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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

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.

Ask a homework question - tutors are online