DC Electrical Circuit Analysis : Practice Problems, Methods, and Solutions [1st ed.] 9783030507107, 9783030507114

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Mehdi Rahmani-Andebili

DC Electrical Circuit Analysis Practice Problems, Methods, and Solutions

DC Electrical Circuit Analysis

Mehdi Rahmani-Andebili

DC Electrical Circuit Analysis Practice Problems, Methods, and Solutions

Mehdi Rahmani-Andebili Department of Engineering Technology The State University of New York, Buffalo State Buffalo, NY, USA

ISBN 978-3-030-50710-7 ISBN 978-3-030-50711-4 https://doi.org/10.1007/978-3-030-50711-4

(eBook)

# Springer Nature Switzerland AG 2020 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Preface

Electrical Circuit Analysis is one of the most fundamental subjects of Electrical Engineering major which is taught in two courses in successive semesters under the names of “Electrical Circuit Analysis I” and “Electrical Circuit Analysis II” or under the names of “DC Electrical Circuit Analysis” and “AC Electrical Circuit Analysis” in universities and colleges all over the world. This textbook includes basic and advanced exercises of DC Electrical Circuit Analysis with very detailed and multiple methods of solutions. The textbook can be used as a practicing textbook by students and as a supplementary teaching source by instructors. To help students study the textbook in the most efficient way, the exercises have been categorized in nine different levels. In this regard, for each problem of the textbook, a difficulty level (easy, normal, or hard) and a calculation amount (small, normal, or large) have been assigned. Moreover, in each chapter, problems have been ordered from the easiest problem with the smallest calculations to the most difficult problems with the largest calculations. Therefore, students are recommended to start studying the textbook from the easiest problems and continue practicing until they reach the normal and then the hardest ones. On the other hand, this classification can help instructors choose appropriate problems to conduct a quiz or a test. Moreover, the classification of computation amount can help students manage their time during future exams and instructors give appropriate problems based on exam duration. Since the problems have very detailed solutions and some of them include multiple methods of solution, the textbook can be useful for the under-prepared students. In addition, the textbook is beneficial for knowledgeable students because it includes advanced exercises. While preparing the problem solutions, the use of typical methods of electrical circuit analysis was attempted to present the textbook as an instructor-recommended one. In other words, the heuristic methods of problem solution have never been used as the first method of problem solution. By considering this key point, the textbook will be aligned to instructors’ lectures, and the instructors will not see any untaught problem solutions in their students’ answer sheets. The Iranian University Entrance Exams for the Master’s and PhD degrees of Electrical Engineering major [1] is the main reference of the textbook; however, all the problem solutions have been provided by me. The Iranian University Entrance Exam is one of the most competitive university entrance exams in the world that allows only 10% of the applicants to get into prestigious and tuition-free Iranian universities. Students and instructors are welcome to email their comments to me if they find any misprint or any other possible mistakes in the textbook. Their name and contribution will be mentioned in the next version of the textbook. [1] Iranian National Organization of Educational Testing, www.sanjesh.org. Buffalo, NY, USA

Mehdi Rahmani-Andebili

v

Contents

1 2

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

33

3

Problems: First-Order Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

4

Solutions of Problems: First-Order Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

5

Problems: Second-Order and Higher-Order Circuits . . . . . . . . . . . . . . . . . . . . 195

6

Solutions of Problems: Second-Order and Higher-Order Circuits . . . . . . . . . . 209

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261

vii

About the Author

Dr. Mehdi Rahmani-Andebili is an Assistant Professor in the Engineering Technology Department at The State University of New York, Buffalo State. As a professor, he has taught many courses such as Essentials of Electrical Engineering Technology, Electrical Circuits Analysis I, Electrical Circuits Analysis II, Electrical Circuits and Devices, Industrial Electronics, and Renewable Distributed Generation and Storage. Dr. Rahmani-Andebili has more than 100 single-author publications including textbooks, monographs, book chapters, journal papers, and conference papers. He received his M.Sc. and Ph.D. degrees in Electrical Engineering (Power System) from Tarbiat Modares University and Clemson University in 2011 and 2016, respectively. Moreover, he was a Postdoctoral Fellow at Sharif University of Technology during 2016–2017

ix

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Abstract

This chapter helps both groups of underprepared and knowledgeable undergraduate students taking courses in electrical circuit analysis. In this chapter, the basic and advanced problems of three important subjects of electrical circuit analysis, that is, circuit components, methods of circuit analysis, and circuit theorems, are presented. The problems of methods of circuit analysis are concerned with mesh analysis, nodal analysis, and heuristic techniques. Moreover, the problems of circuit theorems are related to source transformation theorem, superposition theorem, Thevenin theorem, Norton theorem, and maximum power transform theorem. Like the other chapters of this textbook, the problems are categorized in different levels based on their difficulty levels (easy, normal, or hard) and calculation amounts (small, normal, or large). Additionally, the problems are ordered from the easiest problem with the smallest calculations to the most difficult problems with the largest calculations. 1.1. What is the value of I in the circuit of Fig. 1.1? Difficulty level ● Easy ○ Normal ○ Hard Computation amount ● Small ○ Normal ○ Large 1) 1.5 A 2) 2 A 3) 2.5 A 4)
0.5A

Fig. 1.1 The circuit of problem 1.1 # Springer Nature Switzerland AG 2020 M. Rahmani-Andebili, DC Electrical Circuit Analysis, https://doi.org/10.1007/978-3-030-50711-4_1

1

2

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

1.2. How much are the values of Vx and Vy in the circuit of Fig. 1.2? Difficulty level ● Easy ○ Normal ○ Hard Computation amount ● Small ○ Normal ○ Large 1) Vx ¼ 8 V, Vy ¼ 2 V 2) Vx ¼
8 V, Vy ¼ 2 V 3) Vx ¼ 8 V, Vy ¼
2 V 4) Vx ¼
8 V, Vy ¼
2 V

Fig. 1.2 The circuit of problem 1.2

1.3. Determine the value of Ia in the circuit of Fig. 1.3. Difficulty level ● Easy ○ Normal ○ Hard Computation amount ● Small ○ Normal ○ Large 1) 0.5 A 2) 1 A 3) 2 A 4) 1.5 A

Fig. 1.3 The circuit of problem 1.3

1.4. In the circuit of Fig. 1.4, determine the voltage across the 5 Ω resistor. Difficulty level ● Easy ○ Normal ○ Hard Computation amount ● Small ○ Normal ○ Large 1) 10 V 2) 15 V 3) 20 V 4)
30 V

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Fig. 1.4 The circuit of problem 1.4

1.5. How much is the value of Ix in the circuit of Fig. 1.5? Difficulty level ● Easy ○ Normal ○ Hard Computation amount ○ Small ● Normal ○ Large 1) 40 21 A 2) 40 7 A 80 3) 7 A 4) 80 21 A

Fig. 1.5 The circuit of problem 1.5

1.6. How much should be the value of V in the circuit of Fig. 1.6 to have Vx ¼ 0? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ● Small ○ Normal ○ Large 1) 3 V 2) 6 V 3) 12 V 4) 18 V

Fig. 1.6 The circuit of problem 1.6

3

4

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

1.7. How much is the resistance seen from terminal A-B in the circuit of Fig. 1.7? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ● Small ○ Normal ○ Large 1) 0 2) 0.5 Ω 3) 2 Ω 4) 1

Fig. 1.7 The circuit of problem 1.7

1.8. What must be the value of RL to absorb the maximum power in the circuit of Fig. 1.8? Moreover, how much is the maximum power transferrable to RL? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ● Small ○ Normal ○ Large 1) 5 Ω, 2.5 W 2) 10 Ω, 2.5 W 3) 7.5 Ω, 1.25 W 4) 2.5 Ω, 1.75 W

Fig. 1.8 The circuit of problem 1.8

1.9. In the circuit of Fig. 1.9, which source is absorbing power? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ● Small ○ Normal ○ Large 1) I1 2) V2 3) I3 4) V4

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Fig. 1.9 The circuit of problem 1.9

1.10. Determine the power of the independent current source in the circuit of Fig. 1.10. Difficulty level ○ Easy ● Normal ○ Hard Computation amount ● Small ○ Normal ○ Large 1) 120 W 2)
30 W 3)
120 W 4)
150 W

Fig. 1.10 The circuit of problem 1.10

1.11. The power of some of the components is given in the circuit of Fig. 1.11. Determine the current of I. Difficulty level ○ Easy ● Normal ○ Hard Computation amount ● Small ○ Normal ○ Large 1) 1 A 2) 2 A 3) 0.5 A 4) 0.25 A

Fig. 1.11 The circuit of problem 1.11

5

6

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

1.12. Determine the Thevenin resistance seen from terminal a-b in the circuit of Fig. 1.12. Difficulty level ○ Easy ● Normal ○ Hard Computation amount ● Small ○ Normal ○ Large 1) 1 Ω 2) 1.33 Ω 3) 1.5 Ω 4) 2 Ω

Fig. 1.12 The circuit of problem 1.12

1.13. Determine the Thevenin voltage and Thevenin resistance of the circuit of Fig. 1.13 seen from terminal a-b. Difficulty level ○ Easy ● Normal ○ Hard Computation amount ● Small ○ Normal ○ Large 1) 15 V, 2 Ω 2) 12 V, 4 Ω 3) 12 V, 5 Ω 4) 15 V, 5 Ω

Fig. 1.13 The circuit of problem 1.13

1.14. What is the power of 5 V voltage source in the circuit of Fig. 1.14? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ● Small ○ Normal ○ Large 1) 15 W generation 2) 15 W consumption 3) 20 W consumption 4) 20 W generation

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Fig. 1.14 The circuit of problem 1.14

1.15. What is the Thevenin resistance seen from terminal a-b of the circuit of Fig. 1.15? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ● Small ○ Normal ○ Large 1) 0 2) 1 Ω 3) 1 4) 12 Ω

Fig. 1.15 The circuit of problem 1.15

1.16. How much is Vo in the circuit of Fig. 1.16? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ● Small ○ Normal ○ Large 1) 27 5 V 2) 5 V 3) 24 5 V 4) 6 V

7

8

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Fig. 1.16 The circuit of problem 1.16

1.17. How much is Vab in the circuit of Fig. 1.17? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ● Small ○ Normal ○ Large 1) 1.25 V 2) 1.75 V 3) 2.5 V 4) 3.25 V

Fig. 1.17 The circuit of problem 1.17

1.18. What must be the size of RL that will absorb the maximum power in the circuit of Fig. 1.18? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ● Small ○ Normal ○ Large 1) 5 W 2) 10 W 3) 15 W 4) 20 W

Fig. 1.18 The circuit of problem 1.18

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

9

1.19. What is the Thevenin equivalent circuit seen from terminal a-b in the circuit of Fig. 1.19? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ● Normal ○ Large 1) RTh ¼ 100 Ω, VTh ¼ 10 V 2) RTh ¼ 180 Ω, VTh ¼ 0 V 3) RTh ¼ 120 Ω, VTh ¼ 15 V 4) RTh ¼ 100 Ω, VTh ¼ 0 V

Fig. 1.19 The circuit of problem 1.19

1.20. In the circuit of Fig. 1.20, which matrix shows the right equations of mesh analysis based on the given mesh currents? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ● Normal ○ Large 2 32 3 2 3 9
4 0 i1 7 6 76 7 6 7 1) 4 20
13 3 54 i2 5 ¼ 4 5 5 2
1
1 0 i3 2

9 6 2) 4 20


4
13

2


1

2

32 3 2 3 0 i1 7 76 7 6 7 3 54 i2 5 ¼ 4
3 5
1 0 i3 32

2
1

2 3 3 76 7 6 7 3 54 i2 5 ¼ 4 5 5
3 0 i3

9 6 4) 4 5


4 2

32 3 2 3 0 i1 3 76 7 6 7
3 54 i2 5 ¼ 4 5 5

2


1

9

6 3) 4 5 2 2


4

0

3

i1

i3

3

0

10

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Fig. 1.20 The circuit of problem 1.20

1.21. What is the range of the input resistance of the circuit of Fig. 1.21 seen from terminal a-b? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ● Normal ○ Large 1) 0 < Rin < 1 2) 0.75 < Rin < 1 3) 0 < Rin < 0.83 4) 0.78 < Rin < 0.83

Fig. 1.21 The circuit of problem 1.21

1.22. Determine the value of RL to absorb the maximum power in the circuit of Fig. 1.22. Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ● Normal ○ Large 1) 6 Ω 2) 18 Ω 3) 12 Ω 4) 8 Ω

Fig. 1.22 The circuit of problem 1.22

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

1.23. How much is the value of VAB in the circuit of Fig. 1.23? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ● Normal ○ Large 1) 1 V 2) 2 V 3) 0.5 V 4) 0.25 V

Fig. 1.23 The circuit of problem 1.23

1.24. How much is the input resistance of the circuit of Fig. 1.24 seen from terminal a-b? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ● Normal ○ Large 1) 47 Ω 2) 15 14 Ω 3) 74 Ω 4) 14 15 Ω

Fig. 1.24 The circuit of problem 1.24

1.25. Determine the value of I in the circuit of Fig. 1.25. Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ● Normal ○ Large 1) 5 A 2) 6 A 3)
5 A 4)
6 A

11

12

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Fig. 1.25 The circuit of problem 1.25

1.26. In the circuit of Fig. 1.26, for what value of k is the input resistance seen from terminal a-b zero? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ● Normal ○ Large 1) 52 Ω 2) 25 Ω 4 3) 13 Ω 4) 13 4 Ω

Fig. 1.26 The circuit of problem 1.26

1.27. What is the maximum power that can be transferred to the load in the circuit of Fig. 1.27? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ● Normal ○ Large 1) 2.25 W 2) 4.5 W 3) 6 W 4) 9 W

Fig. 1.27 The circuit of problem 1.27

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

1.28. How much is the Thevenin voltage seen from the indicated terminal of the circuit of Fig. 1.28? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ● Normal ○ Large 1) 3 V 2)
3 V 3) 13 V 4)
13 V

Fig. 1.28 The circuit of problem 1.28

1.29. In the circuit of Fig. 1.29, what is the Thevenin resistance seen from terminal a-b? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ● Normal ○ Large 1) R 2) R(1 + 2αμ)   3) R 1 þ αμ 2 4) R3

Fig. 1.29 The circuit of problem 1.29

1.30. How much is the power absorbed by the 2 Ω resistor in the circuit of Fig. 1.30? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ● Normal ○ Large 1) 0.08 W 2) 0.32 W 3) 0.64 W 4) 3.92 W

13

14

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Fig. 1.30 The circuit of problem 1.30

1.31. Which one of the following circuits shown in Figs. 1.31.1–4 is the Norton equivalent circuit seen from terminal a-b? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ● Normal ○ Large

