Ch03

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: 5 −16 40 v4 | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents CHAPTER 3 Methods of Analysis 87 Using Cramer’s rule, −1 −1 −5 3 =6 6 3 3 =6 6 −2 −2 = −18, −16 1 0 = 80 40 0 −2 80 −2 = −3120, 40 −16 4 −1 −1 −5 3 =6 6 −2 −2 = −480 −16 −1 −1 −5 0 80 = 840 40 Thus, we arrive at the node voltages as = −480 = 26.667 V, −18 v4 = 4 = v3 = 3 = −3120 = 173.333 V −18 840 = −46.667 V −18 and v2 = v1 − 20 = 6.667 V. We have not used Eq. (3.4.5); it can be used to cross check results. @ Network Analysis PRACTICE PROBLEM 3.4 6Ω Find v1 , v2 , and v3 in the circuit in Fig. 3.14 using nodal analysis. Answer: v1 = 3.043 V, v2 = −6.956 V, v3 = 0.6522 V. 10 V v1 +− 5i v2 +− 1 v1 = i 2Ω Figure 3.14 3.4 MESH ANALYSIS v v | e-Text Main Menu 3Ω For Practice Prob. 3.4. Electronic Testing Tutorials Mesh analysis provides another general procedure for analyzing circuits, using mesh currents as the circuit variables. Using mesh currents...
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