tailieunhanh - Bài giải phần giải mạch P2

Chapter 2, Solution 1 v = iR i = v/R = (16/5) mA = mA Chapter 2, Solution 2 p = v2/R → Chapter 2, Solution 3 R = v/i = 120/() = 48k ohms Chapter 2, Solution 4 (a) (b) i = 3/100 = 30 mA i = 3/150 = 20 mA R = v2/p = 14400/60 = 240 ohms Chapter 2, Solution 5 n = 9; l = 7; b = n + l – 1 = 15 Chapter 2, Solution 6 n = 12; l = 8; b = n + l –1 = 19 Chapter 2, Solution 7. | Chapter 2 Solution 1 v iR i v R 16 5 mA mA Chapter 2 Solution 2 p v2 R R v2 p 14400 60 240 ohms Chapter 2 Solution 3 R v i 120 3 48k ohms Chapter 2 Solution 4 a i 3 100 30 mA b i 3 150 20 mA Chapter 2 Solution 5 n 9 l 7 b n l - 1 15 Chapter 2 Solution 6 n 12 l 8 b n l-1 19 Chapter 2 Solution 7 7 elements or 7 branches and 4 nodes as indicated. 4 Chapter 2 Solution 8 a 8 A 12 A w ii i2 b r i3 12 A c 9 A d At node a At node c At node d 8 12 ii 9 8 i2 9 12 i3 11 - 4A 12 1A 13 -3A Chapter 2 Solution 9 Applying KCL 11 1 10 2 ii 11A 1 i2 2 3 i2 4A 12 i3 3 i3 1A Chapter 2 Solution 10 At node 1 At node 3 4 3 i1 ii 7A 3 i2 -2 i2 -5A Chapter 2 Solution 11 Applying KVL to each loop gives -8 vi 12 0 -12-v2 6 0 10 - 6 - V3 0 -V4 8 - 10 0 vi 4v -- V2 -6v v3 4v v4 -2v Chapter 2 Solution 12 15v - 20v v3 For loop 1 For loop 2 For loop 3 -20 -25 10 V1 0 -10 15 -v2 0 -v1 v2 v3 0 v1 35v v2 5v v3 30v Chapter 2 Solution 13 2A I1 .