Fig. 1.31 The circuit of problem 1.31

1.32. What are the parameters of the Thevenin equivalent circuit seen from terminal a-b in the circuit of Fig. 1.32? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ● Normal ○ Large 1) Req ¼ 4 Ω, V oc ¼ 45 I s 2) Req ¼ 6 Ω, V oc ¼ 45 I s 3) Req ¼ 6 Ω, V oc ¼ 54 I s 4) Req ¼ 4 Ω, V oc ¼ 54 I s

Fig. 1.32 The circuit of problem 1.32

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

15

1.33. In the circuit of Fig. 1.33, how much is the Req? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ● Normal ○ Large 1) 13 R 2) 2R 3) 3R 4) 12 R

Fig. 1.33 The circuit of problem 1.33

1.34. In the circuit of Fig. 1.34, for what value of k, the voltage V, resulted from the independent current source, is equal to the half value of the current of the independent current source (V ¼ I2s )? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ● Normal ○ Large 1) 1 2) 0.5 3)
5 4) For no value of k, since the circuit does not have a unique solution

Fig. 1.34 The circuit of problem 1.34

1.35. How much are the Thevenin voltage and the Thevenin resistance seen from the terminal in the circuit of Fig. 1.35? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ● Normal ○ Large 1) 2 V, 5 Ω 2) 5 V, 2 Ω 3) 25 V, 20 Ω 4) 25 V, 25 Ω

16

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Fig. 1.35 The circuit of problem 1.35

1.36. Determine the Thevenin voltage and the Thevenin resistance seen from terminal a-b in the circuit of Fig. 1.36. Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ● Normal ○ Large 1) 40 V, 20 23 Ω 2) 40 V, 40 23 Ω 3) 50 V, 40 23 Ω 4) 60 V, 20 23 Ω

Fig. 1.36 The circuit of problem 1.36

1.37. Determine the Thevenin voltage and the Thevenin resistance seen from terminal a-b in the circuit of Fig. 1.37. Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ● Normal ○ Large 1) 8 V, 6 Ω 2) 12 V, 6 Ω 3) 16 V, 2 Ω 4) 24 V, 4 Ω

Fig. 1.37 The circuit of problem 1.37

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

17

1.38. How much is the power of the dependent current source in the circuit of Fig. 1.38? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ● Normal ○ Large 1)
1.28 W 2) 1.28 W 3) 0.64 W 4)
0.64 W

Fig. 1.38 The circuit of problem 1.38

1.39. What is the relation between Vt and It in the circuit of Fig. 1.39? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ● Normal ○ Large 1) Vt ¼ 3It + 20 2) Vt ¼ 5It + 30 3) Vt ¼ 7It + 10 4) Vt ¼ 7It + 20

Fig. 1.39 The circuit of problem 1.39

1.40. Determine the Thevenin voltage (V) and the Thevenin resistance (Ω), respectively, seen from terminal a-b, in the circuit of Fig. 1.40. Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ● Normal ○ Large 9 1) 24 25 , 25 2)

9 24 25 I s , 25

3)

22 4 231 I s , 231

4)

16 20 77 I s , 77

18

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Fig. 1.40 The circuit of problem 1.40

1.41. In the circuit of Fig. 1.41, how much is the Req seen from terminal a-b? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ● Normal ○ Large 1) 12 Ω 2) 23 Ω 3) 32 Ω 4) 52 Ω

Fig. 1.41 The circuit of problem 1.41

1.42. How much is the power of the independent current source in the circuit of Fig. 1.42? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ● Normal ○ Large 1) It generates 32 W. 2) It consumes 32 W. 3) It generates 67 W. 4) It consumes 64 W.

Fig. 1.42 The circuit of problem 1.42

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

1.43. In the circuit of Fig. 1.43, how much is the Norton resistance seen from terminal a-b? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ● Normal ○ Large 1) 30 Ω 2) 60 Ω 3) 90 Ω 4) 120 Ω

Fig. 1.43 The circuit of problem 1.43

1.44. In the circuit of Fig. 1.44, determine the power of the independent voltage source. Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ● Normal ○ Large 1)
5 W 2)
10 W 3)
40 W 4)
20 W

Fig. 1.44 The circuit of problem 1.44

1.45. In the circuit of Fig. 1.45, how much is the power lost in the vertical 4 Ω resistor? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ● Normal ○ Large 1) 1.44 W 2) 2.44 W 3) 7.2 W 4) 12.96 W

19

20

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Fig. 1.45 The circuit of problem 1.45

1.46. What is the input resistance seen from terminal A-B of the circuit of Fig. 1.46? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ○ Normal ● Large 1) 3R + 3.2 2) 2R
1.5 3) 1.4R + 3 4) 2.4R + 2

Fig. 1.46 The circuit of problem 1.46

1.47. Determine the value of Vx and Vy in the circuit of Fig. 1.47. Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ○ Normal ● Large 1) Vx ¼ 20 V, Vy ¼
7.5 V 2) Vx ¼ 20 V, Vy ¼ 7.5 V 3) Vx ¼ 7.5 V, Vy ¼
15 V 4) Vx ¼ 7.5 V, Vy ¼ 15 V

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Fig. 1.47 The circuit of problem 1.47

1.48. In the circuit of Fig. 1.48, what is the Thevenin equivalent circuit seen from terminal A-B? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ○ Normal ● Large 1) A voltage source with the size of 53 I s 2) A voltage source with the size of 35 I s series with a 45 Ω resistor 3) A voltage source with the size of 35 I s 4) A voltage source with the size of 53 I s series with a 45 Ω resistor

Fig. 1.48 The circuit of problem 1.48

1.49. What is the Thevenin equivalent circuit seen from terminal A-B in the circuit of Fig. 1.49? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ○ Normal ● Large 1) RTh ¼
2 Ω, VTh ¼
36Is 2) RTh ¼
2 Ω, VTh ¼ 36Is 3) RTh ¼ 2 Ω, VTh ¼
36Is 4) RTh ¼ 2 Ω, VTh ¼ 36Is

21

22

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Fig. 1.49 The circuit of problem 1.49

1.50. What is the equivalent resistance seen from the terminal in the circuit of Fig. 1.50? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ○ Normal ● Large 1) 2) 3) 4)

2þβ 2
β RL βþ2 β
2 RL 2
β 2þβ RL β
2 βþ2 RL

Fig. 1.50 The circuit of problem 1.50

1.51. How much is the maximum power that the resistor can absorb in the circuit of Fig. 1.51? Difficulty level ○ Easy ● Normal ○ Hard Computation amount ○ Small ○ Normal ● Large 1) 4.5 W 2) 9 W 3) 13.5 W 4) 18 W

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

23

Fig. 1.51 The circuit of problem 1.51

1.52. In the circuit of Fig. 1.52, all the resistors are 1 Ω. How much is the equivalent resistance between the nodes of “A” and “B”? Difficulty level ○ Easy ○ Normal ● Hard Computation amount ● Small ○ Normal ○ Large 1) 14 Ω 2) 13 Ω 3) 12 Ω 4) 34 Ω

Fig. 1.52 The circuit of problem 1.52

1.53. In the circuit of Fig. 1.53, what is the Thevenin equivalent circuit seen from terminal a-b? Difficulty level ○ Easy ○ Normal ● Hard Computation amount ● Small ○ Normal ○ Large 1) V Th ¼ 56 E, RTh ¼ 56 R. 2) V Th ¼ 56 E, RTh ¼ 5R. 3) V Th ¼ 56 E, RTh ¼ 0. 4) There is no Thevenin equivalent circuit for this circuit.

24

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Fig. 1.53 The circuit of problem 1.53

1.54. The network shown in Fig. 1.54 is resistive, linear, and time-invariant. About 20% of the power of the voltage source is absorbed by the 5 Ω resistor. How to increase the size of the voltage source to transfer 30% of its power to the resistor? Difficulty level ○ Easy ○ Normal ● Hard Computation amount ● Small ○ Normal ○ Large 1) 32 times. 2) 94 times. 3) It is impossible, since the power absorbed by the resistor only depends on its resistance. 4) It is impossible, since the power absorbed by the resistor only depends on its resistance and the network configuration.

Fig. 1.54 The circuit of problem 1.54

1.55. How much is the input resistance seen from terminal a-b in the circuit of Fig. 1.55? Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ● Normal ○ Large 5 1) 11 Ω 2) 1 Ω 3) 3 Ω 4) 39 11 Ω

Fig. 1.55 The circuit of problem 1.55

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

25

1.56. How much is the value of I in the circuit of Fig. 1.56? Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ● Normal ○ Large 1) 13 mA 2) 23 mA 3) 12 mA 4) 43 mA

Fig. 1.56 The circuit of problem 1.56

1.57. In the circuit of Fig. 1.57, suppose that the resistance of each resistor is R. Determine the input resistance of the circuit. Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ● Normal ○ Large 1) R 2) 0.618R 3) 0.5R 4) 0.382R

Fig. 1.57 The circuit of problem 1.57

1.58. In the circuit of Fig. 1.58, the conductance of each resistor is given in mho or siemens. What is the condition on α to adjust the circuit as an amplifier? Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ● Normal ○ Large 1) α > 10. 2) α > 12. 3) α > 14. 4) Because of the presence of a dependent source in the circuit, the circuit will always behave as an amplifier.

26

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Fig. 1.58 The circuit of problem 1.58

1.59. Determine the range of α that results in a negative value for the input resistance of the circuit of Fig. 1.59 seen from terminal a-b. Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ● Normal ○ Large 1) 73 < a < 3 2) 2 < a < 3 3) 73 < a < 4 4) 2 < a < 4

Fig. 1.59 The circuit of problem 1.59

1.60. How much is the value of Isc in the circuit of Fig. 1.60? Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ● Normal ○ Large 1) 1 A 2) 2 A 3) 4 A 4) 8 A

Fig. 1.60 The circuit of problem 1.60

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

27

1.61. How much is the equivalent resistance seen from terminal A-B in the circuit of Fig. 1.61? Assume that each resistor has 1 Ω resistance. Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ● Normal ○ Large 1) 13 Ω 2) 23 Ω 3) 1 Ω 4) 12 Ω

Fig. 1.61 The circuit of problem 1.61

1.62. In the circuit of Fig. 1.62, determine the Norton equivalent circuit seen by the 7 Ω resistor. Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ● Normal ○ Large 1) RN ¼ 2 Ω, I N ¼ 13 A 2) RN ¼ 403 Ω, I N ¼ 13 A 3) RN ¼ 203 Ω, I N ¼ 12 A 4) RN ¼ 403 Ω, I N ¼ 12 A

Fig. 1.62 The circuit of problem 1.62

28

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

1.63. In the circuit of Fig. 1.63, for what value of “m,” the value of resistance, seen from terminal a-b, is negative? Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ● Normal ○ Large 1) 3 2) 4 3) 5 4) 6

Fig. 1.63 The circuit of problem 1.63

1.64. What is the equivalent circuit seen from terminal a-b in the circuit of Fig. 1.64? Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ● Normal ○ Large 6 1) RTh ¼ 13 R, V Th ¼ EþIR 2 2) RTh ¼ 2R, VTh ¼ E 3) RTh ¼ R, V Th ¼ EþIR 4 4) RTh ¼ R2 , V Th ¼ E
IR 4

Fig. 1.64 The circuit of problem 1.64

1.65. In the circuit of Fig. 1.65, if a ! 1, how much will be RTh and VTh seen from terminal a-b? Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ● Normal ○ Large 1) 1 Ω, 2 V 2) 0 Ω,
2 V 3) 0 Ω, 1 V 4) 1 Ω,
1 V

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

29

Fig. 1.65 The circuit of problem 1.65

1.66. In the circuit of Fig. 1.66, for what value of α, the input resistance (Rin), seen from terminal a-b, will be equal to R (Rin ¼ R)? Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ● Normal ○ Large 1) α ¼ 0 2) α ¼ 0.5 3) α ¼ 1 4) Any value of α, but α 6¼ 0

Fig. 1.66 The circuit of problem 1.66

1.67. What are the parameters of the Norton equivalent circuit seen from terminal a-b in the circuit of Fig. 1.67? Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ● Normal ○ Large 1) I N ¼ 1 A, RN ¼ 43 Ω 2) I N ¼ 1 A, RN ¼ 73 Ω 3) I N ¼ 34 A, RN ¼ 43 Ω 4) I N ¼ 34 A, RN ¼ 73 Ω

30

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Fig. 1.67 The circuit of problem 1.67

1.68. In the circuit of Fig. 1.68, the resistance of each resistor is 20 Ω. Determine the equivalent resistance between the nodes of “A” and “O.” Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ○ Normal ● Large 1) 9 Ω 2) 1.8 Ω 3) 2.15 Ω 4) 7.5 Ω

Fig. 1.68 The circuit of problem 1.68

1.69. How much is the input resistance (Rin) in the circuit of Fig. 1.69, when α goes to infinity (α ! 1)? Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ○ Normal ● Large 1) 23 R 2)
12 R 3) 32 R 4)
2R

Fig. 1.69 The circuit of problem 1.69

1

Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

1.70. What are the Thevenin voltage and the Thevenin resistance seen from the terminal in the circuit of Fig. 1.70? Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ○ Normal ● Large 5 1)
10 3 V, 3 Ω 1 2) 2 V, 2 Ω 3) 2 V, 12 Ω 5 4) 10 3 V, 3 Ω

Fig. 1.70 The circuit of problem 1.70

31

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Abstract

In this chapter, the problems of the first chapter are fully solved, in detail, step-by-step, and with multiple methods, by using mesh analysis, nodal analysis, and heuristic techniques. In all the problems’ solutions, Kirchhoff current law (KCL) and Kirchhoff voltage law (KVL) as well as the concepts of mesh current, node voltage, supermesh, and supernode are clearly described. Moreover, in the application of circuit theorems, the concepts of open circuit voltage, short-circuit current, Thevenin voltage, Thevenin resistance, Norton current, and Norton resistance are explained. In addition, this chapter shows how to deal with circuit components like resistors, dependent current and voltage sources, and independent current and voltage sources. 2.1. Applying conventional methods like mesh and nodal analyses will need lots of calculations. By looking at the circuit, it can be noticed that we need to apply one KVL and one KCL, as can be seen in the figure. KVL :
V þ I þ 2
4I ¼ 0 KCL : 1 þ I þ

V ¼ 0 ) V ¼
I
1 1

ð1Þ ð2Þ

put ð2Þ in ð1Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
ð
I
1Þ þ I þ 2
4I ¼ 0 ) 3 ¼ 2I ) I ¼ 1:5 A Choice (1) is the answer.

