the reference node is selected and the node voltages

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Unformatted text preview: or from the right-hand side.) The reference node is selected, and the node voltages v1 and v2 are now to be determined. At node 1, applying KCL and Ohm’s law gives v1 − v 2 v1 − 0 ⇒ 5= + i1 = i2 + i3 4 2 Multiplying each term in the last equation by 4, we obtain 2 6Ω 10 A (a) 5A i1 = 5 i2 i1 = 5 4Ω v1 v2 20 = v1 − v2 + 2v1 i4 = 10 or i2 i 5 i3 6Ω 2Ω 3v1 − v2 = 20 10 A At node 2, we do the same thing and get v 1 − v2 v2 − 0 ⇒ + 10 = 5 + i2 + i4 = i1 + i5 4 6 Multiplying each term by 12 results in 3v1 − 3v2 + 120 = 60 + 2v2 (b) | For Example 3.1: (a) original circuit, (b) circuit for analysis. v v Figure 3.3 | e-Text Main Menu (3.1.1) or −3v1 + 5v2 = 60 | Textbook Table of Contents | (3.1.2) Problem Solving Workbook Contents CHAPTER 3 Methods of Analysis 79 Now we have two simultaneous Eqs. (3.1.1) and (3.1.2). We can solve the equations using any method and obtain the values of v1 and v2 . METHOD 1 Using the elimination technique, we add Eqs. (3.1.1) and (3.1.2). 4v2 = 80 ⇒...
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