Chapt03PP - 2009 CHAPTER 3 P.P.3.1 1A 6 i1 1A 1 i2 i1 2 2...

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© 2009 CHAPTER 3 P.P.3.1 2 Ω 1 A 6 Ω 7 Ω 1 A 4 A 4 A 1 i 1 i 1 i 2 i 3 2 At node 1, 1 = i 1 + i 2 1 = 2 0 v 6 v v 1 2 1 + or 6 = 4v 1 - v 2 (1) At node 2, 3 1 i 4 i + = 7 0 v 4 6 v v 2 2 1 + = or 168 = 7v 1 - 13v 2 (2) Solving (1) and (2) gives v 1 = –2 V , v 2 = –14 V v 3 3 Ω 10 A 2 Ω 4 Ω P.P.3.2 i 1 i x i 2 i 2 4i x i 3 v 2 v 1 6 Ω
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At node 1, 10 = i 1 + i 2 = 3 v v 2 v v 2 1 3 1 + or 60 = 5v 1 - 2v 2 - 3v 3 ( 1 ) At node 2, x x 2 i i 4 i = + 0 4 v 3 3 v v 2 2 1 = + or 4v 1 + 5v 2 = 0 ( 2 ) At node 3, i 1 = i 3 + 4i x 4 v 4 6 0 v 2 v v 2 3 3 1 + = or -3v 1 + 6v 2 + 4v 3 = 0 (3) Solving (1) to (3) gives v 1 = 80 V , v 2 = –64 V , v 3 = 156 V P.P.3.3 At the supernode in Fig. (a), 6 v 2 v 3 v 4 v 21 1 1 + + = or 63 = 7v + 8v 1 (1) Applying KVL to the loop in Fig. (b), – v – 9 + v 1 = 0 v 1 = v + 9 (2) - + 9 V + v - + v 1 - v (a) - + 21V v 1 + v - 4 Ω 3 Ω 2 Ω 6 Ω (b)
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Solving (1) and (2), v = – 600 mV v 1 = v + 9 = 8.4, i 1 = 2 . 4 2 v 1 = i 1 = 4.2 A P.P.3.4 From Fig. (a), 0 3 v 4 v 2 v 3 2 1 = + + 6v 1 + 3v 2 + 4v 3 = 0 (1) From Fig. (b), - v 1 + 10 + v 2 = 0 v 1 = v 2 + 10 (2) - v 2 - 5i + v 3 = 0 v 3 = v 2 + 5i (3) Solving (1) to (3), we obtain v 1 = 3.043V , v 2 = –6.956 V , v 3 = 652.2 mV P.P.3.5 We apply KVL to the two loops and obtain
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Chapt03PP - 2009 CHAPTER 3 P.P.3.1 1A 6 i1 1A 1 i2 i1 2 2...

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