Fig. 2.1 The circuit of solution of problem 2.1 # Springer Nature Switzerland AG 2020 M. Rahmani-Andebili, DC Electrical Circuit Analysis, https://doi.org/10.1007/978-3-030-50711-4_2

33

34

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

2.2. Nodal analysis is the best method to solve this problem, since there are just two independent nodes (nodes 1 and 2) with unknown voltages (e1 and e2), as can be seen in the figure. KCL in the indicated supernode of the circuit of Fig. 2.2.2:
3V y þ

e2 e
10 3 ¼ 0 )
3V y þ e2
9 ¼ 0
4þ 2 2 2 1

ð1Þ

Now, we need to define Vx and Vy based on the voltages of independent nodes (e1 and e2). V x ¼ 10
e2

ð2Þ

V y ¼
e2

ð3Þ

3 9
3ð
e2 Þ þ e2
9 ¼ 0 ) e2 ¼ 9 ) e2 ¼ 2 2 2

ð4Þ

Now, put (3) in (1) and simplify them.

Solving (2) and (4): V x ¼ 10
e2 ¼ 10
2 ¼ 8 V Solving (3) and (4): V y ¼
e2 ¼
2 V Choice (3) is the answer.

Fig. 2.2 The circuits of solution of problem 2.2

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

35

2.3. Nodal analysis is the best method to solve the problem, since the circuit will include the least variables (two independent nodes with unknown voltages, i.e., e1 and e2). As can be seen in the circuit, the dependent voltage source is located between two nodes; thus, we must use KCL for the supernode including these nodes. KCL in the supernode, and simplifying it: e1
4 e1 e2 þ þ
4 ¼ 0 ) e1 þ e2 ¼ 6 2 2 1

ð1Þ

The relation between the voltage of voltage source and node voltages: e2
e1 ¼ 2I a

ð2Þ

Defining Ia based on the node voltages: Ia ¼

e1 2


 ð2Þ, ð3Þ e ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) e2
e1 ¼ 2 1 ¼ e1 ) e2 ¼ 2e1 2

ð3Þ ð4Þ

Solving (1) and (4): ) e1 þ 2e1 ¼ 6 ) e1 ¼ 2 V ð3Þ, ð5Þ e 2 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) I a ¼ 1 ¼ ¼ 1 V 2 2 Choice (2) is the answer.

Fig. 2.3 The circuit of solution of problem 2.3

2.4. Using Ohm’s law in the left-side loop: I1 ¼

2 A 3

Using Ohm’s law in the right-side loop: 2 V 2 ¼ 5I 2 ¼ 5ð6I 1 Þ ¼ 30  ¼ 20 V 3 Choice (3) is the answer.

ð5Þ

36

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Fig. 2.4 The circuit of solution of problem 2.4

2.5. Mesh analysis is the best method to solve this problem. Since there is a current source between meshes 2 and 3, we must consider a supermesh instead of these meshes. KVL in mesh 1:
8 þ 1ði1
i3 Þ þ 2ði1
i2 Þ ¼ 0 ) 3i1
2i2
i3 ¼ 8

ð1Þ

KVL in the supermesh including meshes 2 and 3: 1ði3
i1 Þ þ 4i3 þ 3I x þ 2ði2
i1 Þ ¼ 0 )
3i1 þ 3I x þ 2i2 þ 5i3 ¼ 0

ð2Þ

Using Ix ¼ i1, which is clear in the figure: 2i2 þ 5i3 ¼ 0

ð3Þ

i2 ¼ i3 þ 3

ð4Þ

Auxiliary equation for the supermesh:

Solving (3) and (4): 2ði3 þ 3 Þ þ 5i3 ¼ 0 ) i3 ¼


6 A 7

ð5Þ

Solving (4) and (5): i2 ¼

15 A 7

Solving (1), (5), and (6):  3i1
2

   15 6 80

¼ 8 ) i1 ¼ A 7 7 21

Choice (4) is the answer.

Fig. 2.5 The circuit of solution of problem 2.5

ð6Þ

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

37

2.6. First Method: Based on the given information, Vx ¼ 0. Hence, the voltage of the resistor is zero, and no current will flow through that branch. By applying KCL in the indicated supernode of the circuit of Fig. 2.6.2, we have: I
1 þ 0 þ
2 þ I ¼ 0 ) 1:5I ¼ 3 ) I ¼ 2 A 2 Moreover, we need to write a KVL in the indicated loop.
V þ 3I
0 ¼ 0 ) V ¼ 3I ¼ 3  2 ) V ¼ 6 V Choice (2) is the answer. Second Method: By applying nodal analysis, we will have two unknowns and two equations, since there are four nodes that one of them needs to be considered as the ground and the other one is connected to a grounded voltage source. Third Method: Mesh analysis can be applied in this question, but it will need a lot of calculations. By applying mesh analysis, we will have four unknowns and four equations. Note that there are five meshes, but the current of top mesh is known (1 A).

Fig. 2.6 The circuits of solution of problem 2.6

2.7. We need to apply one KVL in the loop and one KCL in the supernode, as can be seen in Figs. 2.7.1 and 2.7.2, respectively. Then, RAB ¼ VI 11 . KVL in the loop of the circuit of Fig. 2.7.1: 4
V 1 þ 2V 2
3V 1 þ V 2 ¼ 0 ) V 2 ¼ V 1 3

ð1Þ

38

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

KCL in the supernode of the circuit of Fig. 2.7.2:
I 1 þ 3V 2
3I 1
2V 1 ¼ 0 )
4I 1 þ 3V 2
2V 1 ¼ 0

ð2Þ


 ð1Þ, ð2Þ 4 V ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
4I 1 þ 3 V 1
2V 1 ¼ 0 )
4I 1 þ 2V 1 ¼ 0 ) 1 ¼ 2 ) RAB ¼ 2 Ω 3 I1 Choice (3) is the answer.

Fig. 2.7 The circuits of solution of problem 2.7

2.8. To absorb the maximum power, RL must be equal to the Thevenin resistance (RTh) seen by itself. To calculate the Thevenin resistance, we must turn off all the power sources. As can be seen in Fig. 2.8.2, since there is no dependent power source in the circuit, we can use the series-parallel rule to determine the input resistance or equivalent resistance which is equal to the Thevenin resistance.
   RTh ¼ Rin ¼ Req ¼ 6 þ 44 þ 2 ¼ 10 Ω RL ¼ RTh ¼ 10 Ω Based on the maximum power transfer theorem, the maximum power that can be transferred to the load (PL, max) is equal to V 2Th 4RTh. To determine VTh, we need to calculate the open circuit voltage across the load (Voc). Then, VTh ¼ Voc. Note that power sources are not turned off in this part. As can be seen in Fig. 2.8.3, since there is no current flowing through the 6 Ω and 2 Ω resistors (no voltage drop), the Voc will be equal to the voltage of vertical 4 Ω resistor. Therefore, by using voltage division formula, we have: V Th ¼ V oc ¼ PL,max ¼ Choice (2) is the answer.

4  20 ¼ 10 V 4þ4

V 2Th 102 ¼ ¼ 2:5 W 4RTh 4  10

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

39

Fig. 2.8 The circuits of solution of problem 2.8

2.9. A source, or in general a component, absorbs power if the value of multiplication of its voltage and current (VI) is positive, assuming an associated reference direction for its voltage and current (current enters its positive terminal). Now, let us check the value of VI for each component of the circuit. P1 ¼ V 1 I 1 ¼ 2  ð
2Þ < 0 P2 ¼ V 2 I 2 ¼ 8  ð
2Þ < 0 P3 ¼ V 3 I 3 ¼ 10  4 > 0 P4 ¼ V 4 I 4 ¼ 10  ð
5Þ < 0 As can be seen, the third power source is a power absorber. Choice (3) is the answer.

Fig. 2.9 The circuit of solution of problem 2.9

40

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

2.10. To calculate the power of the independent current source, its voltage must be known. To find its voltage, the circuit needs to be analyzed. In this problem, nodal analysis is the best method. KCL in node 1:
6I 4 þ

e1
ð
70Þ e 3e
5 þ 1 ¼ 0 )
6I 4 þ 1 þ 9 ¼ 0 5 10 10 I4 ¼

e1 10

ð1Þ ð2Þ

ð1Þ, ð2Þ e 3e 3e ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
6 1 þ 1 þ 9 ¼ 0 ) 1 ¼ 9 ) e1 ¼ 30 V 10 10 10 P1 ¼ V 1 I 1 ¼ 30  ð
5Þ ¼
150 W Choice (4) is the answer.

Fig. 2.10 The circuit of solution of problem 2.10

2.11. Based on the energy conservation law, the algebraic sum of power of components in any circuit is zero. Herein, we must apply associated reference direction (current enters the positive terminal of the component). Therefore: N X

Pi ¼ 0 ) P1 þ P2 þ P3 þ P4 þ RI 2 þ 16I ¼ 0

i¼1

) 6 þ ð
3Þ þ 1 þ 12 þ 16I 2
32I ¼ 0 ) 16I 2
32I þ 16 ¼ 0 ) I 2
2I þ 1 ¼ 0 ) I ¼ 1 A Choice (1) is the answer.

Fig. 2.11 The circuit of solution of problem 2.11

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

41

2.12. To determine the Thevenin resistance (RTh) of a circuit, first, we need to turn off all the independent sources (independent voltage and current sources). In other words, an independent voltage source must be replaced by a short circuit branch, and an independent current source needs to be replaced by an open circuit branch (see Fig. 2.12.2). Then, if the circuit includes, at least, one dependent source (dependent voltage or dependent current source), we must apply a test source (test voltage or current source with the voltage and current of Vt and It) to analyze the circuit to determine the value of VI tt , which is equal to RTh. Otherwise, if the circuit does not include any dependent source, we can simply apply the series-parallel rule to calculate the Thevenin resistance of the circuit. Although there is a dependent source in the circuit of the problem, we do not need to apply a test source to find the value of VI tt because the circuit is changed to a simple one. As can be seen in Fig. 2.12.3, Ix ¼ 0 after turning off the independent current source. Consequently, the dependent current source is shut down and changed to an open circuit. The remaining circuit includes two 1 Ω resistors. Thus: RTh ¼ 1 þ 1 ¼ 2Ω: Choice (4) is the answer.

Fig. 2.12 The circuits of solution of problem 2.12

42

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

2.13. To find the Thevenin voltage of the circuit (VTh), we need to determine the open circuit voltage (Voc) seen from the indicated terminal (a-b terminal). Moreover, to find the Thevenin resistance of the circuit (RTh), we must turn off all the independent sources and apply a test source (e.g., a test voltage source) to calculate the ratio of VI tt if the circuit includes a dependent source. However, analyzing the circuit to find Voc and VI tt by using the conventional methods (mesh and nodal analyses) needs lots of calculations. The best approach to solve this problem is applying a heuristic technique. As can be seen in Fig. 2.13.2, I ¼ 1 A, since no current flows through the 3 Ω resistor because terminal a-b is electrically open. Thus, by writing a KVL in the indicated mesh, we have:
V oc þ 10 þ 2  1 þ 3  0 ¼ 0 ) V oc ¼ 12 V Additionally, as can be seen in Fig. 2.13.3, the Thevenin resistance of the circuit is a series connection of 2 Ω and 3 Ω resistors. Therefore: RTh ¼ 2 þ 3 ¼ 5 Ω Choice (3) is the answer.

Fig. 2.13 The circuits of solution of problem 2.13

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

43

2.14. To calculate the power of a component, its voltage and current must be known. The current of the 5 V voltage source can be calculated by applying KCL in the indicated supernode of the circuit of Fig. 2.14, as follows:
1 þ I 5V
3 ¼ 0 ) I 5V ¼ 4 A Thus, the power of 5 V voltage source is P ¼ V5VI5V ¼ 5  4 ¼ 20 W, since the current enters the positive terminal of the voltage source. Additionally, since P > 0, the 5 V voltage source is a power consumer component. Choice (3) is the answer.

Fig. 2.14 The circuit of solution of problem 2.14

2.15. The circuit does not include any independent source, but it includes dependent sources. Therefore, we need to apply a test source to calculate the ratio of VI tt to determine the Thevenin resistance of the circuit (RTh). To analyze the circuit of this problem, nodal analysis is applied. The circuit includes four nodes of which one is considered as the ground node; therefore, it includes three independent nodes. However, two of them are connected to the voltage sources. Thus, we need to apply KCL in node 2. Defining Vx based on the node voltage (e2): V x ¼ e2

ð1Þ

Ix ¼ It

ð2Þ

From the circuit:

KCL in node 2:
I x þ

Using ð1Þ, ð2Þ e2 e
2V x ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
3I t þ 2e2
2e2 ¼ 0
2I x þ 2 ¼ 0 )
3I x þ 2e2
2V x ¼ 0 ¼ 1 1 )
3I t ¼ 0 ) I t ¼ 0 )

Choice (3) is the answer.

Vt ¼ 1 ) RTh ¼ 1 It

44

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Fig. 2.15 The circuits of solution of problem 2.15

2.16. To solve this problem, application of nodal and mesh analyses will result in a four-equation-four-unknown problem that needs lots of calculations. Herein, a heuristic method can simply solve the problem, as is shown in Fig. 2.16.2. As can be seen, by using KCL, the current of resistors in the horizontal branches is determined, and then KVL is applied in the top loop. KVL in the top loop: 3ð3
I Þ þ 2ð5
I Þ þ ð6
I Þ
1:5 I ¼ 0 )
7:5I þ 25 ¼ 0 ) I ¼ Using Ohm’s law: V o ¼ 1:5I ¼ 1:5  Choice (2) is the answer.

10 ¼5V 3

10 A 3

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

45

Fig. 2.16 The circuit of solution of problem 2.16

2.17. Since the circuit does not include any dependent source, we can apply current division formula to find the voltage nodes “a” and “b” and then calculate Vab. I¼

12  12 12 ¼ ¼1A ¼ 124  9 þ3 9 þ ð4 þ 8Þð3 þ 1Þ 9 þ 12þ4 I 4,8 ¼

ð3 þ 1 Þ 4 ¼ 0:25 A 1¼ 16 ð 3 þ 1Þ þ ð 4 þ 8Þ

I 3,1 ¼

ð4 þ 8 Þ 12 ¼ 0:75 A 1¼ 16 ð 3 þ 1Þ þ ð 4 þ 8Þ

V ab ¼ V a
V b ¼ 8  I 4,8
1  I 3,1 ¼ 8  0:25
1  0:75 ¼ 2
0:75 ¼ 1:25 V Choice (1) is the answer.

46

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Fig. 2.17 The circuit of solution of problem 2.17

2.18. To transfer the maximum power to RL, it must be equal to the Thevenin resistance (RTh) of the circuit. To determine the Thevenin resistance, we must turn off all the power sources. Since the circuit includes at least one dependent source, we must apply a test source at the terminal and determine the value of VI tt to find the Thevenin resistance of the circuit. In the circuit of Fig. 2.18.2, it is clear that: Ix ¼ It

ð1Þ

Applying KVL in the only loop of the circuit: Using ð1Þ
V t þ 5I x þ 10I x ¼ 0 )
V t þ 15I x ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
V t þ 15I t ¼ 0 ) V t ¼ 15I t ) RTh ¼

Vt ¼ 15 It

RL ¼ RTh ¼ 15 Ω Choice (3) is the answer.

