# It is a shortcut approach based on mere inspection of

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Unformatted text preview: o But io = −i4 , hence, i2 = i3 − 3i4 (3.7.3) Applying KVL in mesh 4, 2i4 + 8(i4 − i3 ) + 10 = 0 or 5i4 − 4i3 = −5 (3.7.4) From Eqs. (3.7.1) to (3.7.4), i1 = −7.5 A, i2 = −2.5 A, i3 = 3.93 A, i4 = 2.143 A PRACTICE PROBLEM 3.7 2Ω 6V + − i1 3A 1Ω i3 2Ω Use mesh analysis to determine i1 , i2 , and i3 in Fig. 3.25. Answer: i1 = 3.474 A, i2 = 0.4737 A, i3 = 1.1052 A. 4Ω i2 8Ω @ | Network Analysis For Practice Prob. 3.7. v v Figure 3.25 | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents CHAPTER 3 † Methods of Analysis 95 3.6 NODAL AND MESH ANALYSES BY INSPECTION This section presents a generalized procedure for nodal or mesh analysis. It is a shortcut approach based on mere inspection of a circuit. When all sources in a circuit are independent current sources, we do not need to apply KCL to each node to obtain the node-voltage equations as we did in Section 3.2. We can obtain the equations by mere inspection of the circuit. As an example, let us reexamine the circuit in...
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