# 321 answer 5 a 8 2 i2 10io for practice prob 36 35

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Unformatted text preview: = | v v − − − | 12 0 0 12 0 −5 19 −1 −5 19 −6 −2 2 −6 −2 11 −5 −1 11 −5 12 0 0 12 0 −6 −2 2 −6 −2 e-Text Main Menu + + + + + + = 456 − 24 = 432 = 24 + 120 = 144 | Textbook Table of Contents | Problem Solving Workbook Contents 4io 92 PART 1 DC Circuits 3 = − − − 11 −5 −1 11 −5 −5 19 −1 −5 19 12 0 0 12 0 + + + = 60 + 228 = 288 We calculate the mesh currents using Cramer’s rule as i1 = 1 = 432 = 2.25 A, 192 i3 = 3 = i2 = 2 = 144 = 0.75 A 192 288 = 1.5 A 192 Thus, io = i1 − i2 = 1.5 A. PRACTICE PROBLEM 3.6 6Ω io 4Ω 20 V + − Figure 3.21 i1 i3 Using mesh analysis, ﬁnd io in the circuit in Fig. 3.21. Answer: −5 A. 8Ω 2Ω i2 – + 10io For Practice Prob. 3.6. 3.5 MESH ANALYSIS WITH CURRENT SOURCES Electronic Testing Tutorials Applying mesh analysis to circuits containing current sources (dependent or independent) may appear complicated. But it is actually much easier than what we encountered in the previous section, because...
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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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