Fig. 2.18 The circuits of solution of problem 2.18

2.19. Quickly, we can conclude that VTh ¼ 0, since there is no independent source in the circuit. To find the Thevenin equivalent resistance (RTh), we need to connect a test voltage source (Vt) to the terminal, as is shown in Fig. 2.19.2, and then RTh ¼ VI tt . To analyze the circuit, it is better to apply the nodal analysis method, since we will have fewer unknown variables. Herein, there are two independent nodes, one of which is connected to a grounded voltage source. Thus, we need to write one KCL in the supernode. Note that node “b” is assigned as the ground. KCL in the supernode of the circuit of Fig. 2.19.2:
I t þ I x þ

V x V t
1220I x þ ¼0 50 20

We need to define Ix based on the favorable variables (Vt, It to find RTh ¼ VI tt ); thus by using Ohm’s law:

ð1Þ

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Ix ¼

Vt 1000

47

ð2Þ

And, as is clear: Vx ¼ Vt

ð3Þ

Vt put ð2Þ, ð3Þ in ð1Þ V t
1220 1000 V V ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
I t þ t þ t þ ¼0 1000 50 20  1  1 1 1
1220 1000 )
It þ V t þ þ ¼0 1000 50 20

)
It þ V t





 1 1 V ¼ 0 ) Vt ¼ I t ) t ¼ 100 ) RTh ¼ 100 Ω 100 100 It

Choice (4) is the answer.

Fig. 2.19 The circuits of solution of problem 2.19

2.20. We just need to apply KVL in the indicated meshes and organize the resultant equations based on the mesh currents. However, since there is a current source between meshes 2 and 3, we need to apply KVL for the indicated supermesh. KVL in mesh 1: 2ði1
2Þ
3 þ 4ði1
i2 Þ þ 3i1 ¼ 0 ) 9i1
4i2 ¼ 7 KVL in the supermesh including meshes 2 and 3: 3i2
2V 2 þ 4ði2
i1 Þ þ 3 þ 2i3 þ 5V 1 þ i3 ¼ 0

ð1Þ

48

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

)
4i1 þ 7i2 þ 3i3
2V 2 þ 5V 1 þ 3 ¼ 0

ð2Þ

The relation between the current of current source and the related mesh currents: 2I o ¼ i2
i3

ð3Þ

Defining V1 based on the mesh currents, by using Ohm’s law for the 4 Ω resistor: V 1 ¼ 4ð i 1
i 2 Þ

ð4Þ

Defining V2 based on the mesh currents, by using Ohm’s law for the 2 Ω resistor: V 2 ¼ 2ð 2
i 1 Þ

ð5Þ

I o ¼ i1
i3

ð6Þ


4i1 þ 7i2 þ 3i3
2ð2ð2
i1 ÞÞ þ 5ð4ði1
i2 ÞÞ þ 3 ¼ 0 ) 20i1
13i2 þ 3i3 ¼ 5

ð7Þ

Defining Io based on the mesh currents:

Solving (2), (4), and (5):

Solving (3) and (6): 2ði1
i3 Þ ¼ i2
i3 ) 2i1
i2
i3 ¼ 0 Equations (1), (7), (8) can be written in the form of matrices, as is seen in the following: 2 9
4 ð4Þ, ð5Þ, ð6Þ 6 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 4 20
13 2
1

32

2 3 7 76 7 6 7 3 5 4 i2 5 ¼ 4 5 5
1 0 i3 0

i1

3

Choice (1) is the answer.

Fig. 2.20 The circuits of solution of problem 2.20

ð8Þ

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

49

2.21. Since the circuit includes only resistors, we can use the series-parallel rule to calculate the input resistance of the circuit. 

 5Rþ26
   3R þ 12 5R þ 26     5R þ 26 Rþ7 ¼ 6Rþ33 Rin ¼ 1 2 þ 3ð4 þ RÞ ¼ 1 2 þ ¼ ¼ 1 Rþ7 Rþ7 6R þ 33 Rþ7 This is a rational function and it always has an ascending or descending trend. Therefore, to find its range, we need to calculate its value for the lowest and highest value of its parameter (R). Note: The range of the resistance R is [0, 1). lim R!0 Rin ¼ lim R!0

5R þ 26 26 ¼ ¼ 0:78 Ω 6R þ 33 33

lim R!1 Rin ¼ lim R!1

5R þ 26 5 ¼ ¼ 0:83 Ω 6R þ 33 6

Thus, 0.78 < Rin < 0.83. Choice (4) is the answer.

Fig. 2.21 The circuit of solution of problem 2.21

2.22. To transfer the maximum power to the load, RL needs to be equal to the Thevenin resistance (RTh) of the rest of the circuit. To determine the Thevenin resistance of the circuit, all the power sources need to be turned off. In this problem, since the circuit includes a dependent power source, we cannot use the series-parallel rule to calculate the Thevenin resistance of the circuit. Instead, we must connect a test source (e.g., test voltage source) to the circuit, as can be seen in the circuit of Fig. 2.22.2, and then analyze the circuit to determine the value of VI tt , which is equal to the Thevenin resistance. In this problem, mesh analysis is the best method to solve the problem, since the equations will include the least variables. Herein, the current of the right-side mesh is assigned by using the available variable, that is, It, to minimize the number of variables. KVL in the right-side mesh:
V t þ 4ðI t þ I 1 Þ ¼ 0

ð1Þ

6I 1
2V ab þ 4ðI t þ I 1 Þ ¼ 0 ) 10I 1
2V ab þ 4I t ¼ 0

ð2Þ

KVL in the left-side mesh:

As can be seen in the circuit, the 4 Ω resistor and the test source are in parallel. Thus:

Calculate I1 from (4) and put it in (1):

V ab ¼ V t

ð3Þ

ð2Þ, ð3Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 10I 1
2V t þ 4I t ¼ 0

ð4Þ

50

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

)
V t þ 4ðI t þ ð0:2V t
0:4I t ÞÞ ¼ 0 )
0:2V t þ 2:4I t ¼ 0 ) RTh ¼

Vt ¼ 12 Ω It

) RL ¼ RTh ¼ 12 Ω Choice (3) is the answer.

Fig. 2.22 The circuits of solution of problem 2.22

2.23. Mesh and nodal analyses are the worst methods to solve this problem, since the circuit includes many nodes and meshes. This problem can be easily solved by using the source transformation theorem, as can be seen in Figs. 2.23.1–2.23.7. From Fig. 2.23.1 to Fig. 2.23.2: In the left-side dashed box, the source transformation theorem is applied: I¼

V 3 ¼ ¼ 3 A ðUpwardÞ R 1

In the right-side dashed box, the source transformation theorem is used: V ¼ IR ¼ 5  1 ¼ 5 V From Fig. 2.23.2 to Fig. 2.23.3: In the left-side dashed box, the parallel current sources are combined: I ¼ 3 þ 2 ¼ 5 A ðUpwardÞ In the right-side dashed box, the series resistors are combined: R¼3þ1¼4Ω From Fig. 2.23.3 to Fig. 2.23.4: In the left-side dashed box, the source transformation theorem is applied: V ¼ IR ¼ 5  1 ¼ 5 V In the right-side dashed boxes, the source transformation theorem is applied twice: I1 ¼

V1 1 ¼ ¼ 0:5 A ðDownward Þ R1 2

I2 ¼

V2 5 ¼ ¼ 1:25 A ðUpwardÞ R2 4

From Fig. 2.23.4 to Fig. 2.23.5: In the left-side dashed box, the resistors are combined: R¼1þ3¼4Ω

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

In the right-side dashed boxes, the parallel resistors and the parallel current sources are combined, respectively: R¼3þ1¼4Ω I ¼ 1:25
0:5 ¼ 0:75 A From Fig. 2.23.5 to Fig. 2.23.6: In the dashed box, the source transformation theorem is applied: I¼

V 5 ¼ ¼ 1:25 A ðUpward Þ R 4

From Fig. 2.23.6 to Fig. 2.23.7: The parallel resistors and the parallel current sources are combined, respectively: 1 1 1 1 1þ4þ3 ¼ 2 ) Req ¼ 0:5 Ω ¼ þ þ ¼ Req 4 1 43 4 I ¼ 1:25 þ 0:75 ¼ 2 A In Fig. 2.23.7: V ab ¼ 2  0:5 ¼ 1 V Choice (1) is the answer.

Fig. 2.23 The circuits of solution of problem 2.23

51

52

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Fig. 2.23 (continued)

2.24. Since the circuit includes a dependent source, we must apply a test source to the terminal to calculate the input resistance of the circuit (Rin ¼ VI tt ). If the circuit had independent sources, we would need to turn them off. To analyze the circuit, nodal analysis is the best method. To minimize the computations, it is better to use Vt instead of e3.

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

53

KCL in node 1 of the circuit of Fig. 2.24.2: e1 e1
e2 e 1
V t þ þ ¼ 0 ) 2:5e1
0:5e2
V t ¼ 0 1 2 1

ð1Þ

Defining Vx based on the node voltages: V x ¼ V t
e1

ð2Þ

Using ð2Þ e2 ¼ 2V x ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) e2 ¼ 2ðV t
e1 Þ

ð3Þ

From the circuit:

KCL in node 3:
I t þ

V t V t
e2 V t
e1 þ þ ¼ 0 )
I t
e1
e2 þ 2:5V t ¼ 0 2 1 1

ð4Þ

Solving (3) and (1): 4 2:5e1
0:5ð2ðV t
e1 ÞÞ
V t ¼ 0 ) e1 ¼ V t 7

ð5Þ


I t
e1
2ðV t
e1 Þ þ 2:5V t ¼ 0 )
I t þ e1 þ 0:5V t ¼ 0

ð6Þ

Solving (3) and (4):

Solving (5) and (6): 4 15 V 14 14
I t þ V t þ 0:5V t ¼ 0 )
I t þ V t ¼ 0 ) t ¼ ) Rin ¼ Ω 7 14 15 15 It Choice (4) is the answer.

Fig. 2.24 The circuits of solution of problem 2.24

54

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

2.25. Nodal analysis is the best approach to solve the problem, since the circuit includes only one independent node (supernode) with unknown voltage (e1 ¼ e2). From the circuit: e2 ¼ e1

ð1Þ

KCL in the supernode including nodes 1 and 2: Using ð1Þ 15 þ 12 þ 10 e1
75 e1 e
13
450
312
52 þ
13 þ 2 ¼0¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) e1 þ ¼0 4 6 60 24 5 )

37 814 e ¼ ) e1 ¼ 55 V 60 1 24

ð2Þ

e1
75 e1 þ þI ¼0 4 5

ð3Þ

KCL in node 1:

Solving (2) and (3): 55
75 55 þ þ I ¼ 0 ) I ¼ 5
11 ¼
6 A 4 5 Choice (4) is the answer.

Fig. 2.25 The circuit of solution of problem 2.25

2.26. To calculate the equivalent resistance seen from terminal A-B, a test source must be connected to the terminal, since the circuit includes a dependent source. In addition, the circuit does not include any independent source to be turned off. The value of VI tt will be the equivalent resistance seen from the terminal (RAB). In this problem, mesh analysis will result in the least equations. Herein, the current of left-side mesh has been assigned by using the available variable (It) to minimize the number of variables. Using Ohm’s law for the 2 Ω resistor to define V based on mesh current: V ¼ 2i1 KVL in the left-side mesh:
V t þ 4I t þ 3ðI t
i1 Þ þ kV ¼ 0 )
V t þ 7I t
3i1 þ kV ¼ 0

ð1Þ

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

55

Using ð1Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
V t þ 7I t þ ð2k
3Þi1 ¼ 0

ð2Þ

KVL in the right-side mesh:
kV þ 3ði1
I t Þ þ 2i1 ¼ 0 )
kV þ 5i1
3I t ¼ 0 Using ð1Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) ð5
2kÞi1
3I t ¼ 0 ) i1 ¼

3I t 5
2k

ð3Þ

Solving (2) and (3):
V t þ 7I t þ ð2k
3Þ  )
V t þ I t )

  3ð2k
3Þ 3I t ¼0 ¼ 0 )
V t þ It 7 þ 5
2k 5
2k

35
14k þ 6k
9 5
2k



 ¼ 0 )
V t þ It

26
8k 5
2k

 ¼0

V t 26
8k 8k
26 V 8k
26 26 13 ¼ ¼ ) RAB ¼ t ¼ ¼0)k¼ ¼ 5
2k 2k
5 2k
5 8 4 It It

Choice (4) is the answer.

