Charles k alexander,matthew n o sadiku fundamentals of electric circuits instructor solutions manual mcgraw hill (2013)

W ith its objective to present circuit analysis in a manner that is clearer, more interesting, and easier to understand than other texts, Fundamentals of Electric Circuits by Charles Alexander and Matthew Sadiku has become the student choice for introductory electric circuits courses FiFth Edition FiFt h Edition Building on the success of the previous editions, the fifth edition features the latest updates and advances in the field, while continuing to present material with an unmatched pedagogy and communication style Fundamentals of Pedagogical Features Matched Example Problems and Extended Examples Each illustrative example is immediately followed by a practice problem and answer to test understanding of the preceding example one extended example per chapter shows an example problem worked using a detailed outline of the six-step method so students can see how to practice this technique Students follow the example step-by-step to solve the practice problem without having to flip pages or search the end of the book for answers ■ Comprehensive Coverage of Material not only is Fundamentals the most comprehensive text in terms of material, but it is also self-contained in regards to mathematics and theory, which means that when students have questions regarding the mathematics or theory they are using to solve problems, they can find answers to their questions in the text itself they will not need to seek out other references ■ Computer tools PSpice® for Windows is used throughout the text with discussions and examples at the end of each appropriate chapter MAtLAB® is also used in the book as a computational tool ■ new to the fifth edition is the addition of 120 national instruments Multisim™ circuit files Solutions for almost all of the problems solved using PSpice are also available to the instructor in Multisim ■ We continue to make available KCidE for Circuits (a Knowledge Capturing integrated design Environment for Circuits) ■ An icon is used to identify homework problems that either should be solved or are more easily solved using PSpice, Multisim, and/or KCidE Likewise, we use another icon to identify problems that should be solved or are more easily solved using MAtLAB Teaching Resources McGraw-hill Connect® Engineering is a web-based assignment and assessment platform that gives students the means to better connect with their coursework, with their instructors, and with the important concepts that they will need to know for success now and in the future Contact your McGraw-hill sales representative or visit www connect.mcgraw-hill.com for more details Electric Circuits INSTRUCTOR SOLUTIONS MANUAL MD DALIM 1167970 10/30/11 CYAN MAG YELO BLACK ■ Fundamentals of Problem-Solving Methodology A six-step method for solving circuits problems is introduced in Chapter and used consistently throughout the book to help students develop a systems approach to problem solving that leads to better understanding and fewer mistakes in mathematics and theory Electric Circuits ■ the text also features a website of student and instructor resources Check it out at www.mhhe.com/alexander Alexander Sadiku Charles K Alexander | Matthew n o Sadiku Chapter 1, Solution (a) q = 6.482x1017 x [-1.602x10-19 C] = –103.84 mC (b) q = 24x1018 x [-1.602x10-19 C] = –198.65 mC (c) q = 2.46x1019 x [-1.602x10-19 C] = –3.941 C (d) q = 1.628x1020 x [-1.602x10-19 C] = –26.08 C