# 2 v 14 a v i 7v 3 figure 311 for practice prob

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Unformatted text preview: ng: (a) KCL to the supernode, (b) KVL to the loop. e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents CHAPTER 3 Methods of Analysis 85 PRACTICE PROBLEM 3.3 Find v and i in the circuit in Fig. 3.11. 3V 4Ω +− Answer: −0.2 V, 1.4 A. + v − i 7V + − 3Ω Figure 3.11 For Practice Prob. 3.3. 2Ω EXAMPLE 3.4 Find the node voltages in the circuit of Fig. 3.12. 3Ω + vx − 20 V 1 +− 2Ω Figure 3.12 6Ω 2 3vx 3 +− 4Ω 10 A 4 1Ω For Example 3.4. Solution: Nodes 1 and 2 form a supernode; so do nodes 3 and 4. We apply KCL to the two supernodes as in Fig. 3.13(a). At supernode 1-2, i3 + 10 = i1 + i2 Expressing this in terms of the node voltages, v 1 − v4 v1 v3 − v 2 + 10 = + 6 3 2 or 5v1 + v2 − v3 − 2v4 = 60 (3.4.1) At supernode 3-4, i1 = i3 + i4 + i5 ⇒ v 1 − v4 v3 − v2 v4 v3 = + + 3 6 1 4 or | v v 4v1 + 2v2 − 5v3 − 16v4 = 0 | e-Text Main Menu | Textbook Table of Contents | (3.4.2) Problem Solving Workbook Contents 6Ω 86 PART 1 DC Circuits 3Ω 3Ω + vx − + vx − i1 6Ω v2 v1 i2...
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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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