Unformatted text preview: o
But io = −i4 , hence,
i2 = i3 − 3i4
Applying KVL in mesh 4,
2i4 + 8(i4 − i3 ) + 10 = 0
5i4 − 4i3 = −5
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
− 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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