Solucionário Fundamentos Circuitos Elétricos-Sadiku 5ed PDF

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Summary

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 ...

Description

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.

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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.

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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.

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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.

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We continue to make available KCidE for Circuits (a Knowledge Capturing integrated design Environment for Circuits).

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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

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Fundamentals of

Problem-Solving Methodology. A six-step method for solving circuits problems is introduced in Chapter 1 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

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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 1 (a) q = 6.482x1017 x [-1.602x10-19 C] = –103.84 mC (b) q = 1. 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 2

(a) (b) (c) (d) (e)

i = dq/dt = 3 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 3 (a) q(t)   i(t)dt  q(0)  (3t  1) C (b) q(t)   (2t  s) dt  q(v)  (t 2  5t) mC

(c) q(t)   20 cos 10t   / 6   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 4 q = it = 7.4 x 20 = 148 C

Chapter 1, Solution 5 10

1 t 2 10 q   idt   tdt   25 C 2 4 0 0

Chapter 1, Solution 6 (a) At t = 1ms, i 

dq 30   15 A dt 2

(b) At t = 6ms, i 

dq  0A dt

(c) At t = 10ms, i 

dq  30   –7.5 A dt 4

Chapter 1, Solution 7 25A, dq  i  - 25A, dt   25A,

0t2 2t6 6t8

which is sketched below:

Chapter 1, Solution 8

q   idt 

10  1  10  1  15 μC 2

Chapter 1, Solution 9 1

(a) q   idt   10 dt  10 C 0

3 5 1  q   idt  10  1  10    5 1 0 (b) 2    15  7.5  5  22.5C 5

(c) q   idt  10  10  10  30 C 0

Chapter 1, Solution 10 q = it = 10x103x15x10-6 = 150 mC

Chapter 1, Solution 11 q= it = 90 x10-3 x 12 x 60 x 60 = 3.888 kC E = pt = ivt = qv = 3888 x1.5 = 5.832 kJ

Chapter 1, Solution 12 For 0 < t < 6s, assuming q(0) = 0, t

t





0

0

q (t )  idt  q (0 )  3tdt  0  1.5t 2

At t=6, q(6) = 1.5(6)2 = 54 For 6 < t < 10s, t

t





6

6

q (t )  idt  q (6 )  18 dt  54  18 t  54

At t=10, q(10) = 180 – 54 = 126 For 10> I=inv(Z)*V I= –1.3750 –10.0000 17.8750 I o = I 1 – I 2 = –5 – 1.375 = –6.375 A. Check using the super mesh (equation (3)): –2I 2 + 6 I 3 + 6I 4 = 2.75 – 60 + 107.25 = 50!

Chapter 3, Solution 39 Using Fig. 3.50 from Prob. 3.1, design a problem to help other students to better understand mesh analysis. Solution Given R 1 = 4 kΩ, R 2 = 2 kΩ, and R 3 = 2 kΩ, determine the value of I x using mesh analysis. R1

R2 Ix

12 V

+ 

I2

I1 R3

9V

+ 

Figure 3.50 For Prob. 3.1 and 3.39. For loop 1 we get –12 +4kI 1 + 2k(I 1 –I 2 ) = 0 or 6I 1 – 2I 2 = 0.012 and at loop 2 we get 2k(I 2 –I 1 ) + 2kI 2 + 9 = 0 or –2I 1 + 4I 2 = –0.009. Now 6I 1 – 2I 2 = 0.012 + 3[–2I 1 + 4I 2 = –0.009] leads to, 10I 2 = 0.012 – 0.027 = –0.015 or I 2 = –1.5 mA and I 1 = (–0.003+0.012)/6 = 1.5 mA. Thus, I x = I 1 –I 2 = (1.5+1.5) mA = 3 mA.

Chapter 3, Solution 40 2 k

6 k

6 k 56V

+

–

i2 2 k

i1 i3

4 k

4 k

Assume all currents are in mA and apply mesh analysis for mesh 1. –56 + 12i 1 – 6i 2 – 4i 3 = 0 or 6i 1 – 3i 2 – 2i 3 = 28

(1)

for mesh 2, –6i 1 + 14i 2 – 2i 3 = 0 or –3i 1 + 7i 2 – i 3

=0

(2)

=0

(3)

for mesh 3, –4i 1 – 2i 2 + 10i 3 = 0 or –2i 1 – i 2 + 5i 3

Solving (1), (2), and (3) using MATLAB, we obtain, i o = i 1 = 8 mA.

