Download Chapter 19-Electrical Circuit Analysis-Problem Solutions and more Exercises Electrical Circuit Analysis in PDF only on Docsity! Figure 19.65 For Prob. 19.1 and 19.28. docsity.com Ω=++== 4)24(||61 1 1 11 I V z 1o 2 1 II = , 1o2 2 IIV == Ω== 1 1 2 21 I V z To get 22z and 12z , consider the circuit in Fig. (b). Ω=+== 667.1)64(||2 2 2 22 I V z 22 ' o 6 1 102 2 III = + = , 2o1 '6 IIV == Ω== 1 2 1 12 I V z Hence, =][z Ω⎥⎦ ⎤ ⎢⎣ ⎡ 667.11 14 4 Ω + V1 − + V2 − I1 = 0 1 Ω (b) Io' I2 = 04 Ω + V1 − 1 Ω I1 (a) Io + V2 − docsity.com Chapter 19, Problem 3. Find the z parameters of the circuit in Fig. 19.67. Figure 19.67 For Prob. 19.3. Chapter 19, Solution 3. 12 216z j z= = 11 12 11 124 4 4 6 z z z z j− = ⎯⎯→ = + = + Ω 22 12 22 1210 10 4 z z j z z j j− = − ⎯⎯→ = − = − Ω 4 6 6 [ ] 6 4 j j z j j +⎡ ⎤ = Ω⎢ ⎥−⎣ ⎦ = Ω⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − + 4j6j 6j6j4 docsity.com Chapter 19, Problem 4. Calculate the z parameters for the circuit in Fig. 19.68. Figure 19.68 For Prob. 19.4. Chapter 19, Solution 4. Transform the Π network to a T network. 5j12 120j 5j10j12 )10j)(12( 1 + = −+ =Z j512 j60- 2 + =Z 5j12 50 3 + =Z The z parameters are j4.26--1.775 25144 j5)--j60)(12( 22112 =+ === Zzz 26.4j775.1 169 )5j12)(120j( 1212111 +=+ − =+= zzZz 739.5j7758.1 169 )5j12)(50( 2121322 −=+ − =+= zzZz Thus, =][z Ω⎥⎦ ⎤ ⎢⎣ ⎡ −− −+ 739.5j775.126.4j775.1- 26.4j775.1-26.4j775.1 Z3 Z1 docsity.com Chapter 19, Problem 5. Obtain the z parameters for the network in Fig. 19.69 as functions of s. Figure 19.69 For Prob. 19.5. Chapter 19, Solution 5. Consider the circuit in Fig. (a). s 1 s1 1s 1 s 1 s1 1s 1 s 1 s1|| s 1 1 s 1 s 1 s1|| s 1 ||111 +++⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++ + =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++=z 1s3s2s 1ss 23 2 11 +++ ++ =z 1 2 11o 1ss 1s s 1s s s 1 s1 1s 1 1s 1 s 1 s1 s 1 ||1 s 1 ||1 IIII +++ + += +++ + += +++ = 123o 1s3s2s s II +++ = 1s3s2ss 1 23 1 o2 +++ == I IV 1s3s2s 1 23 1 2 21 +++ == I V z I2 = 0 + V1 − 1 I1 (a) + V2 − s 1/s Io docsity.com 1 1 11 1 1 (20 5) 25 V Iz I I + = = = Ω 1 1 20 20 25o V V I= = 2 2 2 1 1 1 14 0 4 20 4 24o oV I V V V I I I I− − + = ⎯⎯→ = + = + = 2 21 1 24 Vz I = = Ω To find z12 and z22, consider the circuit below. I1=0 5Ω 10Ω 4I1 I2 – + + V1 20 Ω V2 – 2 2 2(10 20) 30V I I= + = 2 22 1 30 Vz I = = Ω 1 220V I= 1 12 2 20 Vz I = = Ω Thus, 25 20 [ ] 24 30 z ⎡ ⎤= Ω⎢ ⎥ ⎣ ⎦ + _ docsity.com Figure 19.71 For Prob. 19.7 and 19.80. Chapter 19, Solution 7. To get z11 and z21, we consider the circuit below. I2=0 I1 20Ω 100Ω + + + vx 50Ω 60 Ω V1 - V2 - - 12vx - + docsity.com 1x xxxx1 V 121 40V 160 V12V 50 V 20 VV =⎯→⎯ + += − 88.29 I Vz) 20 V( 121 81 20 VVI 1 1 11 1x1 1 ==⎯→⎯= − = 37.70 I VzI37.70 I 81 121x20) 121 40( 8 57V) 121 40( 8 57V 8 57V12) 160 V13(60V 1 2 211 11xx x 2 −==⎯→⎯−= −=−=−=−= To get z12 and z22, we consider the circuit below. I2 I1=0 20Ω 100Ω + + + vx 50Ω 60 Ω V1 - V2 - - 12vx - + 2 x22 222x V09.060 V12V 150 VI,V 3 1V 50100 50V =++== + = 11.1109.0/1 I Vz 2 2 22 === 704.3 I VzI704.3I 3 11.11V 3 1VV 2 1 12222x1 ==⎯→⎯==== Thus, Ω⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = 11.1137.70 704.388.29 ]z[ docsity.com