Fig. 2.26 The circuits of solution of problem 2.26 V2

2.27. First Method: The maximum power transferrable to the load (Pmax) is 4RThTh, 14 V Th I N , or 14 RN I 2N . In these three formulas, the Thevenin resistance and Norton resistance are equal (RTh ¼ RN), and VTh and IN are the Thevenin voltage and Norton current that are equal to the open circuit voltage (Voc) and the short-circuit current (Isc), respectively. As can be noticed from the three power formulas, two out of three quantities (VTh, RTh, IN) are needed to determine the value of maximum transferrable power. In this problem, calculating the value of VTh and IN will result in the least calculations. To determine the value of VTh, we need to determine the value of the open circuit voltage (Voc). Mesh analysis is the best method to analyze the circuit. As is shown in Fig. 2.27.1, the mesh currents have been assigned based on the available variables (Ix and Ix) to minimize the number of variables. KVL in the right-side mesh: 1ðI x
2I x Þ þ 6 þ I x þ 2I x ¼ 0 ) 2I x þ 6 ¼ 0 ) I x ¼
3 A

ð1Þ

V oc ¼ 2I x ¼
6 V

ð2Þ

56

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

To determine the value of IN, we need to calculate the value of the short-circuit current (Isc). As can be seen in Fig. 2.27.2, the two ends of the 2 Ω resistor are now equipotential. Thus, no current will flow from the resistor (Ix ¼ 0). Consequently, the dependent current source is turned off by being an open circuit branch (see Fig. 2.27.3). KVL in the indicated loop of the circuit of Fig. 2.27.3: I sc þ 6 þ I sc ¼ 0 ) I sc ¼
3 A

ð3Þ

Therefore, 1 1 Pmax ¼ V Th I N ¼  ð
6Þ  ð
3Þ ¼ 4:5 W 4 4 Choice (2) is the answer. Second and Third Methods: In these methods, we need to determine the value of the Thevenin resistance or Norton resistance of the circuit (RTh ¼ RN). Additionally, we need to calculate one of the Thevenin voltage and Norton current which is like the first method. Then: Pmax ¼

V 2Th 1 or Pmax ¼ RN I 2N 4 4RTh

To calculate RTh or RN, all the independent sources (independent voltage and current sources) are turned off (independent voltage sources are changed to short-circuit branches and independent current sources are converted to open circuit branches). Then, a test source (with the voltage and current of Vt and It, respectively) is connected to the terminal to find Vt Vt I t (RTh ¼ RN ¼ I t ) if the circuit includes dependent sources; otherwise, they can be calculated by using the regular series-parallel rules. In this problem, there is one independent voltage source and one dependent current source. Therefore, the independent voltage source must be short-circuited, and a test source is needed to be connected to the terminal to find VI tt . Herein, nodal analysis is applied to analyze the circuit because it needs the least variables and calculations. The voltage of the right-side node is defined based on the available variable, that is, Vt. Using Ohm’s law for the 2 Ω resistor to define Ix based on the node voltage: Ix ¼

Vt 2

ð4Þ

KCL in node 1:
2I x þ


 Using ð4Þ e1 e1
V t V ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
2 t þ 2e1
V t ¼ 0 þ ¼ 0 )
2I x þ 2e1
V t ¼ 0 ¼ 1 1 2 )
2V t þ 2e1 ¼ 0 ) e1 ¼ V t

ð5Þ

KCL in the right-side node:
I t þ

V t V t
e1 e þ ¼ 0 )
It þ V t
1 ¼ 0 2 2 2

Solving (5) and (6):
I t þ V t


Vt V V V ¼ 0 )
I t þ t ¼ 0 ) t ¼ I t ) t ¼ 2 ) RTh ¼ RN ¼ 2 Ω 2 2 2 It

ð6Þ

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Pmax ¼

57

V 2Th ð
6Þ2 ¼ ¼ 4:5 W 4RTh 4  2

or, 1 1 Pmax ¼ RN I 2N ¼ ð2Þ  ð
3Þ2 ¼ 4:5 W 4 4 Fourth Method: In this method, the values of VTh and RTh are simultaneously calculated. In this method, a test source (with the voltage of Vt and the current of It) is applied to the terminal, but the independent sources are not shut down. Then, one needs to find a relation between Vt and It in the form of Vt ¼ αIt + β. Based on this method, RTh ¼ α and VTh ¼ β. Mesh analysis is the best approach to analyze this circuit, since it will result in the least variables and calculations. Herein, the currents of the left-side and right-side meshes have been assigned by using the available variables (Ix and It). From the circuit, we see that: I x ¼ i1 þ I t

ð7Þ

KVL in the middle mesh: 1ði1
2I x Þ þ 6 þ i1 þ 2ði1 þ I t Þ ¼ 0 ) 4i1
2I x þ 2I t þ 6 ¼ 0 Using ð7Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 4i1
2ði1 þ I t Þ þ 2I t þ 6 ¼ 0 ) 2i1 þ 6 ¼ 0 ) i1 ¼
3

ð8Þ

KVL in the right-side mesh:
V t þ 2ði1 þ I t Þ ¼ 0

ð9Þ

Solving (8) and (9):
V t þ 2ð
3 þ I t Þ ¼ 0 ) V t ¼ 2I t
6 ) RTh ¼ 2 Ω, V Th ¼
6 V Pmax ¼

V 2Th ð
6Þ2 ¼ ¼ 4:5 W 4RTh 4  2

Fifth Method: In this method, the values of IN and RN are simultaneously calculated. Like the previous method, in this method, a test source (with the voltage and current of Vt and It, respectively) is applied to the terminal without turning off the independent sources. After that, a relation between Vt and It in the form of I t ¼ α1 V t
γ is sought. Based on this method, RN ¼ α and IN ¼ γ. Solving (8) and (9): 1
V t þ 2ð
3 þ I t Þ ¼ 0 )
V t
6 þ 2I t ¼ 0 ) I t ¼ V t
ð
3Þ ) RN ¼ 2 Ω, I N ¼
3 A 2 1 Pmax ¼  2  ð
3Þ2 ¼ 4:5 W 4

58

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Fig. 2.27 The circuits of solution of problem 2.27

2.28. First Method: The Thevenin voltage of a circuit is equal to the open circuit voltage of the terminal (VTh ¼ Voc). Therefore, we need to determine the value of Voc. To analyze the circuit, nodal analysis is suggested, since it will result in the least unknown variables and equations. From the circuit of Fig. 2.28.1, we can see that:
3V
e2 ¼ V ) e2 ¼
4V

ð1Þ

Using ð1Þ e2
ð
3V Þ V e2
e3 þ þ ¼0¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 2ð
4V Þ þ 3:5V
e3 ¼ 0 ) e3 ¼
4:5V 1 2 1

ð2Þ

KCL in node 2:

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

59

KCL in node 3:


Using ð1Þ, ð2Þ V e3
e2 V þ
1¼0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
þ ð
4:5V Þ
ð
4V Þ ¼ 1 ) V ¼
1 2 2 1

ð3Þ

As can be seen in the circuit of Fig. 2.28.1, Voc ¼
3V. Hence: V oc ¼
3V ¼
3  ð
1Þ ¼ 3 V V Th ¼ V oc ¼ 3 V Choice (1) is the answer. Second Method: In this problem, the application of a heuristic (non-conventional) method can minimize the calculations. By applying KCL in the indicated supernode shown in the circuit of Fig. 2.28.2, the current of 1 Ω resistor in the horizontal branch is identified (1 A). Therefore: V ¼
1  1 ¼
1 V As can be seen in the circuit of Fig. 2.28.2, Voc ¼
3V. Hence: V oc ¼
3V ¼
3  ð
1Þ ¼ 3 V V Th ¼ V oc ¼ 3 V Choice (1) is the answer.

Fig. 2.28 The circuits of solution of problem 2.28

60

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

2.29. To find the input resistance of the circuit in this problem, we just need to apply a test voltage source and determine the ratio of VI tt (as RTh) by analyzing the circuit. Nodal analysis is the best method to solve this problem, since the circuit includes only two independent nodes. As can be seen in the circuit of Fig. 2.29.2, the node voltages have been defined based on the available variables (Vt and V2) to decrease the number of variables. KCL in the left-side supernode:
I t þ I 1 ¼ 0 ) I 1 ¼ I t

ð1Þ

KCL in the right-side supernode: αI 1 þ

V2 V2 2V RαI 1 Using ð1Þ RαI t þ ¼ 0 ) αI 1 þ 2 ¼ 0 ) V 2 ¼
¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) V 2 ¼
R R R 2 2

ð2Þ

Defining I1 based on the node voltages: I1 ¼

V t
ð
μV 2 Þ Using ð1Þ V þ μV 2 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) I t ¼ t R R

Solving (2) and (3): It ¼


 t V t þ μ
RαI μRαI t μα 2 ) Vt
¼ RI t ) V t ¼ I t R 1 þ R 2 2

 V μα μα ) t ¼R 1þ ) RTh ¼ R 1 þ 2 2 It

Choice (3) is the answer.

Fig. 2.29 The circuits of solution of problem 2.29

ð3Þ

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

61

2.30. Source transformation technique is the best way to analyze this circuit. First, the two series voltage sources are combined considering their polarities, as can be seen in Fig. 2.30.2. Then, the 3 V voltage source in series with the 1 Ω resistor is converted to the 1 Ω resistor in parallel with the 3 A current source, by using source transformation, as can be seen in Fig. 2.30.3. Next, the two current sources (3 A and 4 A in opposite directions) are combined considering their polarities, as can be seen in Fig. 2.30.4. After that, the source transformation technique is applied, and the circuit is simplified, as is shown in Fig. 2.30.5. Finally, the series resistors (1 Ω and 0.8 Ω) are combined, as is illustrated in Fig. 2.30.6. Now, by using the voltage division formula:

V 2Ω

  23 1:2 1:2   ð
1Þ ¼
¼
¼
0:4 V ¼  1:2 þ 1:8 3 2 þ 1:8 2

P2Ω ¼

V 2 ð
0:4Þ ¼ ¼ 0:08 W R 2

Choice (1) is the answer.

Fig. 2.30 The circuits of solution of problem 2.30

62

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

2.31. First Method: To find the Norton equivalent circuit, we need to determine the value of IN (Isc flowing through a-b terminal) and RN (Req seen from a-b terminal). To find RN, the independent sources are turned off (herein, the independent voltage source is replaced by a shortcircuit branch), as can be seen in Fig. 2.31.2. Moreover, a test source (herein, a test voltage source with the voltage and current of Vt and It) is connected to the terminal, and then the value of VI tt will be the Norton resistance of the circuit. Mesh analysis is the best technique to solve this problem. As can be seen in Fig. 2.31.2, mesh currents have been defined based on the available variables (0.1Vab and It) to simplify the equations. In Fig. 2.31.2, it is clear that: V ab ¼ V t

ð1Þ

KVL in mesh 2: 2ði2
0:1V ab Þ þ i2
0:3V ab þ ði2 þ I t Þ ¼ 0 ) 4i2
0:5V ab þ I t ¼ 0 Using ð1Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 4i2
0:5V t þ I t ¼ 0

ð2Þ


V t þ ði2 þ I t Þ ¼ 0 ) i2 ¼ V t
I t

ð3Þ

KVL in the right-side mesh:

ð2Þ, ð3Þ V 6 6 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 4ðV t
I t Þ
0:5V t þ I t ¼ 0 ) 3:5V t
3I t ¼ 0 ) t ¼ ) RN ¼ Req ¼ Ω 7 7 It Figure 2.31.3 illustrates the circuit to calculate Isc. As can be seen, since the a-b terminal is now short-circuited, Vab ¼ 0. Therefore, the current of the dependent current source and the voltage of dependent voltage source are zero. Hence, the dependent current and voltage sources of the original circuit (Fig. 2.31.1) must be replaced by the open circuit and shortcircuit branches, respectively, as is shown in Fig. 2.31.3. As can be noticed, the left-side branch, including 2 Ω and 1 Ω resistors, is eliminated due to the circuit branch. Thus: I sc ¼

2 ¼ 2 A ) I N ¼ I sc ¼ 2 A 1

Therefore, the equivalent Norton circuit is a 67 Ω resistor in parallel to a 2 A current source (with the upward current direction), as can be seen in Fig. 2.31.5 and Fig. 2.31.6. Choice (1) is the answer. Second Method: In this method, the parameters of the Thevenin equivalent circuit (VTh and RTh) are determined, and then the parameters of Norton equivalent circuit (IN and RN) are calculated by using (1) and (2). V Th RTh

ð1Þ

RN ¼ RTh

ð2Þ

IN ¼

In this method, RN is determined like the previous method. Thus, RN ¼ 67 Ω. To find VTh: KVL in mesh 2 in Fig. 2.31.4: 2ði2
0:1V ab Þ þ i2
0:3V ab þ i2 þ 2 ¼ 0 ) 4i2
0:5V ab þ 2 ¼ 0

ð3Þ

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

63

Applying KVL in the right-side mesh: V ab ¼ i2 þ 2 ) i2 ¼ V ab
2 ð3Þ, ð4Þ 12 V ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 4ðV ab
2Þ
0:5V ab þ 2 ¼ 0 ) 3:5V ab ¼ 6 ) V ab ¼ 7 V Th ¼ V oc ¼ V ab ¼

12 V 7

12 Using ð1Þ V ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) I N ¼ Th ¼ 67 ¼ 2 A RN 7

Fig. 2.31 The circuits of solution of problem 2.31

ð4Þ

64

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Fig. 2.31 (continued)

2.32. First, we should combine the parallel 4 Ω and 1 Ω resistors to simplify the analysis of the circuit, as is shown in Fig. 2.32.2. The parameters of the Thevenin equivalent circuit include the Thevenin resistance (RTh) and Thevenin voltage (VTh). To calculate the Thevenin voltage, we need to determine the value of the open circuit voltage (VTh ¼ Voc) seen from a-b terminal, as can be seen in Fig. 2.32.2. Moreover, to calculate the Thevenin resistance, since the circuit includes a dependent source (dependent current source), we must connect a test source (with the voltage and current of Vt and It) to the terminal and analyze the circuit to determine the value of VI tt, while all the independent sources are shut down (an independent voltage source is replaced by the short-circuit branch and an independent current source is replaced by the open circuit branch), as is shown in Fig. 2.32.3. To calculate the Thevenin voltage by using mesh analysis (see Fig. 2.32.2): KCL in the supernode: e 5
I þ 4
4I
I s ¼ 0 ) e
5I
I s ¼ 0 4 5

ð1Þ

Since the terminal is an open circuit, I¼0

ð2Þ

ð1Þ, ð2Þ 5 4 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) e
I s ¼ 0 ) e ¼ I s 4 5

ð3Þ

KVL in the left-side mesh:
V oc þ 2I þ e ¼ 0 ) V oc ¼

4 4 I ) V Th ¼ I s 5 s 5

To calculate the Thevenin resistance by using mesh analysis (see Fig. 2.32.3):

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

65

KCL in the supernode: e 5
I þ 4
4I ¼ 0 ) e
5I ¼ 0 4 5

ð4Þ

I ¼ It

ð5Þ

ð4Þ, ð5Þ 5 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) e
5I t ¼ 0 ) e ¼ 4I t 4

ð6Þ

As can be seen in the circuit,

KCL in node “a”:
I t þ

Using ð6Þ Vt
e V
4I t V ¼0¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
I t þ t ¼ 0 )
6I t þ V t ¼ 0 ) t ¼ 6 ) RTh ¼ 6 Ω 2 2 It

Choice (2) is the answer.

Fig. 2.32 The circuits of solution of problem 2.32

2.33. To calculate the equivalent resistance of the circuit, we must turn off all the independent sources and connect a test source (with the voltage and current of Vt and It) to the terminal, since the circuit includes a dependent source. Then, we need to determine the value of VI tt which is equal to RTh. In this problem, the application of nodal analysis will need the least calculations. As can be seen in the circuit, it has been tried to define the node voltages based on the available variable (Vt). KCL in the supernode:
I t þ I þ

V t
e1 ¼ 0 )
I t þ I þ V t
e1 ¼ 0 1

ð1Þ

66

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

KCL in node 1:
I þ

e1
V t e1
V t þ ¼ 0 )
I þ 2e1
2V t ¼ 0 1 1

ð2Þ


I þ


 V t
e1 V t 1 þ ¼ 0 )
I
e1 þ 1 þ V t ¼ 0 R 1 R

ð3Þ

KCL in the right-side node:

Solving (1) and (2) to eliminate I:
I t þ ð2e1
2V t Þ þ V t
e1 ¼ 0 )
I t þ e1
V t ¼ 0 ) e1 ¼ I t þ V t

ð4Þ

Solving (2) and (3) to eliminate I:


 1 1 1 V
ð2e1
2V t Þ
e1 þ 1 þ V t ¼ 0 )
3e1 þ 3 þ V t ¼ 0 ) e1 ¼ 1 þ R R 3R t

ð5Þ

Solving (4) and (5) to eliminate e1:
 1 1 V V ) It ¼ V ) t ¼ 3R ) RTh ¼ 3R It þ V t ¼ 1 þ 3R t 3R t It Choice (3) is the answer.