Chapter 3, Solution 41 10 

i1 6V

2

+– 1

i2 4

5

i3 8V

+

– i i2

i3 0

For loop 1, 6 = 12i 1 – 2i 2

3 = 6i 1 – i 2

(1)

For loop 2, -8 = – 2i 1 +7i 2 – i 3

(2)

For loop 3, -8 + 6 + 6i 3 – i 2 = 0

2 = – i 2 + 6i 3

We put (1), (2), and (3) in matrix form, 6  1 0  i1   3  2  7 1  i    8    2    0  1 6 i 3  2 6

1 0

6 3 0

  2  7 1  234,  2  2 8 1  240 0

1 6

0 2 6

(3)

6

1 3

 3  2  7 8  38 0 1 2

At node 0, i + i 2 = i 3 or i = i 3 – i 2 =

3  2  38  240  = 1.188 A   234

Chapter 3, Solution 42 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 Determine the mesh currents in the circuit of Fig. 3.88.

Figure 3.88 Solution For mesh 1,  12  50I 1  30I 2  0   12  50I1  30I 2 For mesh 2,  8  100 I 2  30 I 1  40 I 3  0   8  30 I 1  100 I 2  40 I 3 For mesh 3,  6  50 I 3  40 I 2  0   6  40 I 2  50 I 3 Putting eqs. (1) to (3) in matrix form, we get 0  I 1  12   50  30        30 100  40  I 2    8   0  40 50  I 3   6  

 

AI  B

Using Matlab,  0.48    I  A B   0.40   0.44    1

i.e. I 1 = 480 mA, I 2 = 400 mA, I 3 = 440 mA

(1) (2) (3)

Chapter 3, Solution 43 20  a 80 V

+

i1

–

30 

+ 30 

i3 20  80 V

+

i2

–

30 

20 

V ab

– b

For loop 1, 80 = 70i 1 – 20i 2 – 30i 3

8 = 7i 1 – 2i 2 – 3i 3

(1)

80 = 70i 2 – 20i 1 – 30i 3

8 = -2i 1 + 7i 2 – 3i 3

(2)

0 = -30i 1 – 30i 2 + 90i 3

0 = i 1 + i 2 – 3i 3

For loop 2,

For loop 3,

(3) Solving (1) to (3), we obtain i 3 = 16/9 I o = i 3 = 16/9 = 1.7778 A V ab = 30i 3 = 53.33 V.

Chapter 3, Solution 44 90 V +

2

4

i3

i2

1 180V +

– 5

i1 45 A i1

i2

Loop 1 and 2 form a supermesh. For the supermesh, 6i 1 + 4i 2 – 5i 3 + 180 = 0

(1)

For loop 3,

–i 1 – 4i 2 + 7i 3 + 90 = 0

(2)

Also,

i 2 = 45 + i 1

(3)

Solving (1) to (3), i 1 = –46, i 3 = –20;

i o = i 1 – i 3 = –26 A

Chapter 3, Solution 45 4

30V

+

–

8

i3

i4

2

6

i1

3

i2

1

For loop 1,

30 = 5i 1 – 3i 2 – 2i 3

(1)

For loop 2,

10i 2 - 3i 1 – 6i 4 = 0

(2)

For the supermesh,

6i 3 + 14i 4 – 2i 1 – 6i 2 = 0

(3)

But

i 4 – i 3 = 4 which leads to i 4 = i 3 + 4

Solving (1) to (4) by elimination gives i = i 1 = 8.561 A.

(4)

Chapter 3, Solution 46 For loop 1,  12  3i1  8(i1  i 2 )  12  11i1  8i 2  0 For loop 2, 8(i 2  i1 )  6i 2  2v o  8i1  14i 2  2v o  0 But vo  3i1 ,

 

11i1  8i 2  12

 8i1  14i2  6i1  0   i1  7i2 Substituting (2) into (1), 77i2  8i2  12   i 2  0.1739 A and i1  7i2  1.217 A

(1)

(2)

Chapter 3, Solution 47 First, transform the current sources as shown below. - 6V +

2

8

V1

4

V2

I3 V3

4 2

I1

8 I2

+ 20V -

+ 12V -

For mesh 1,

 20  14 I 1  2 I 2  8I 3  0 For mesh 2,

10  7 I 1  I 2  4 I 3

(1)

 6   I1  7 I 2  2I 3

(2)

 6  14 I 3  4 I 2  8I 1  0   3  4 I 1  2 I 2  7 I 3 Putting (1) to (3) in matrix form, we obtain

(3)

12  14 I 2  2 I 1  4 I 3  0 For mesh 3,

 

 

 7  1  4  I 1   10         1 7  2  I 2     6    4  2 7  I   3   3     Using MATLAB,  2  I  A 1 B  0.0333 1.8667  But

 

AI  B

  I 1  2.5, I 2  0.0333, I 3  1.8667

I1 

20  V 4

 

V1  20  4 I 1  10 V

V 2  2( I 1  I 2 )  4.933 V

Also, I2 

V3  12 8

 

V3  12  8I 2  12.267 V.

