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Bài giải phần giải mạch P19

Chapter 19, Solution 1. To get z 11 and z 21 , consider the circuit in Fig. (a). 1Ω + I1 V1 − 4Ω Io 6Ω 2Ω I2 = 0 + V2 − (a) z 11 = V1 = 1 + 6 || (4 + 2) = 4 Ω I1 Io = z 21 = 1 I , 2 1 V2 = 1Ω I1 V2 = 2 I o = I 1 To get z 22 and z 12 , consider the circuit in Fig. (b). I1 = 0 + V1 − 1Ω Io ' 4Ω + 2Ω V2 − 6Ω I2 (b) z 22 = V2 = 2 || (4 + 6) = 1.667 Ω I2 Io' = z 12 = 2 1. | Chapter 19 Solution 1. To get zn and z 21 consider the circuit in Fig. a . Ii Vi 1 n Wv 4 n AMr w Io 6 n 2 n I2 0 V2 a V1 z z11 -p 1 6 4 2 4 Q I o 1 2 Ii V2 21 o I1 Z 21 V2 - 1 q I1 To get z 22 and z 12 consider the circuit in Fig. b . ii 0 1 n 4 n v Io Vi 6 n 2 n V2 I2 b V2 z22 2 4 6 1.667 Q 12 I o 2 _1 2 1012 612 V1 61 o 12 Z12 Vl 1 Q 12 Hence z 4 1 1 1.667 n Chapter 19 Solution 2. Consider the circuit in Fig. a to get z 11 and z 21. I1 1 n i 1 n 1 n 1 n l2 0 - VA-T- WV- VA T va-- Io v1 1 n 1 n 1 n v2 W ---------------- 1 n 1 n 1 n 1 n a V1 _ z11 - I - 2 1II 2 1II 2 1 I1 3 1 11 4 11 z - 2 1 12 I-2 - 2 - 2.733 11 4 1 114 15 Io 1 3 Io 4Io 1 _ _4_ 1 o - 1 1 411 - 15 11 1 4 1 1 o - 4 1511 - 15 11 1 V2- 1 o-1711 v2 1 z21 - -- -z12 -0.06667 21 1 15 12 To get z 22 consider the circuit in Fig. b . 1 Q Ii 0 1 Q WV 1 Q VA 1 Q VA Vi I2 VA 1 Q VA 1 Q -J wv 1 Q b 1 Q 1 Q 1 Q 1 Q v2 Z 22 . 2 1 2 1 3 z n 2.733 12 Thus z 2.733 0.06667 0.06667 2.733 Q Chapter 19 Solution 3. a To find z11 and z 21 consider the circuit in Fig. a . I1 j Q 1 Q a I Io I2 0 V2 V j 1 - j z r jii 1 -j 4 1 j By current division I o