Chapter 1, Solution (a) (b) (c) (d) (e) i = dq/dt = mA i = dq/dt = (16t + 4) A i = dq/dt = (-3e-t + 10e-2t) nA i=dq/dt = 1200 cos 120 t pA i =dq/dt =  e 4t (80 cos 50 t  1000 sin 50 t )  A Chapter 1, Solution (a) q(t)   i(t)dt  q(0)  (3t  1) C (b) q(t)   (2t  s) dt  q(v)  (t  5t) mC (c) q(t)   20 cos 10t   /   q(0)  (2sin(10t   / 6)  1)  C 10e -30t q(t)   10e sin 40t  q(0)  ( 30 sin 40t - 40 cos t) (d) 900  1600   e - 30t (0.16cos40 t  0.12 sin 40t) C -30t Chapter 1, Solution q = it = 7.4 x 20 = 148 C Chapter 1, Solution 10 t 10 q   idt   tdt   25 C 0 Chapter 1, Solution (a) At t = 1ms, i  dq 30   15 A dt (b) At t = 6ms, i  dq  0A dt (c) At t = 10ms, i  dq  30   –7.5 A dt Chapter 1, Solution 25A, dq  i  - 25A, dt   25A, 0t2 2t6 6t8 which is sketched below: Chapter 1, Solution q   idt  10   10   15 μC Chapter 1, Solution (a) q   idt   10 dt  10 C 1  q   idt  10   10    1 (b)    15  7.5   22.5C (c) q   idt  10  10  10  30 C 5s  5s ZC      s L3  YC 3.4s 3.4 3.4s s C4 i.e an inductor in series with a capacitor L3   1.471 H and 3.4 C4  3.4  0.425 F Thus, the LC network is shown in Fig (c) 425 mF 1.471 H 1H 200 mF (c) Chapter 19, Solution 96 This is a fourth order network which can be realized with the network shown in Fig (a) L1 L3 C2 1 C4 (a)  (s)  (s  3.414s  1)  (2.613s  2.613s) 2.613s  2.613s H(s)  s  3.414s  1 2.613s  2.613s which indicates that -1 2.613s  2.613s s  3.414s   2.613s  2.613s y 21  y 22 We seek to realize y 22 By long division, 2.414s  y 22  0.383s   s C  YA 2.613s  2.613s i.e C  0.383 F and 2.414s  YA  2.613s  2.613s as shown in Fig (b) L1 YA L3 C2 C4 (b) y 22 ZA  2.613s  2.613s  YA 2.414s  By long division, Z A  1.082s  i.e 1.531s  s L3  Z B 2.414s  L  1.082 H and ZB  1.531s 2.414s  as shown in Fig.(c) L1 ZB L3 C2 C4 (c) YB  i.e 1  1.577s   s C2  s L1 1.531s ZB C  1.577 F and L1  1.531 H Thus, the network is shown in Fig (d) 1.531 H 1.577 F 1.082 H 0.383 F (d) 1 Chapter 19, Solution 97 s3 s s  12s H(s)   6s  24 (s  12s)  (6s  24) 1 s  12s Hence, y 22 6s  24    ZA s  12s s C (1) where Z A is shown in the figure below C1 C3 L2 ZA y 22 We now obtain C and Z A using partial fraction expansion Let 6s  24 A Bs  C   2 s (s  12) s s  12 6s  24  A (s  12)  Bs  Cs Equating coefficients : 24  12A   A  s0 : 0C s : s : 6 AB   B  Thus, 6s  24 4s   2 s (s  12) s s  12 (2) Comparing (1) and (2), 1 C3   F A But s  12   s ZA s 4s (3) 1  sC1  s L2 ZA (4) Comparing (3) and (4), C1  F and L2  H Therefore, C1  250 mF , L  333.3 mH , C  500 mF Chapter 19, Solution 98  h   0.8  0.2    h / h 21  h11 / h 21    0.001 [Ta ]  [Tb ]    6  h 22 / h 21  / h 21   2.5x10  10   0.005 2.6x105 0.06  [T]  [Ta ][Tb ]    8 5x105  1.5x10 We now convert this to z-parameters A / C  T / C 1.733x103 [z]     / C D / C  6.667 x10 1000 I1 0.0267   3.33x103  z 11 + z 22 + I2 + + Vs z 12 I ZL z 21 I - - Vo - - Vs  (1000  z11)I1  z12 I (1) Vo  z 22 I  z 21I1 (2) But Vo  I ZL  I  Vo / ZL (3) Substituting (3) into (2) gives   z  22  I1  Vo   z 21 z 21ZL  We substitute (3) and (4) into (1) (4)   z z  22  Vo  12 Vo Vs  (1000  z 11 ) ZL  z 11 z 21 Z L   7.653x10   2.136x10   744V Chapter 19, Solution 99 Z ab  Z1  Z  Z c || (Z b  Z a ) Z c (Z a  Z b ) Z1  Z  Za  Zb  Zc Z cd  Z  Z  Z a || (Z b  Z c ) Z a (Z b  Z c ) Z2  Z3  Za  Zb  Zc Z ac  Z1  Z  Z b || (Z a  Z c ) Z b (Z a  Z c ) Z1  Z  Za  Zb  Zc (1) (2) (3) Subtracting (2) from (1), Z1  Z  Z b (Z c  Z a ) Za  Zb  Zc (4) Adding (3) and (4), Z1  ZbZc Za  Zb  Zc (5) Subtracting (5) from (3), Z2  ZaZb Za  Zb  Zc (6) Subtracting (5) from (1), Z3  ZcZa Za  Zb  