Chapter 3, Solution 48

We apply mesh analysis and let the mesh currents be in mA. 3k  I4 4k 

2k 

Io

1k 

I3 I2

I1 + 6V -

5k 

+ 4V -

10k 

For mesh 1,  6  8  5I1  I 2  4I 4  0   2  5I1  I 2  4I 4 For mesh 2,  4  13I 2  I1  10I 3  2I 4  0   4  I1  13I 2  10I 3  2I 4 For mesh 3,  3  15I 3  10I 2  5I 4  0   3  10I 2  15I 3  5I 4 For mesh 4,  4 I 1  2 I 2  5I 3  14 I 4  0 Putting (1) to (4) in matrix form gives 1  4  I1   2  0  5        1 13  10  2  I 2   4    AI B  0  10 15  5  I    3  3        4  2  5 14  I   0    4    Using MATLAB,

 3.608     4.044  1 IA B 0.148 3.896     3    The current through the 10k  resistor is I o = I 2 – I 3 = 148 mA.

3V +

(1)

(2)

(3) (4)

Chapter 3, Solution 49 3

i3 2

1

2

i1

27 V

i2

+

–

2i 0 i1

i2

0 (a)

2

1

2

i1

+

+

v0 or

v0

–

i2

27V +

–

(b)

For the supermesh in figure (a), 3i 1 + 2i 2 – 3i 3 + 27 = 0

(1)

At node 0,

i 2 – i 1 = 2i 0 and i 0 = –i 1 which leads to i 2 = –i 1

(2)

For loop 3,

–i 1 –2i 2 + 6i 3 = 0 which leads to 6i 3 = –i 1

(3)

Solving (1) to (3), i 1 = (–54/3)A, i 2 = (54/3)A, i 3 = (27/9)A i 0 = –i 1 = 18 A, from fig. (b), v 0 = i 3 –3i 1 = (27/9) + 54 = 57 V.

Chapter 3, Solution 50

i1

4

2

i3

10  8 35 V

+

–

i2 3i 0 i2

For loop 1,

i3

16i 1 – 10i 2 – 2i 3 = 0 which leads to 8i 1 – 5i 2 – i 3 = 0

(1)

For the supermesh, –35 + 10i 2 – 10i 1 + 10i 3 – 2i 1 = 0 or

–6i 1 + 5i 2 + 5i 3 = 17.5

Also, 3i 0 = i 3 – i 2 and i 0 = i 1 which leads to 3i 1 = i 3 – i 2 Solving (1), (2), and (3), we obtain i 1 = 1.0098 and i 0 = i 1 = 1.0098 A

(2) (3)

Chapter 3, Solution 51 5A

i1 8 2

i3

1

+

i2 40 V

4

+

v0

20V

–

+



–

For loop 1,

i 1 = 5A

(1)

For loop 2,

-40 + 7i 2 – 2i 1 – 4i 3 = 0 which leads to 50 = 7i 2 – 4i 3

(2)

For loop 3,

-20 + 12i 3 – 4i 2 = 0 which leads to 5 = - i 2 + 3 i 3

(3)

Solving with (2) and (3), And,

i 2 = 10 A, i 3 = 5 A

v 0 = 4(i 2 – i 3 ) = 4(10 – 5) = 20 V.

Chapter 3, Solution 52

+ v0 2 

i2

– VS

+

–

8

3A

i2

i1

i3 4

i3

+ –

2V 0

For mesh 1, 2(i 1 – i 2 ) + 4(i 1 – i 3 ) – 12 = 0 which leads to 3i 1 – i 2 – 2i 3 = 6

(1)

For the supermesh, 2(i 2 – i 1 ) + 8i 2 + 2v 0 + 4(i 3 – i 1 ) = 0 But v 0 = 2(i 1 – i 2 ) which leads to -i 1 + 3i 2 + 2i 3 = 0 (2) For the independent current source, i 3 = 3 + i 2 Solving (1), (2), and (3), we obtain,

i 1 = 3.5 A, i 2 = -0.5 A, i 3 = 2.5 A.

