Practice problem 39 v v by inspection obtain the mesh

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Unformatted text preview: is 5 by 5. The diagonal terms, in ohms, are: R33 R22 = 2 + 4 + 1 + 1 + 2 = 10 R11 = 5 + 2 + 2 = 9, = 2 + 3 + 4 = 9, R44 = 1 + 3 + 4 = 8, R55 = 1 + 3 = 4 The off-diagonal terms are: R13 = −2, R12 = −2, R21 = −2, R23 = −4, R24 R31 = −2, R32 = −4, R41 = 0, R42 = −1, R43 R51 = 0, R52 = −1, R53 R14 = 0 = R15 = −1, R25 = −1 R34 = 0 = R35 = 0, R45 = −3 = 0, R54 = −3 The input voltage vector v has the following terms in volts: v2 = 10 − 4 = 6 v1 = 4, v3 = −12 + 6 = −6, v 4 = 0, v5 = −6 Thus the mesh-current equations are: 4 9 −2 −2 0 0 i1 −2 10 −4 −1 −1 i2 6 −2 −4 9 0 0 i3 = −6 0 −1 0 8 −3 i4 0 −6 i5 0 −1 0 −3 4 From this, we can obtain mesh currents i1 , i2 , i3 , i4 , and i5 . PRACTICE PROBLEM 3.9 | v v By inspection, obtain the mesh-current equations for the circuit in Fig. 3.30. | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents CHAPTER 3 Methods of Analysis 50 Ω 40 Ω 30 Ω i2 10 Ω 24 V + − + 12 V − i3 20 Ω i1 i4 80 Ω Figure 3.30 i5 − + 10 V...
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This note was uploaded on 07/16/2012 for the course KA KA 2000 taught by Professor Bkav during the Spring '12 term at Cambridge.

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