Zc (7) Using (5) to (7) Z a Z b Z c (Z a  Z b  Z c ) (Z a  Z b  Z c ) Za ZbZc Z1Z  Z Z  Z Z1  Za  Zb  Zc Z1Z  Z Z  Z Z1  Dividing (8) by each of (5), (6), and (7), (8) Za  Z1Z  Z Z  Z Z1 Z1 Zb  Z1Z  Z Z  Z Z1 Z3 Zc  Z1Z  Z Z  Z Z1 Z2 as required Note that the formulas above are not exactly the same as those in Chapter because the locations of Z b and Z c are interchanged in Fig 18.122 Below are answers for the Network Analysis Tutorials Some of the tutorial pages have random parameters For these pages, there are no fixed right answers, and formulas are provided instead Introductory Tutorial (Tut22) Orange 12 3.14159 The Physics of Electricity (Tut1A) 36000 32.04 60 4.32 18000 Basic Elements and Circuit Laws (Tut1) -2 50 50 -60 Answerless page Resistors in Series and Parallel (Tut2) 10 17 11 Formula: R1×R2/(R1 + R2) + (R3 + R4)×R5/(R3 + R4 + R5) + R7 Voltage Dividers and Current Dividers (Tut2A) 40 160 12 20 Circuit Solving with Kirchhoff's Laws (Tut3) 2 3 5 6 I6 V9 Answerless page The Node Voltage Method (Tut4) a g -6 5 g Ve 10 Answerless page 11 -2 12 13 Answerless page 14 I4 15 I4 16 Answerless page The Mesh Current Method (Tut5) c -2 -4 -2 Ib 3A Ia Answerless page Thevenin Laboratory (Tut6) Formula: 1000×Voc×Vr/[R×(Voc - Vr)] Maximum Power Transfer (DC) (Tut6A) Formula: R2×R3/(R2 + R3) Formula: E×R3/(R2 + R3) Formula: VT2/(4×RT) where VT = E×R3/(R2 + R3) Superposition (Tut6B) 2 -3 4 Inductors and Capacitors (Tut7) 0.8 0.2 12t -20e-4t 16cos(8t) -80 9.79992 3t 10 2t3 11 -1.25e-4t 12 -0.25cos[8t] 13 1.5 14 29.532 15 Answerless page First Order Systems (Tut8) 120 600 120 34.016 90 10 450 11 0.45 12 -200 13 -200 14 40 15 450 16 240 17 Answerless page Second Order Systems (Tut9) -80 capacitor 80 20000 series 6000 10 5000 11 over 12 -2683 13 A1 14 -A2 15 s 16 3.0148 17 188 18 1.297 19 Answerless page The Properties of Sinusoids (Tut10) 86 -170 220 23.4 -220 440 155.6 -67 10 50 11 7.958 12 125.7 13 24.2 Root-mean-square (Tut10A) 212 50 40 10 2500 10 -150 11 25000 12 450000 13 900000 14 466667 15 32.27 Complex Numbers (Tut10B) -1 -1 -j 85.9 383 321 655 -459 85 10 -47.86 11 138 12 141.5 13 92 14 35 15 -15 16 73 17 17.0 18 14.13 19 Formula: The equation is in the form M/θ = (-5 + jA)(B/-152°) - (C - j100)/(2/D°) The answer is √(X2 + Y2)where X = √(52 + A2)×B×cos[arctan(-A/5) + 28°] - ( √(C2 + 1002)/2)×cos[arctan(-100/C) - D] Y = √(52 + A2)×B×sin[arctan(-A/5) + 28°] - ( √(C2 + 1002)/2)×sin[arctan(-100/C) - D] AC Circuits (Tut11) 98 120 140 -125 25 30 14.97 200 AC Power (Tut12) 378 466 441 B 466 578 .779 3373 10 D 11 B 12 166 13 2732 Maximum Power Transfer (AC) (Tut12A) Formula: |Z1|2×R/(R2 + |Z1|2) Formula: -(R2×|Z1| - R2×|Z2| - |Z1|2×|Z2|)/(R2 + |Z1|2) Formula: |E|×R/√ R2 + |Z1|2 Formula: Divide the result from page by twice the result from page Formula: Square the result from page and multiply by the result from page Balanced Three-Phase Circuits (Tut13) 133.3 866 866 173.2 200 346.4 400 ... two resistors and one voltage source Hint, you could use both resistors at once or one at a time, it is up to you Be creative Although there is no correct way to work this problem, this is an... on the same kind of problem asked in the third edition Problem Use KCL to obtain currents i 1, i 2, and i in the circuit shown in Fig 2.72 Solution 12 A a i1 b 8A i2 i3 12 A c At node a, At node... branches and nodes Chapter 2, Solution Design a problem, complete with a solution, to help other students to better understand Kirchhoff’s Current Law Design the problem by specifying values of i a