(3)

Chapter 3, Solution 53 Applying mesh analysis leads to; –12 + 4kI 1 – 3kI 2 – 1kI 3 = 0 –3kI 1 + 7kI 2 – 4kI 4 = 0 –3kI 1 + 7kI 2 = –12 –1kI 1 + 15kI 3 – 8kI 4 – 6kI 5 = 0 –1kI 1 + 15kI 3 – 6k = –24 I 4 = –3mA –6kI 3 – 8kI 4 + 16kI 5 = 0 –6kI 3 + 16kI 5 = –24 Putting these in matrix form (having substituted I 4

(1) (2) (3) (4) (5) = 3mA in the above),

 4  3  1 0   I1   12   3 7 0 0  I 2    12    k   1 0 15  6  I 3   24       0  6 16   I 5   24 0

ZI = V Using MATLAB,

>> Z = [4,-3,-1,0;-3,7,0,0;-1,0,15,-6;0,0,-6,16] Z= 4 -3 -1 0 -3 7 0 0 -1 0 15 -6 0 0 -6 16 >> V = [12,-12,-24,-24]' V= 12 -12 -24 -24 We obtain, >> I = inv(Z)*V

I= 1.6196 mA –1.0202 mA –2.461 mA 3 mA –2.423 mA

Chapter 3, Solution 54

Let the mesh currents be in mA. For mesh 1,  12  10  2 I 1  I 2  0   2  2I1  I 2 For mesh 2,  10  3I 2  I 1  I 3  0   10   I 1  3I 2  I 3 For mesh 3,  12  2 I 3  I 2  0   12   I 2  2 I 3 Putting (1) to (3) in matrix form leads to  2  1 0  I 1   2         1 3  1 I 2   10   0  1 2  I  12   3     Using MATLAB,

 5.25  I  A B   8.5  10.25 1

 

AI  B

  I 1  5.25 mA, I 2  8.5 mA, I 3  10.25 mA

I 1 = 5.25 mA, I 2 = 8.5 mA, and I 3 = 10.25 mA.

(1)

(2)

(3)

Chapter 3, Solution 55 10 V

I2

b

i1 4A

c

+

1A

I2

6

1A

2

i2 i3

I4 4A

I3

d

I1

12 

4 +–

a

I3

0

I4 8V

It is evident that I 1 = 4 For mesh 4,

(2)

12(I 4 – I 1 ) + 4(I 4 – I 3 ) – 8 = 0

For the supermesh

At node c,

(1)

6(I 2 – I 1 ) + 10 + 2I 3 + 4(I 3 – I 4 ) = 0 or -3I 1 + 3I 2 + 3I 3 – 2I 4 = -5

I2 = I3 + 1

(3) (4)

Solving (1), (2), (3), and (4) yields, I 1 = 4A, I 2 = 3A, I 3 = 2A, and I 4 = 4A At node b,

i 1 = I 2 – I 1 = -1A

At node a,

i 2 = 4 – I 4 = 0A

At node 0,

i 3 = I 4 – I 3 = 2A

Chapter 3, Solution 56 + v1 – 2 2

i2

2

2 12 V

+

i1

–

2

i3

+ v2

–

For loop 1, 12 = 4i 1 – 2i 2 – 2i 3 which leads to 6 = 2i 1 – i 2 – i 3

(1)

For loop 2, 0 = 6i 2 –2i 1 – 2 i 3 which leads to 0 = -i 1 + 3i 2 – i 3

(2)

For loop 3, 0 = 6i 3 – 2i 1 – 2i 2 which leads to 0 = -i 1 – i 2 + 3i 3

(3)

In matrix form (1), (2), and (3) become,  2  1  1  i1  6   1 3  1 i   0  2       1  1 3  i 3  0 2

1 1

2

6 1

 =  1 3  1  8,  2 =  1 3  1  24 1 1 3 1 0 3 2

1 6

 3 =  1 3 0  24 , therefore i 2 = i 3 = 24/8 = 3A, 1 1 0 v 1 = 2i 2 = 6 volts, v = 2i 3 = 6 volts

Chapter 3, Solution 57 Assume R is in kilo-ohms. V2  4kx15mA  60V, V1  90  V2  90  60  30V Current through R is 3 3   30  (15)R iR  i o, V1  i R R 3 R 3 R This leads to R = 90/15 = 6 kΩ.

Chapter 3, Solution 58 30 

i2 30 

10 

i1

10 

30 

i3

+

– 120 V

For loop 1, 120 + 40i 1 – 10i 2 = 0, which leads to -12 = 4i 1 – i 2

(1)