fundamentals-of-electric-circuits-alexander-5e-solution-chapter-10

Chapter 10, Solution 1. We first determine the input impedance. 1 1 10 10 H j  L j x j 1 1 0 1 10 1 F j j C j x    .1 1 1 11 1 1.0101 0.1 1.015 5.653 10 0.1 1 o Z j in j j                 2 0 1.9704 5.653 1.015 5.653 o o o I       i(t) = 1.9704cos(10t+5.65˚) A Downloaded by Brian Muchoki () lOMoARcPSD| Chapter 10, Solution 2. Using Fig. 10.51, design a problem to help other students better understand nodal analysis. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Solve for Vo in Fig. 10.51, using nodal analysis. 2  + j4  Vo 40 o V –j5  – + _ Figure 10.51 For Prob. 10.2. Solution Consider the circuit shown below. 2 Vo –j5 j4 40 o VAt the main node, 4 40 (10 ) 2 5 4 ooo o VVV V j j j      + _ Vo = 40/(10–j) = (40/10.05)5.71˚ = 3.985.71˚ V Downloaded by Brian Muchoki () lOMoARcPSD| Chapter 10, Solution 3.   4 2 cos( )t4  20 16sin( )t4  16 - 90  -j16 2 H  jL  8j 3j1)(4(j )12 1 j C 1 1 12 F     The circuit is shown below. 4  Applying nodal analysis, 1 6 8j 2 4 3j 16jo o o       V V V o 6 8j 1 4 3j 1 2 1 4 3j 16j- V                        .3 835 - 02.35 .1 2207 88.1 .4 682 - 15.33 22.1 04.0j 92.3 56.2j Vo Therefore, o )t(v  3.835cos(4t – 35.02) V j8  1  -j3  6  Vo -j16 V + 20 A  Downloaded by Brian Muchoki () lOMoARcPSD| Chapter 10, Solution 4. I j1 –j1 Ω V1 x Step 1. Convert the circuit into the frequency domain and solve for the node voltage, V1, using analysis. The find the current IC = V1/[1+(1/(j4x0.25)] which then produces Vo = 1xIC. Finally, convert the capacitor voltage back into the time domain. 160º V 0.5Ix Vo Note that we represented 16sin(4t–10º) volts by 160º V. That will make our calculations easier and all we have to do is to offset our answer by a –10º. Our node equation is [(V1–16)/j] – (0.5Ix) + [(V1–0)/(1–j)] = 0. We have two unknowns, therefore we need a constraint equation. Ix = [(16–V1)/j] = j(V1–16). Once we have V1, we can find Io = V1 /(1–j) and Vo = 1xIo. r 1 = j24/(–0.5+j) = (2490º)/(1.118116.57º) 21.47–26.57º V. 26.57º) (0.707145º) = 15.18118.43º A and o = 1xIo = 15.18118.43º V or vo(t) = 15.181sin(4t–10º+18.43º) = 15.181sin(4t–8.43º) volts. Step 2. Now all we need to do is to solve our equations. [(V1–16)/j] – [0.5j(V1–16] + [(V1–0)/(1–j)] = [–j–j0.5+0.5+j0.5]V1 +j16+j8 = 0 o [0.5–j]V1 = –j24 or V = Finally, Ix = V1/(1–j) = (21.47– V Downloaded by Brian Muchoki () lOMoARcPSD| Chapter 10, Solution 5. 3 0.25 0. H j   L j x x j 3 6 1 2 1 1 4 10 2 10 F j j C jx xx       25 Consider the circuit as shown below. Io 2000 Vo -j125 250 o V j1000 + 10Io – + _ At node Vo, 1( V)14j j160I 25 V 25 V2j 16j V j160I 0 0 j125 V I10 j1000 V 0 2000 V 25 o o o o o o o o o o                But Io = (25–Vo)/2000                .1 7768 37.81 .14 115 94.58 08.25 57.4 1 08.14j 25 2j V 1( V)14j 2j 08.0j V 25 o o o Now to solve for io,           .12 398 06.4 .12 367 .0j 8784mA 2000 25 .0 2666 .1j 7567 2000 25 V I o o io = 12.398cos(4x103 t + 4.06˚) mA. Downloaded by Brian Muchoki () lOMoARcPSD| Chapter 10, Solution 6. Let Vo be the voltage across the current source. Using nodal analysis we get: 0 20 10j V 3 20 V V4 o x o      where x Vo 20 10j 20 V   Combining these we get: 0 1( 5.0j V)3 60 30j 20 10j V 3 20 10j V4 20 V o o o o                    2 5.0j )3(20 or V 2 5.0j 60 30j Vo x 29.11–166˚ V.