# 20 or 5i1 19i2 2i3 0 for example 36 362 for mesh

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Unformatted text preview: For Practice Prob. 3.5. v v 12 Ω i2 4Ω | 9Ω | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents CHAPTER 3 Methods of Analysis 91 EXAMPLE 3.6 Use mesh analysis to ﬁnd the current io in the circuit in Fig. 3.20. i1 Solution: We apply KVL to the three meshes in turn. For mesh 1, i2 io i2 10 Ω −24 + 10(i1 − i2 ) + 12(i1 − i3 ) = 0 24 V or 11i1 − 5i2 − 6i3 = 12 A + − (3.6.1) i1 12 Ω 24 Ω 4Ω i3 For mesh 2, + − 24i2 + 4(i2 − i3 ) + 10(i2 − i1 ) = 0 Figure 3.20 or −5i1 + 19i2 − 2i3 = 0 For Example 3.6. (3.6.2) For mesh 3, 4io + 12(i3 − i1 ) + 4(i3 − i2 ) = 0 But at node A, io = i1 − i2 , so that 4(i1 − i2 ) + 12(i3 − i1 ) + 4(i3 − i2 ) = 0 or −i1 − i2 + 2i3 = 0 (3.6.3) In matrix form, Eqs. (3.6.1) to (3.6.3) become 11 −5 −6 i1 12 −5 19 −2 i2 = 0 −1 −1 2 0 i3 We obtain the determinants as 11 −5 −5 19 −1 −1 11 −5 −5 19 = −6 −2 2 −6 −2 + − + − + − = 418 − 30 − 10 − 114 − 22 − 50 = 192 1 = − − − 2...
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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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