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# 347 v1 and v2 are related as a v1 6i 8 v2 b v1

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Unformatted text preview: Fig. 3.46, applying KCL gives: v1 v1 − v2 12 − v1 = + (a) 2 + 3 6 4 v1 v2 − v1 v1 − 12 = + (b) 2 + 3 6 4 v v 3.1 | e-Text Main Menu | Textbook Table of Contents | 0 − v1 v1 − v 2 12 − v1 = + 3 6 4 v1 − 12 0 − v1 v2 − v 1 (d) 2 + = + 3 6 4 (c) 2 + Problem Solving Workbook Contents + 10 V − 108 PART 1 DC Circuits 3.6 3.7 4Ω v1 1 12 V + − In the circuit in Fig. 3.49, current i1 is: (a) 4 A (b) 3 A (c) 2 A (d) 1 A 8Ω 2A 3Ω The loop equation for the circuit in Fig. 3.48 is: (a) −10 + 4i + 6 + 2i = 0 (b) 10 + 4i + 6 + 2i = 0 (c) 10 + 4i − 6 + 2i = 0 (d) −10 + 4i − 6 + 2i = 0 v2 2 6Ω 6Ω 2Ω Figure 3.46 3.2 For Review Questions 3.1 and 3.2. In the circuit in Fig. 3.46, applying KCL at node 2 gives: v2 − v1 v2 v2 (a) + = 4 8 6 v 1 − v2 v2 v2 (b) + = 4 8 6 v 1 − v2 12 − v2 v2 (c) + = 4 8 6 v2 − v1 v2 − 12 v2 (d) + = 4 8 6 For the circuit in Fig. 3.47, v1 and v2 are related as: (a) v1 = 6i + 8 + v2 (b) v1 = 6i − 8 + v2 (c) v1 = −6i + 8 + v2 (d) v1 = −6i −...
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