# 311 gives 3v1 20 20 v1 40 1333 v 3 m e t h

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Unformatted text preview: v2 = 20 V Substituting v2 = 20 in Eq. (3.1.1) gives 3v1 − 20 = 20 ⇒ v1 = 40 = 13.33 V 3 M E T H O D 2 To use Cramer’s rule, we need to put Eqs. (3.1.1) and (3.1.2) in matrix form as 3 −3 −1 5 20 v1 = 60 v2 (3.1.3) The determinant of the matrix is 3 −1 = = 15 − 3 = 12 −3 5 We now obtain v1 and v2 as v1 = 1 = 20 −1 60 5 = 3 20 −3 60 100 + 60 = 13.33 V 12 180 + 60 = 20 V 12 giving us the same result as did the elimination method. v2 = 2 = = If we need the currents, we can easily calculate them from the values of the nodal voltages. v1 − v2 v1 i2 = = −1.6667 A, i3 = = 6.666 i1 = 5 A, 4 2 v2 i4 = 10 A, = 3.333 A i5 = 6 The fact that i2 is negative shows that the current ﬂows in the direction opposite to the one assumed. PRACTICE PROBLEM 3.1 Obtain the node voltages in the circuit in Fig. 3.4. 6Ω 1 2 Answer: v1 = −2 V, v2 = −14 V. 1A | v v Figure 3.4 | e-Text Main Menu | Textbook Table of Contents | 2Ω 7Ω For Practice Prob. 3.1. Problem Solving Workbook Contents 4A 80 PART 1 DC Circuits EXA...
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