Fig. 2.33 The circuits of solution of problem 2.33

2.34. Since we are interested in determining the effect of the independent current source on the indicated voltage (V), we can turn off other independent sources (the independent voltage source is replaced by a short-circuit branch), based on superposition theorem, as is illustrated in Fig. 2.34.2. Now, we can analyze the circuit to calculate the voltage. In this circuit, nodal analysis is the best approach to solve the problem. Based on the given information: V¼

Is 2

ð1Þ

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

67

KCL in node 1:
I s þ

Using ð1Þ V V
e2 5 e 5I e þ ¼ 0 )
Is þ V
2 ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
I s þ s
2 ¼ 0 2 6 3 3 12 3 )


7I s e2 7
¼ 0 ) e2 ¼
I s 4 12 3

ð2Þ

Defining the voltage of the dependent voltage source based on the node voltages:   Using ð1Þ, ð2Þ 7 k k 7 e2
e3 ¼ kV ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
I s
e3 ¼ I s ) e3 ¼
þ I s 4 2 2 4 KCL in the supernode:     Using ð1Þ, ð2Þ, ð3Þ
74 I s
I2s e2
V e3 e3 k 7 7 1 k 7 þ þ ¼0¼ þ Is ¼ 0 )
þ þ þ Is ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
3 2 2 3 2 4 12 6 2 4 )

7 1 k 7 30 k þ þ þ ¼0) þ ¼0)k¼
5 12 6 2 4 12 2

Choice (3) is the answer.

Fig. 2.34 The circuits of solution of problem 2.34

ð3Þ

68

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

2.35. Mesh analysis and a heuristic technique are the best approaches to calculate the Thevenin resistance and the Thevenin voltage of the circuit, respectively, as can be seen in Figs. 2.35.2 and 2.35.3. To find the Thevenin resistance of the circuit, the independent current source is shut down (open-circuited), and a test voltage source is applied in the terminal to find the value of VI tt . In addition, to find the Thevenin voltage of the circuit, the circuit is analyzed to determine the open circuit voltage of the terminal. To find the Thevenin resistance of the circuit (see Fig. 2.35.2): From the circuit, it is clear that: I2 ¼ It

ð1Þ

Defining the current of the dependent current source based on the mesh currents: Using ð1Þ 3I 2 ¼ i1
I t ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 3I t ¼ i1
I t ) i1 ¼ 4I t

ð2Þ

Using Ohm’s law for the bottom 1 Ω resistor to define V1 based on mesh currents: V 1 ¼ 1  i1 ¼ i1

ð3Þ

KVL in the supermesh including both meshes: Using ð1Þ, ð3Þ
V t þ 5V 1 þ I 2 þ V 1 ¼ 0 )
V t þ 6V 1 þ I 2 ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
V t þ 6i1 þ I t ¼ 0 Using ð2Þ V V ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
V t þ 6ð4I t Þ þ I t ¼ 0 )
V t þ 25I t ¼ 0 ) t ¼ 25 ) RTh ¼ t ¼ 25 Ω It It To find the Thevenin voltage of the circuit (see Fig. 2.35.3): By writing KVL in the left-side mesh:
V oc þ 5V 1 þ I 2  1 þ V 1 ¼ 0 ) V oc ¼ 6V 1 þ I 2

ð1Þ


1 þ I 2 ¼ 0 ) I 2 ¼ 1 A

ð2Þ

KCL in the supernode:

KCL for the node with the node voltage of V1:
I 2
3I 2 þ

Using ð2Þ V1 ¼ 0 ) V 1 ¼ 4I 2 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) V 1 ¼ 4  1 ¼ 4 V 1

Solving (1), (2), (3): V oc ¼ 6ð4Þ þ 1 ¼ 25 V ) V Th ¼ V oc ¼ 25 V Choice (4) is the answer.

ð3Þ

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

69

Fig. 2.35 The circuits of solution of problem 2.35

2.36. Since the circuit does not include any dependent source, the Thevenin resistance of the circuit can be simply calculated by using the series-parallel rule, while all the independent sources are turned off. As is illustrated in Fig. 2.36.2, the independent voltage sources are replaced by the short-circuit branches, and the independent current sources are replaced by the open circuit branches. Moreover, to determine the Thevenin voltage of the circuit, the open circuit voltage of the terminal is calculated by using nodal analysis, as can be seen in Fig. 2.36.3. To calculate the Thevenin resistance (see Fig. 2.36.2):   
 5  8 40 
  
 2  10 5 40   8    3 3 Req ¼ 5 1 þ 210 ¼ 5 1 þ ¼ 5 1 þ ¼ 5 ¼ Ω ¼ 23 ¼ 8 2 þ 10 3 3 23 5þ3 3 RTh ¼ Req ¼

40 Ω 23

To calculate the Thevenin voltage (see Fig. 2.36.3): Finding a relation between the node voltages by defining the voltage of the 30 V independent voltage source based on the node voltages: V oc
e1 ¼ 30 ) e1 ¼ V oc
30 KCL in the supernode:

ð1Þ

70

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems


1 þ

e1
e2 V oc
e2 V oc
50 1 6 3 ¼ 0 ) e1
e2 þ V oc
11 ¼ 0 þ þ 2 10 5 2 10 10

Using ð1Þ 1 6 3 6 8 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) ðV oc
30Þ
e2 þ V oc
11 ¼ 0 )
e2 þ V oc
26 ¼ 0 2 10 10 10 10

ð2Þ

KCL in node 2: e2
e1 e2 e
V oc 1 16 1 þ
7þ 2 ¼ 0 )
e1 þ e2
V oc
7 ¼ 0 2 10 10 2 1 10 Using ð1Þ 1 16 1 16 6 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
ðV oc
30Þ þ e2
V oc
7 ¼ 0 ) e
V þ8¼0 2 10 10 10 2 10 oc

ð3Þ

Solving (2) and (3):    6 10 6 8 23
þ V oc
26 ¼ 0 ) V
8 V
23 ¼ 0 ) V oc ¼ 40 10 16 10 oc 10 40 oc ) V Th ¼ 40 V Choice (2) is the answer.

Fig. 2.36 The circuits of solution of problem 2.36

2.37. To determine the Thevenin voltage of the circuit, the open circuit voltage is calculated by using nodal analysis, as can be seen in Fig. 2.37.2. In addition, to calculate the Thevenin resistance of the circuit, a test voltage source is connected to the terminal to calculate the value of VI tt by using nodal analysis, as can be seen in Fig. 2.37.3.

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

71

To calculate the Thevenin voltage (see Fig. 2.37.2): As can be seen in the circuit, the node with the node voltage of e1 has been grounded. Thus: e1 ¼ 0 V

ð1Þ

Defining the voltage of dependent voltage source (in the horizontal branch) based on the node voltages: Using ð1Þ V 2I ¼ V oc
e1 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 2I ¼ V oc ) I ¼ oc 2

ð2Þ

KCL in the supernode:
6 þ

Using ð1Þ, ð2Þ e1
2I e 1 V þI ¼0)
6þ 1þ I ¼0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
6 þ oc ¼ 0 ) V oc ¼ 24 4 4 2 4 ) V Th ¼ 24 V

To calculate the Thevenin resistance (see Fig. 2.37.3): Defining the voltage of the dependent voltage source (in the horizontal branch) based on the node voltages: Using ð1Þ V 2I ¼ V t
e1 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) I ¼ t 2 KCL in the supernode:
I t þ

Using ð1Þ, ð3Þ e1
2I e 1 V V þ I ¼ 0 )
It þ 1 þ I ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
I t þ t ¼ 0 ) t ¼ 4 4 4 2 4 It ) RTh ¼ 4 Ω

Choice (4) is the answer.

Fig. 2.37 The circuits of solution of problem 2.37

ð3Þ

72

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

2.38. To calculate the power of a component, we need to know its voltage and current considering the conventional polarity for the voltage and current of the component (associate reference direction). Nodal analysis is the best approach to solve the problem, since there are only two independent nodes in the circuit. Defining I0 based on the node voltages (see Fig. 2.38.2): I0 ¼

e1 ) I 0 ¼ e1 1

ð1Þ

KCL in node 1:
2 þ

Using ð1Þ e1 e1
e2 e e
e2 þ þ 2I 0 ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
2 þ 1 þ 1 þ 2e1 ¼ 0 1 2 1 2 7 1
2 þ e1
e2 ¼ 2 2 2

ð2Þ

KCL in node 2: Using ð1Þ e2 e2 e2
e1 e2 e2 e2
e1 5 5 þ þ
2I 0 ¼ 0 ¼ þ þ
2e1 ¼ 0 )
e1 þ e2 ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 2 4 2 4 2 2 4 2 ) e2 ¼ 2e1

ð3Þ

7 1 5 4
2 þ e1
ð2e1 Þ ¼ 0 )
2 þ e1 ¼ 0 ) e1 ¼ 2 2 2 5

ð4Þ

Using ð3Þ, ð4Þ 8 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) e2 ¼ 5

ð5Þ

Using ð1Þ 4 8 I ¼ 2I 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) I ¼ 2e1 ¼ 2  ¼ A 5 5

ð6Þ

Solving (2) and (3):

The current of the independent source is:

The voltage of the independent source is: Using ð4Þ, ð5Þ 4 8 4 V ¼ e1
e2 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) V ¼
¼
V 5 5 5 Using ð6Þ, ð7Þ 4 8 32 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) P ¼ VI ¼
 ¼
W ¼
1:28 W 5 5 25 Choice (1) is the answer.

ð7Þ

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

73

Fig. 2.38 The circuits of solution of problem 2.38

2.39. Determining the relation between Vt and It is equivalent to finding the Thevenin equivalent circuit. Herein, mesh analysis is the best approach to analyze the circuit. Using Ohm’s law for the 2 Ω resistor to define V1 based on mesh currents (see Fig. 2.39.2): V 1 ¼
2i1

ð1Þ

Using ð1Þ
20 þ 2i1 þ 3V 1 ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
20 þ 2i1 þ 3ð
2i1 Þ ¼ 0 )
20
4i1 ¼ 0 ) i1 ¼
5

ð2Þ

KVL in the supermesh:

KVL in the right-side mesh: Using ð1Þ Using ð2Þ
V t þ 5I t þ 3V 1 ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
V t þ 5I t þ 3ð
2i1 Þ ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
V t þ 5I t
6ð
5Þ ¼ 0 ) V t ¼ 5I t þ 30 Choice (2) is the answer.

74

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Fig. 2.39 The circuits of solution of problem 2.39

2.40. Instead of individually determining the Thevenin voltage (VTh) and the Thevenin resistance (RTh) of the circuit, we can calculate them simultaneously. Hence, we need to connect a test source (with the voltage and current of Vt and It, respectively) to the terminal, but we do not turn off the independent sources. Next, we try to find a relation between Vt and It in the form of Vt ¼ αIt + β, where RTh ¼ α and VTh ¼ β. In this problem, nodal analysis is the best technique to apply, since it will result in the least variables and calculations. From the circuit of Fig. 2.40.2, it is clear that: V ¼ Vt

ð1Þ

Using Ohm’s law for the 3 Ω resistor to define I based on node voltages: I¼

e1
V t 3

ð2Þ

KCL in the left-side supernode: 4V þ

Using ð1Þ e1 e
Vt 5 V 5 V
Is þ 1 ¼ 0 ) 4V þ e1
I s
t ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 4V t þ e1
I s
t ¼ 0 6 6 2 3 3 3 )

11 5 V þ e
Is ¼ 0 3 t 6 1

ð3Þ

KCL in the right-side supernode:
I t þ

Vt V
e1 7 e
Iþ t ¼ 0 )
It þ V t
I
1 ¼ 0 12 4 3 3

Using ð2Þ 7 e
V t e1 11 2 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
I t þ V t
1
¼ 0 )
I t þ V t
e1 ¼ 0 12 12 3 3 3

ð4Þ

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

75

Solving (3) and (4):
 11 5 3 11 11 5 55 V t þ 
I t þ V t
I s ¼ 0 ) V
I þ V
Is ¼ 0 3 6 2 12 3 t 4 t 48 t )

77 5 V
I
Is ¼ 0 16 t 4 t

) Vt ¼

20 16 I þ I 77 t 77 s

20 Therefore, V Th ¼ 16 77I s and RTh ¼ 77.

Choice (4) is the answer.

Fig. 2.40 The circuits of solution of problem 2.40

2.41. To determine the equivalent resistance of the circuit, we must apply a test source at the terminal and determine the value of VI tt , since the circuit includes dependent sources. Defining Vx based on the node voltages (see Fig. 2.41.2): V x ¼ e1

ð1Þ

Defining I based on the node voltages: I¼

V t
2V x Using ð1Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) I ¼ V t
2e1 1

ð2Þ

76

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

KCL in the supernode: Using ð2Þ
I t þ I þ I ¼ 0 )
I t þ 2I ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
I t þ 2ðV t
2e1 Þ ¼ 0 )
I t þ 2V t
4e1 ¼ 0

ð3Þ

KCL in node 1:
I
I þ

Using ð2Þ e1 ¼ 0 )
2I þ e1 ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
2ðV t
2e1 Þ þ e1 ¼ 0 )
2V t þ 5e1 ¼ 0 1 2 ) e1 ¼ V t 5

ð4Þ

Solving (3) and (4):
I t þ 2V t
4


 2 2 V 5 5 V t ¼ 0 )
I t þ V t ¼ 0 ) t ¼ ) RTh ¼ 5 5 2 2 It

Choice (4) is the answer.

Fig. 2.41 The circuits of solution of problem 2.41

2.42. To determine the power of a component, we need to calculate its voltage and current. In this regard, we need to consider the polarities of the voltage and the current of the component. In this problem, nodal analysis is the best approach to apply. Using Ohm’s law to define I based on the node voltages (see Fig. 2.42.2): I¼

V2 2

ð1Þ

Defining the voltage of the dependent voltage source based on the node voltages: Using ð1Þ
V 2  1 2I ¼
V 2
V 1 ¼ ¼
V 2
V 1 ) V 2 ¼
V 1 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 2 2 2

ð2Þ

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

77

KCL in the supernode:
4 þ

V1
V 2
V 2
2V 2
3V 1 þ þ ¼ 0 )
4
2V 1
3V 2 ¼ 0 1 2 2


 Using ð2Þ 1 1 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
4
2V 1
3
V 1 ¼ 0 )
4
V 1 ¼ 0 ) V 1 ¼
8 2 2 P ¼ VI ¼ V 1  ð
4Þ ¼ ð
8Þ  ð
4Þ ¼ 32 W As can be seen, the independent current source is consuming power, since its power is positive, considering conventional polarity for the voltage and current (associate reference direction). Choice (2) is the answer.

