sadiku 5ª edição circuitos eletricos e extras - chapt03pp 110626, Notas de estudo de Engenharia Elétrica

Baixe sadiku 5ª edição circuitos eletricos e extras - chapt03pp 110626 e outras Notas de estudo em PDF para Engenharia Elétrica, somente na Docsity! Saturday, June 25, 2011 CHAPTER 3 P.P.3.1 6  3 A At node 1, –3 + i1 + i2 = 0 or 2 0v 6 vv 121   = 3 or 4v1 – v2 = 18 (1) At node 2, –i1 + i3 + 12 = 0 or i1 = 12 + i3 or 7 0v 12 6 vv 221   or 7v1 – 13v2 = 504 (2) Solving (1) and (2) gives v1 = –6 V, v2 = –42 V P.P.3.2 2  3 A i1 i1 12A 2 1 i2 i3 7  12A 2  i1 v3 3  4 A 4  ix i2 4ix i2 i3 v2 v1 6  At node 1, –4 + i1 + i2 = 0 = –4 + 3 vv 2 vv 2131   or 5v1 – 2v2 – 3v3 = 24 (1) At node 2, –i2 + ix – 4ix = 0 = –i2 – 3ix = 0 where ix = [(v2–0)/4] or 0 4 v 3 3 vv 221   which leads to 4v1 + 5v2 = 0 (2) At node 3, –i1 + i3 + 4ix = 0 = 4 v 4 6 0v 2 vv 2313     or –3v1 + 6v2 + 4v3 = 0 (3) Solving (1) to (3) gives v1 = 32 V, v2 = –25.6 V, v3 = 62.4 V P.P.3.3 – At the supernode in Fig. (a), 6 v 2 v 3 v 4 v14 11   or 42 = 7v + 8v1 (1) Applying KVL to the loop in Fig. (b), – v – 6 + v1 = 0 v1 = v + 6 (2) –+ 6 V + v – + v1 – v (a) – + 14V v1 + v – 4  3  2  6  (b) 1128032 051 813 042 051 813 3        i1 = ,571.2 28 721      i2 = 714.7 28 2162      , i3 = A4 28 1123      Io = i3 = –4 A P.P.3.7 For the supermesh, –8 + 2i1 – 2i3 + 12i2 – 4i3 = 0 or i1 + 6i2 – 3i3 = 4 (1) For mesh 3, 8i3 – 2i1 – 4i2 = 0 or –i1 – 2i2 + 4i3 = 0 (2) At node 0 in Fig. (a), i1 = 4 + i2 i1 – i2 = 4 Solving (1) to (3) yields i1 = 4.632 A, i2 = 631.6 mA, i3 = 1.4736 A (b) 4  8  – + 2  2  i3 i1 4 A i2 i1 – +8 V 1  2  2  4  8  3 A i2 i2 i1 i3 (a) 0 P.P.3.8 G11 = 1/(1) + 1/(20) + 1/(5) = 1.25, G12 = –1/(5) = –0.2, G33 = 1/(4) + 1 = 1.25, G44 = 1/(1) + 1/(4) = 1.25, G12 = –1/(5) = –0.2, G13 = –1, G14 = 0, G21 = –0.2, G23 = 0 = G26, G31 = –1, G32 = 0, G34 = –1/4 = –0.25, G41 = 0, G42 = 0, G43 = 0.25, i1 = 0, i2 = 3+2 = 5, i3 = –3, i4 = 2. Hence,                 25.125.000 25.025.101 002.02.0 012.025.1                           2 3 5 0 4 3 2 1 v v v v P.P.3.9 R11 = 50 + 20 + 80 = 150, R22 = 20 + 30 + 15 = 65, R33 = 30 + 20 = 50, R44 = 15 + 80 = 95, R55 = 20 + 60 = 80, R12 = –40, R13 = 0, R14 = –80, R15 = 0, R21 = –40, R23 = –30, R24 = –15, R25 = 0, R31 = 0, R32 = –30, R34 = 0, R35 = –20, R41 = –80, R42 = –15, R43 = 0, R45 = 0, R51 = 0, R52 = 0, R53 = –20, R54 = 0, v1 = 30, v2 = 0, v3 = –12, v4 = 20, v5 = –20 Hence the mesh-current equations are =                      8002000 09501580 20050300 015306540 080040150                 5 4 3 2 1 i i i i i                   20 20 12 0 30 P.P.3.10 The schematic is shown below. It is saved and simulated by selecting Analysis/Simulate. The results are shown on the viewpoints: v1 = –10 V, v2 = 14.286 V, v3 = 50 V P.P.3.11 The schematic is shown below. After saving it, it is simulated by choosing Analysis/Simulate. The results are shown on the IPROBES. i 1 = –428.6 mA, i2 = 2.286 A, i3 = 2 A 2.000E+00 2.286E+00 -4.286E-01 50 V -10.0000 50.0000 14.2858 500 mA