0.3, 5 10 G22 = 111 + + = 0.5, 884 The off-diagonal terms are G33 = G12 = − G21 = −0.2, G31 = 0, G44 = 1 = −0.2, 5 G23 = − 111 + + = 1.625 821 G13 = G14 = 0 1 = −0.125, 8 G32 = −0.125, G41 = 0, 111 + + = 1.325 581 G24 = − G34 = − G42 = −1, 1 = −1 1 1 = −0.125 8 G43 = −0.125 The input current vector i has the following terms, in amperes: i1 = 3, i2 = −1 − 2 = −3, i 3 = 0, i4 = 2 + 4 = 6 Thus the node-voltage equations are 3 0.3 −0.2 0 0 v1 −0.2 v2 −3 1.325 −0.125 −1 = 0 −0.125 0.5 −0.125 v3 0 6 v4 0 −1 −0.125 1.625 which can be solved to obtain the node voltages v1 , v2 , v3 , and v4 . PRACTICE PROBLEM 3.8 1Ω By inspection, obtain the node-voltage equations for the circuit in Fig. 3.28. Answer: 1.3 −0.2 −1 −0.2 0.2 0 −1 0 1.25 0 0 −0.25 4Ω v3 v4 1A 0 0 v1 0 v2 3 = −0.25 v3 −1 3 v4 0.75 v1 10 Ω Figure 3.28 5Ω v2 2Ω 2A For Practice Prob. 3.8. EXAMPLE 3.9 | v v By inspection, write the mesh-current equations for the circuit in Fig. 3.29. | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents 3A 98 PART 1 DC Circuits 5Ω i1 4V 2Ω 2Ω +− 2Ω i2 10 V + − 4Ω Figure 3.29 1Ω i4 4Ω 3Ω 1Ω 3Ω i5 i3 + − 6V + 12 V − For Example 3.9. Solution: We have five meshes, so the resistance matrix...
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