# 321 and 323 v v 5v1 5v2 12 e text main

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Unformatted text preview: e the equations in two ways. METHOD 1 Using the elimination technique, we add Eqs. (3.2.1) and (3.2.3). | v v 5v1 − 5v2 = 12 | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents CHAPTER 3 Methods of Analysis or 12 = 2.4 5 v 1 − v2 = (3.2.4) Adding Eqs. (3.2.2) and (3.2.3) gives −2v1 + 4v2 = 0 ⇒ v1 = 2v2 (3.2.5) Substituting Eq. (3.2.5) into Eq. (3.2.4) yields 2v2 − v2 = 2.4 ⇒ v2 = 2.4, v1 = 2v2 = 4.8 V From Eq. (3.2.3), we get v3 = 3v2 − 2v1 = 3v2 − 4v2 = −v2 = −2.4 V Thus, v1 = 4.8 V, METHOD 2 v2 = 2.4 V, v3 = −2.4 V To use Cramer’s rule, we put Eqs. (3.2.1) to (3.2.3) in matrix form. −2 7 −3 3 −4 2 −1 v1 12 −1 v2 = 0 1 0 v3 From this, we obtain v1 = 1 , v2 = 2 , v3 = 3 where , 1 , 2 , and 3 are the determinants to be calculated as follows. As explained in Appendix A, to calculate the determinant of a 3 by 3 matrix, we repeat the ﬁrst two rows and cross multiply. 3 −2 3 −2 −1 −4 7 = −4 7 −1 = 2 −3 2 −3 1 3 −2 − −4 7 − − = 21 − 12 + 4 + 14 − 9 − 8 = 10 −1 −1 1 −1 −1 + + + Similarly, we obtain 1 = | v v − − − | 12 0 0 12 0 −2 7 −3 −2 7 −1 −1 1 −1 −1 e-Text Main Menu + + + = 84 + 0 + 0 − 0 − 36 − 0 = 48 | Textbook Table of Contents | Problem Solving Workbook Contents 81 82 PART 1 2 DC Circuits = − − − 3 = − −...
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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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