Fig. 2.42 The circuits of solution of problem 2.42

2.43. As we know, the Norton resistance is equal to the equivalent resistance seen by the terminal. As can be seen in Fig. 2.43.2, to calculate the equivalent resistance, since the circuit includes a dependent source, we must apply a test source to determine the value of VI tt . Defining I based on the node voltages (see Fig. 2.43.2): I¼


e1 10

ð1Þ

78

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

KCL in the left-side supernode:
 Using ð1Þ 7 e1 e
Vt 11 V e V e1 þ 2I
t ¼ 0 ¼ e1 þ 2
1
t ¼ 0 þ 2I þ 1 ¼0) ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 20 20 10 4 4 10 4 )

3 V 5 e
t ¼ 0 ) e1 ¼ V t 20 1 4 3

ð2Þ

KCL in the right-side supernode:
I t þ

  Using ð2Þ V t V t
e1 9 1 9 1 5 Vt ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
I t þ V t
þ ¼ 0 )
I t þ V t
e1 ¼ 0 ¼ 20 4 20 4 3 5 4 )
It þ

1 V V ¼ 0 ) Req ¼ t ¼ 30 ) RN ¼ Req ¼ 30 Ω 30 t It

Choice (1) is the answer.

Fig. 2.43 The circuits of solution of problem 2.43

2.44. To calculate the power of a component, we need to multiply its voltage and current considering the voltage and current polarities. Mesh analysis is the best method to solve this problem. KVL in mesh 1 (see Fig. 2.44.2):
2 þ I x  1 þ V x
I x ¼ 0 )
2 þ V x ¼ 0 ) V x ¼ 2

ð1Þ

Using Ohm’s law for the 1 Ω resistor in the vertical branch, to define Vx based on the mesh current: V x ¼ 1ð i 1
8Þ ) V x ¼ i 1
8

ð2Þ

) i1
8 ¼ 2 ) i1 ¼ 10

ð3Þ

Solving (1) and (2):

Using ð3Þ P ¼ VI ¼
ð2Þði1 Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) P ¼
ð2Þð10Þ ¼
20 W The minus sign was applied in the power formula because the current is leaving the positive terminal of the voltage of the component. Choice (4) is the answer.

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

79

Fig. 2.44 The circuits of solution of problem 2.44

2.45. We need to calculate the voltage and current of the resistor, and then multiply them to determine its lost power considering its voltage and current polarities. Nodal analysis is the best technique to solve this problem. Defining Ix based on the node voltages (see Fig. 2.45.2): Ix ¼

e1
e2 2

ð1Þ

KCL in node 1:
3 þ I x þ

Using ð1Þ 3 1 1 e1
e3 e
e2 e1
e3 ¼ 0 ) e1
e2
e3 ¼ 3 ¼0¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
3 þ 1 þ 4 2 4 4 4 2 3e1
2e2
e3 ¼ 12

ð2Þ

e2
e1 e2 e2
e3 1 7 1 þ þ ¼ 0 )
e1 þ e2
e3 ¼ 0 )
4e1 þ 7e2
e3 ¼ 0 2 8 8 2 4 8

ð3Þ

KCL in node 2:

80

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

KCL in node 3: Using ð1Þ e3
e2 e3
e1 e3
e2 e3
e1 e
e2 þ þ 2I x ¼ 0 ¼ þ þ2 1 ¼0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 8 4 8 4 2 )

3 9 3 e
e þ e ¼ 0 ) 6e1
9e2 þ 3e3 ¼ 0 4 1 8 2 8 3

ð4Þ

Since we are interested in knowing the value of e2, we should eliminate e1 and e3 by solving equations (2), (3), and (4). Solving (2) and (3) to eliminate e1: 1 21 3 13 7 3  ð7e2
e3 Þ
2e2
e3 ¼ 12 ) e
e
2e2
e3 ¼ 12 ) e
e ¼ 12 4 4 2 4 3 4 2 4 3

ð5Þ

Solving (3) and (4) to eliminate e1: 1 3 3 6 ð7e2
e3 Þ
9e2 þ 3e3 ¼ 0 ) e2 þ e3 ¼ 0 ) e3 ¼
e2 4 2 2

ð6Þ

Solving (5) and (6) to eliminate e3: 13 7 12 e
ð
e2 Þ ¼ 12 ) 5e2 ¼ 12 ) e2 ¼ 4 2 4 5

ð7Þ

Therefore, the voltage of 4 Ω resistor in the vertical branch is V ¼ 12 5
0 ¼ 2:4 V. P¼

V 2 2:42 ¼ ¼ 1:44 W R 4

Choice (1) is the answer.

Fig. 2.45 The circuits of solution of problem 2.45

2.46. Since the circuit includes dependent sources, we need to connect a test source to terminal A-B to determine the value of VI tt to calculate the input resistance of the circuit. In addition, the circuit does not include any independent source to shut down. To analyze the circuit, mesh analysis is the best one because it needs the least calculations. As can be seen in Fig. 2.46.2, the available variables (I0and It) have been used to choose the mesh currents. To avoid being involved with the unknown voltages of independent current sources, we need to apply KVL for the supermesh. Defining the current of the right-side dependent current source based on the mesh currents: 11I 0 ¼
i2
I t ) i2 ¼
I t
11I 0

ð1Þ

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

81

Defining the current of the left-side dependent current source based on the mesh currents: Using ð1Þ 13I 0 ¼ i1
i2 ) i1 ¼ 13I 0 þ i2 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) i1 ¼ 13I 0 þ ð
I t
11I 0 Þ ) i1 ¼ 2I 0
I t

ð2Þ

KVL in the supermesh: Ri1 þ 3ði2 þ I 0 Þ þ 7ð
I t þ I 0 Þ þ V t ¼ 0 ) Ri1 þ 3i2 þ 10I 0
7I t þ V t ¼ 0 Using ð1Þ&ð2Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) Rð2I 0
I t Þ þ 3ð
I t
11I 0 Þ þ 10I 0
7I t þ V t ¼ 0 ) I 0 ð2R
23Þ þ I t ð
R
10Þ þ V t ¼ 0

ð3Þ

KVL in the top mesh: 3ði2 þ I 0 Þ þ 7ð
I t þ I 0 Þ þ 3I 0 ¼ 0 ) 3i2 þ 13I 0
7I t ¼ 0 Using ð1Þ I ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 3ð
I t
11I 0 Þ þ 13I 0
7I t ¼ 0 )
10I t
20I 0 ¼ 0 ) I 0 ¼
t 2 Solving (3) and (4) result in:





 It V ð2R
23Þ þ I t ð
R
10Þ þ V t ¼ 0 ) I t ð
2R þ 1:5Þ þ V t ¼ 0 ) t ¼ 2R
1:5 2 It ) RAB ¼ 2R
1:5 Ω

Choice (2) is the answer.

Fig. 2.46 The circuits of solution of problem 2.46

ð4Þ

82

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

2.47. Mesh analysis is the best method to solve this problem because we will need to solve the least equations. As can be seen in the circuit of Fig. 2.47, there is one current source between meshes 1 and 2 as well as between meshes 1 and 3. Therefore, we must apply a supermesh for meshes 1, 2, and 3. Using Ohm’s law for the 5 Ω resistor to define Vy based on the mesh current: V y ¼
5i2

ð1Þ

Using Ohm’s law for the 10 Ω resistor to define Vx based on the mesh currents: V x ¼ 10ði2
i3 Þ

ð2Þ


110 þ 20i1 þ 5i2 þ 15i3 ¼ 0

ð3Þ

KVL in the supermesh:

Defining the current of the lower dependent current source based on the mesh currents: Using ð1Þ 1:6V y ¼ i1
i3 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 1:6ð
5i2 Þ ¼ i1
i3 ) i1 þ 8i2
i3 ¼ 0

ð4Þ

Defining the current of the upper dependent current source based on the mesh currents: Using ð2Þ 0:5V x ¼ i1
i2 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 0:5ð10ði2
i3 ÞÞ ¼ i1
i2 ) i1
6i2 þ 5i3 ¼ 0

ð5Þ

Solving (4) and (5): 3
ð8i2
i3 Þ
6i2 þ 5i3 ¼ 0 )
14i2 þ 6i3 ¼ 0 ) i2 ¼ i3 7

ð6Þ

Solving (3), (4), and (6):


 
 3 3
110 þ 20
8 i3
i3 þ 5 i3 þ 15i3 ¼ 0 7 7 )
110


340 15 220 7 i þ i3 þ 15i3 ¼ 0 )
110
i ¼ 0 ) i3 ¼
7 3 7 7 3 2

ð7Þ

Solving (6) and (7): i2 ¼


 3 7 3
¼
7 2 2

Solving (1) and (8):
 3 V y ¼
5
¼ 7:5 V 2 Solving (2), (7), and (8):

 3 7 V x ¼ 10


¼ 20 V 2 2 Choice (2) is the answer.

ð8Þ

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

83

Fig. 2.47 The circuit of solution of problem 2.47

2.48. To determine the Thevenin equivalent circuit, we need to calculate the Thevenin resistance and Thevenin voltage of the circuit. To find the Thevenin resistance of the circuit, the only independent source (independent current source) of the circuit is shut down, and a test source (test voltage source) is connected to the input terminal, as can be seen in Fig. 2.48.2. Mesh analysis is applied to calculate the Thevenin resistance of the circuit. To deal with the fewest variables, the currents of the second and third meshes are defined based on the available variables (It and 2Vo). As can be seen in the circuit: Vo ¼ Vt

ð1Þ

Using Ohm’s law for the 2 Ω resistor to define Vx based on the mesh current: V x ¼
2i1

ð2Þ

KVL in mesh 1: 2:5V x þ 2i1 þ 3ði1
2V o Þ þ 4ði1 þ I t Þ ¼ 0 ) 2:5V x þ 9i1
6V o þ 4I t ¼ 0 Using ð1Þ, ð2Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 2:5ð
2i1 Þ þ 9i1
6V t þ 4I t ¼ 0 ) 4i1
6V t þ 4I t ¼ 0

ð3Þ

KVL in the right-side mesh:
V t þ 4ði1 þ I t Þ ¼ 0

ð4Þ

Solving (3) and (4):
V t þ 4ð1:5V t
I t þ I t Þ ¼ 0 )
V t þ 6V t ¼ 0 ) 5V t ¼ 0 ) V t ¼ 0 )

Vt ¼ 0 ) RTh ¼ 0 It

To determine the Thevenin voltage of the circuit, we need to calculate the open circuit voltage of the terminal, as can be seen in Fig. 2.48.3. Again, mesh analysis is the best approach to solve the problem. To work with the smallest number of variables, the currents of the second and third meshes are defined based on the available variables and parameters (Is and 2Vo). Using Ohm’s law for the 4 Ω resistor to define Vo based on the mesh current: V oc ¼ V o ¼ 4i1

ð1Þ

Using Ohm’s law for the 2 Ω resistor to define Vx based on the mesh currents: V x ¼ 2ðI s
i1 Þ

ð2Þ

84

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

KVL in mesh 1: 2:5V x þ 2ði1
I s Þ þ 3ði1
2V o Þ þ 4i1 ¼ 0 ) 2:5V x þ 9i1
2I s
6V o ¼ 0 Using ð1Þ, ð2Þ 3 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 2:5ð2ðI s
i1 ÞÞ þ 9i1
2I s
6ð4i1 Þ ¼ 0 ) 3I s
20i1 ¼ 0 ) i1 ¼ I s 20 Solving (1) and (3): V oc ¼ V o ¼ 4i1 ¼ 4




 3 3 Is ¼ Is 20 5

3 V Th ¼ V oc ¼ I s 5 Therefore, the Thevenin equivalent circuit is a single voltage source with the size of 35 I s . Choice (3) is the answer.

Fig. 2.48 The circuits of solution of problem 2.48

ð3Þ

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

85

2.49. To find the Thevenin equivalent circuit, we need to calculate the Thevenin resistance (RTh) and the Thevenin voltage (VTh). Since the circuit includes a dependent source (dependent current source), we need to connect a test source (with the voltage and current of Vt and It) to the A-B terminal to find the Thevenin resistance (RTh ¼ VI tt ), while all the independent sources are turned off. In addition, VTh will be the same as the open circuit voltage (Voc). As can be seen in Fig. 2.49.2, nodal analysis is applied to find RTh. Defining Ix based on the node voltages in the circuit of Fig. 2.49.2: Ix ¼

e1 8

ð1Þ

KCL in node 1: Using ð1Þ e1 e
e2 e e
e2 e e
2I x þ 1 ¼0¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
1 þ 1 ¼ 0 )
1
2 ¼ 0 ) e1 ¼
4e2 8 10 8 10 40 10

ð2Þ

KCL in node 2: Using ð1Þ e2
e1 e2 e2
V t e1 e2
e1 e2 e2
V t ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) þ þ þ 2I x ¼ 0 ¼ þ þ þ ¼0 10 3 4 10 3 4 4 )

Using ð2Þ 41 41 3 V 3 V 5 V e2 þ e1
t ¼ 0 ¼ e þ ð
4e2 Þ
t ¼ 0 ) e
t¼0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 60 20 60 2 20 60 2 4 4 4 ) e2 ¼ 3V t

ð3Þ

Defining It based on the node voltages: It ¼

V t
e2 Using ð3Þ V
3V t 1 V ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) I t ¼ t ¼
V t ) t ¼
2 2 4 4 It RTh ¼

Vt ¼
2Ω It

Nodal analysis is applied to find VTh, as can be seen in Fig. 2.49.3. Defining Ix based on the node voltages: Ix ¼

e1 8

ð4Þ

KCL in node 1:
I s þ

Using ð1Þ e1 e
e2 e e
e2 e e
2I x þ 1 ¼0¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)
I s
1 þ 1 ¼ 0 )
Is
1
2 ¼ 0 8 10 8 10 40 10

ð5Þ

KCL in node 2: Using ð1Þ e2
e1 e2 e1 e2
e1 e2 13e2 3e1 26 þ þ 2I x ¼ 0 ¼ þ þ ¼0) þ ¼ 0 ) e1 ¼
e2 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 9 10 3 10 3 4 30 20 Solving (5) and (6):
I s



26 e 1 9 e2
2 ¼ 0 )
I s
e2 ¼ 0 ) e2 ¼
36I s 36 40 10

)Voc ¼ e2 ¼
36Is, because there is no voltage drop on the 4 Ω resistor, since the terminal is open circuit. V Th ¼
36I s Choice (1) is the answer.

