Chapter 10 Part 2-Electrical Circuit Analysis-Problem Solutions, Exercises of Electrical Circuit Analysis

Download Chapter 10 Part 2-Electrical Circuit Analysis-Problem Solutions and more Exercises Electrical Circuit Analysis in PDF only on Docsity! Chapter 10, Problem 1. Determine i in the circuit of Fig. 10.50. Figure 10.50 For Prob. 10.1. Chapter 10, Solution 1. We first determine the input impedance. 1 1 10 10H j L j x jω⎯⎯→ = = 111 0.1 10 1 F jj C j xω ⎯⎯→ = = − 1 1 1 11 1.0101 0.1 1.015 5.653 10 0.1 1 o inZ jj j − ⎛ ⎞ = + + + = − = < −⎜ ⎟−⎝ ⎠ 2 0 1.9704 5.653 1.015 5.653 o o oI < = = < < − ( ) 1.9704cos(10 5.653 ) Aoi t t= + = 1.9704cos(10t+5.65˚) A docsity.com Figure 10.51 For Prob. 10.2. Chapter 10, Solution 2. Consider the circuit shown below. 2 Vo –j5 j4 4∠0o V- At the main node, 4 40 (10 ) 2 5 4 o o o o V V V V j j j − = + ⎯⎯→ = + − 40 3.98 5.71 A 10 o oV j = = < − + _ docsity.com Find oi in the circuit of Fig. 10.54. Figure 10.54 For Prob. 10.5. docsity.com Chapter 10, Solution 5. 30.25 0.25 4 10 1000H j L j x x jω⎯⎯→ = = 3 6 112 125 4 10 2 10 F jj C j x x x µ ω −⎯⎯→ = = − Consider the circuit as shown below. Io 2000 Vo -j125 25∠0o V j1000 + 10Io – At node Vo, 25I160jV)14j1( 0I160jV16jV2j25V 0 125j I10V 1000j 0V 2000 25V oo oooo oooo =−+ =−+−− = − − + − + − But Io = (25–Vo)/2000 °−∠ °∠ °∠ = + + = =+−+ 37.817768.1 94.58115.14 57.408.25 08.14j1 2j25V 25V08.0j2jV)14j1( o oo Now to solve for io, °∠= += +− = − = 06.4398.12 mA8784.0j367.12 2000 7567.1j2666.025 2000 V25 I oo io = 12.398cos(4x103t + 4.06˚) mA + _ docsity.com Figure 10.55 For Prob. 10.6. Chapter 10, Solution 6. Let Vo be the voltage across the current source. Using nodal analysis we get: 0 10j20 V 3 20 V4V oxo = + +− − where ox V10j20 20V + = Combining these we get: 30j60V)35.0j1(0 10j20 V 3 10j20 V4 20 V o ooo +=−+→= + +− + − = +− = +− + = 5.0j2 )3(20Vor 5.0j2 30j60V xo 29.11∠–166˚ V. docsity.com 40 VV 100j V 20 V V1.0156 21111 o −+ − +=+∠ or 21 025.0)01.0025.0(5529.17955.5 VVjj −+−=+ (1) At node 2, 21 2 1 21 V)2j1(V30 20j V V1.0 40 VV −+=⎯→⎯+= − (2) From (1) and (2), BAVor 0 )5529.1j7955.5( V V )2j1(3 025.0)01.0j025.0( 2 1 =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − −+− Using MATLAB, V = inv(A)*B leads to 09.1613.110,23.12763.70 21 jVjV +−=−−= o21 o 17.82276.740 VV I −∠= − = Thus, A )17.82t200cos(276.7)t(i oo −= docsity.com Use nodal analysis to find ov in the circuit of Fig. 10.58. Figure 10.58 For Prob. 10.9. docsity.com Chapter 10, Solution 9. 33 10,010)t10cos(10 =ω°∠⎯→⎯ 10jLjmH10 =ω⎯→⎯ 20j- )1050)(10(j 1 Cj 1 F50 6-3 =× = ω ⎯→⎯µ Consider the circuit shown below. At node 1, 20j-2020 10 2111 VVVV −+= − 21 j)j2(10 VV −+= (1) At node 2, 10j3020 )4( 20j- 2121 + += − VVVV , where 20 1 o V I = has been substituted. 21 )8.0j6.0()j4-( VV +=+ 21 j4- 8.0j6.0 VV + + = (2) Substituting (2) into (1) 22 jj4- )8.0j6.0)(j2( 10 VV − + ++ = or 2.26j6.0 170 2 − =V °∠= − ⋅ + = + = 26.70154.6 2.26j6.0 170 j3 3 10j30 30 2o VV Therefore, =)t(vo 6.154 cos(10 3 t + 70.26°) V j10 Ω 20 Ω -j20 Ω20 Ω 10∠0° V + − 4 Io V1 V2 + Vo − 30 Ω Io docsity.com