ð6Þ

86

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Fig. 2.49 The circuits of solution of problem 2.49

2.50. The circuit includes a dependent source. Hence, we must apply a test source (with the voltage and current of Vt and It) at the terminal, while all the independent sources are shut down. After that, we need to determine the quantity of VI tt (Req ¼ VI tt ). Herein, nodal analysis is applied to solve the problem that the node voltages are defined based on the available variable (Vt). KCL in the left-side node:
I t þ I x þ

V t
e1 V e ¼ 0 )
It þ Ix þ t
1 ¼ 0 R R R

ð1Þ

KCL in node 1: e1
V t e1
V t 2 2 þ ¼ 0 ) βI x þ e1
V t ¼ 0 R R R R

ð2Þ

  V t
e1 V t 1 1 1 þ ¼ 0 )
I x
e1 þ þ V ¼0 R R RL t R RL

ð3Þ

βI x þ KCL in the right-side node:
I x þ

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Solving (1) and (2) to eliminate Ix:       2 2 V e 2 1 2 1 V t þ t
1 ¼ 0 )
It þ þ Vt þ

e ¼0
I t þ
e1 þ Rβ Rβ Rβ R Rβ R 1 R R
 2 I t
Rβ þ R1 V t RβI
ð2 þ βÞV Rβ t t
 ¼ I þ Vt ) e1 ¼ ) e1 ¼
2 þβ t
ð 2 þ β Þ 2 1
Rβ þ R Solving (2) and (3) to eliminate Ix:         1 1 1 2 2 2
β β
2 β ) β
e1 þ þ e1 þ þ V þ e1
V t ¼ 0 ) V ¼0 R R RL t R R R R RL t Solving (4) and (5) to eliminate e1:      2
β Rβ β
2 β
It þ V t þ þ V ¼0 R 2þβ R RL t     2
β Rβ 2
β β
2 β )
 I þ þ þ V ¼0 R 2þβ t R R RL t V R  2þβ ) t ¼ 2
β β
2 It þ þ 2
β

R

Rβ

R

) Req ¼

β RL

¼

βð2
βÞ 2þβ β RL

¼

2
β R 2þβ L

Vt 2
β R ¼ 2þβ L It

Choice (3) is the answer.

Fig. 2.50 The circuits of solution of problem 2.50

87

ð4Þ

ð5Þ

88

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

2.51. To calculate the maximum power absorbable by the variable resistor, we need to determine the Thevenin voltage and Thevenin resistance seen by that resistor. As can be seen in Fig. 2.51.2, to determine the Thevenin voltage of the circuit, the open circuit voltage is calculated by using nodal analysis. Moreover, as can be seen in Fig. 2.51.3, to calculate the Thevenin resistance of the circuit, a test voltage source is applied to calculate the value of VI tt by using mesh analysis. To determine the Thevenin voltage of the circuit (see Fig. 2.51.2): Using Ohm’s law for the 3 Ω resistor to define I1 based on the node voltages: I1 ¼

e2 3

ð1Þ

Defining the voltage of the dependent voltage source based on the node voltages: Using ð1Þ 5 5I 1 ¼ e1
V oc ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) e1 ¼ e2 þ V oc 3

ð2Þ

KCL in the supernode: e1
e2 V oc
e2 þ ¼ 0 ) e1
3e2 þ 2V oc
6 ¼ 0 2 1

ð3Þ

e2
e1 e2
V oc e2 þ þ ¼ 0 )
3e1 þ 11e2
6V oc ¼ 0 2 1 3

ð4Þ


3 þ KCL in node 1:

Solving (2) and (3) to eliminate e1: 5 4 e þ V oc
3e2 þ 2V oc
6 ¼ 0 )
e2 þ 3V oc
6 ¼ 0 3 2 3

ð5Þ

Solving (3) and (4) to eliminate e1:
3ð3e2
2V oc þ 6Þ þ 11e2
6V oc ¼ 0 ) 2e2
18 ¼ 0 ) e2 ¼ 9

ð6Þ

Solving (5) and (6) to eliminate e2: 4
ð9Þ þ 3V oc
6 ¼ 0 )
18 þ 3V oc ¼ 0 ) V oc ¼ 6 V ) V Th ¼ 6 V 3 To determine the Thevenin resistance of the circuit (see Fig. 2.51.3): From the circuit, it is clear that: I1 ¼ It

ð7Þ

Using ð7Þ 2i2 þ 5I 1 þ 1ði2 þ I t Þ ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 3i2 þ 6I t ¼ 0 ) i2 ¼
2I t

ð8Þ

KVL in the top mesh:

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

89

KVL in the bottom mesh: Using ð7Þ Using ð8Þ 3I 1
V t þ 1ðI t þ i2 Þ ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 4I t
V t þ i2 ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 4I t
V t þ ð
2I t Þ ¼ 0 Using ð8Þ V ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 2I t
V t ¼ 0 ) t ¼ 2 ) RTh ¼ 2 Ω It 2

Based on maximum power transfer theorem, the maximum transferable power can be calculated by using Pmax ¼ V4RThTh . Thus: Pmax ¼

V Th 2 62 36 ¼ 4:5 W ¼ ¼ 8 4RTh 4  2

Choice (1) is the answer.

Fig. 2.51 The circuits of solution of problem 2.51

90

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

2.52. Solving the problem will be easy if we consider the symmetry of the circuit that exists between the nodes of “A” and “B.” In this circuit, since calculating the equivalent resistance between the nodes of “A” and “B” is of interest, the nodes of “C1,” “C2,” “C3,” and “C4” will have equal electrical potential, as can be seen in Fig. 2.52.1. Therefore, these nodes can be short-circuited. Hence, the resistors, indicated by “f,” are removed. Similarly, the nodes of “D1,” “D2,” “D3,” and “D4” have equal electrical potential, and hence the resistors, indicated by “h,” are eliminated. On the other hand, the short-circuited nodes cause the resistors (indicated by “e”) to be in parallel, as can be seen in Fig. 2.52.2. Likewise, the resistors, indicated by “g,” as well as the resistors, indicated by “i,” will be in parallel (see Fig. 2.52.2). Therefore, the equivalent resistance seen from terminal A-B is: 1 1 1 3 RAB ¼ þ þ ¼ Ω 4 4 4 4 Choice (4) is the answer.

Fig. 2.52 The circuits of solution of problem 2.52

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

91

2.53. To calculate the Thevenin voltage of the circuit, we need to calculate the open circuit voltage of terminal a-b, as is illustrated in Fig. 2.53.2. KVL in the loop of the circuit of Fig. 2.53.2: 1
5V x
V x þ E ¼ 0 ) V x ¼ E 6

ð1Þ

V oc ¼ 5V x

ð2Þ

Also, as can be seen in the circuit:

Solving (1) and (2): 5 5 V oc ¼ E ) V Th ¼ E 6 6 In addition, to find the Thevenin resistance of the circuit, we must apply a test source to determine the value of RTh ¼ VI tt, as can be seen in Fig. 2.53.3. Moreover, we must turn off all the independent sources. Thus, the independent voltage source is changed to a short-circuit branch, and consequently the parallel resistor is eliminated, as can be seen in Fig. 2.53.4. KVL in the loop of the circuit of Fig. 2.53.4:
5V x
V x ¼ 0 ) V x ¼ 0

ð3Þ

Also, voltage sources are in parallel, thus: V t ¼ 5V x

ð4Þ

Solving (3) and (4): V t ¼ 0 ) RTh ¼

Vt ¼0 It

Choice (3) is the answer.

Fig. 2.53 The circuits of solution of problem 2.53

92

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

2.54. As is shown in Fig. 2.54.2, the part of the circuit seen by the 5 Ω resistor is replaced by its Thevenin equivalent circuit. The power of each resistor can be calculated by using the voltage division and power formulas and presented in (1) and (2).  2 V 2R 1 R PR ¼ V ¼ R R þ RTh Th R PRTh

ð1Þ

 2 V 2RTh 1 RTh ¼ ¼ V RTh RTh R þ RTh Th

ð2Þ

The percentage of powers absorbed by the resistors is presented in (3) and simplified in (4) by using (1) and (2). Power percentage ¼

PR  100 PRTh

Using ð1Þ, ð2Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) Power percentage ¼

1 R




1 RTh

) Power percentage ¼

R RþRTh




ð3Þ 2 V Th

RTh RþRTh

2  100

V Th

RTh ðRÞ2 R  100 ¼  100 RTh RðRTh Þ2

ð4Þ

As can be seen in (4), the power percentage does not depend on the size of the power source (herein, the voltage source), and it only depends on the ratio of resistors. Therefore, by changing the size of the voltage source, the power absorbed by the 5 Ω resistor will not be affected. Choice (4) is the answer.

Fig. 2.54 The circuits of solution of problem 2.54

2.55. As is shown in the circuit of Fig. 2.55.2, we need to turn off all the independent sources and connect a test source with the voltage and current of Vt and It, respectively. Next, we need to analyze the circuit to calculate the value of VI tt, which is the same as the equivalent resistance of the circuit (Req). To analyze this circuit, mesh analysis is the best approach. To minimize the number of variables, the current of the test voltage source (It) and the current of the independent current source (3V ) are considered as the currents of mesh 1 and mesh 3, respectively. Defining V1 based on the mesh current: V 1 ¼ 4  ð
3V Þ ¼
12V

ð1Þ

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

93

KVL in mesh 1: 7 7
V t þ 1ðI t
3V Þ þ 1  I t
V 1 ¼ 0 )
V t þ 2I t
3V
V 1 ¼ 0 6 6

ð2Þ

Solving (1) and (2): 7
V t þ 2I t
3V
ð
12V Þ ¼ 0 )
V t þ 2I t þ 11V ¼ 0 6

ð3Þ

KVL in mesh 2:
1  I t þ 4ð
3V Þ þ V ¼ 0 )
I t
11V ¼ 0 ) V ¼


1 I 11 t

Solving (3) and (4):
 1 V )
V t þ 2I t þ 11
I t ¼ 0 )
V t þ I t ¼ 0 ) t ¼ 1 11 It ) Req ¼

Vt ¼1 It

Choice (2) is the answer.

Fig. 2.55 The circuits of solution of problem 2.55

ð4Þ

94

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

2.56. Note: In this circuit, it is convenient to consider the units of kΩ, mA, and V for the resistances, the currents, and the voltages, respectively. First Method (Nodal Analysis): As can be seen in Fig. 2.56.2, there are four nodes in the circuit; however, one of them is dependent (ground zone) and one of them has a known voltage (1 V). The voltages of other nodes are indicated by e1 and e2, as can be seen in the circuit. To solve the problem by using nodal analysis, we need to apply KCL in each independent node. Note: Since there is a voltage source between two nodes, we must apply KCL for the supernode including these two nodes. KCL in the supernode of the circuit of Fig. 2.56.2: e1
1 e
3I x þ 2
2 ¼ 0 1 0:5

ð1Þ

Also, we need to identify the relation between the voltage of the voltage source and these node voltages (e1 and e2). e1
e2 ¼ 3V x

ð2Þ

As can be seen in the equations, other variables have appeared in them; thus, we need to define them based on the node voltages. Using Ohm’s laws for the 1kΩ resistor, to define Ix based on the node voltages: Ix ¼

1
e1 1

ð3Þ

Also, it is clear that: V x ¼ e2

ð4Þ

e1
1 1
e1 e2
3 þ
2 ¼ 0 ) e1
1
3 þ 3e1 þ 2e2
2 ¼ 0 ) 4e1 þ 2e2 ¼ 6 1 1 0:5

ð5Þ

Put (3) in (1) and simplify it:

Put (4) in (2) and simplify it: e1
e2 ¼ 3e2 ) e1 ¼ 4e2 Solving (5) and (6): 4ð4e2 Þ þ 2e2 ¼ 6 ) e2 ¼

1 V 3

As can be noticed from the circuit: 1

I¼ Choice (2) is the answer.

V e2 2 )I¼ 3 ¼ mA 0:5 0:5 kΩ 3

ð6Þ

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

95

Second Method (Mesh Analysis): As can be seen in Fig. 2.56.3, although the circuit includes three meshes, the current of the right-side mesh is known (2 mA). i3 ¼ 2 mA

ð7Þ

To solve the problem by using mesh analysis, we need to apply KVL in each independent mesh. However, since there is a current source between the two meshes, we must apply KVL for the supermesh including these two meshes. KVL for the supermesh of the circuit of Fig. 2.56.3:
1 þ I x  1 þ 3V x þ V x ¼ 0 )
1 þ I x þ 4V x ¼ 0

ð8Þ

Moreover, we need to determine the relation between the current of the current source and these mesh currents (i1 and i2). i2
i1 ¼ 3I x

ð9Þ

As can be seen in equations (8) and (9), they are not purely based on the independent variables (i1 and i2). Therefore, we need to define Vx and Ix based on them. Using Ohm’s laws for the 500 Ω resistor to define Vx based on the mesh currents: Using ð7Þ V x ¼ 0:5ði2 þ i3 Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) V x ¼ 0:5ði2 þ 2Þ

ð10Þ

I x ¼ i1

ð11Þ


1 þ i1 þ 4ð0:5ði2 þ 2ÞÞ ¼ 0 ) i1 þ 2i2 þ 3 ¼ 0

ð12Þ

Additionally, it is clear that:

Put (10) and (11) in (8) and simplify it:

Put (11) in (9) and simplify it: i2
i1 ¼ 3i1 ) i1 ¼

i2 4

Solving (13) and (12): i2 9 4 þ 2i2 þ 3 ¼ 0 ) i2 ¼
3 ) i2 ¼
mA 4 3 4 As can be seen in the circuit: 4 2 I ¼ i2 þ i3 ¼
þ 2 ¼ mA 3 3 Choice (2) is the answer.

ð13Þ

96

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

Fig. 2.56 The circuits of solution of problem 2.56

2.57. As can be noticed from the circuit, there are some symmetries and specific characteristics in that. The blue-color boundary in Fig. 2.57.1 shows a parallel circuit with infinite number of components. Therefore, its equivalent resistance will be zero, as can be seen in the following:     R R     RBlue ¼ RR . . . RR ¼ ¼ ¼0 n 1 |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl}

ð1Þ

n times

Thus, this part of the circuit can be replaced by a short-circuit branch, as can be seen in Fig. 2.57.2. Figure 2.57.3 is the simplified version of Fig. 2.57.2. This type of circuit is called infinite ladder circuit. As can be seen in Fig. 2.57.4, the equivalent resistance of the red-color boundary is called RRed. Therefore, Rin is just a parallel connection of R and RRed.  RRRed  ð2Þ Rin ¼ RRRed ¼ R þ RRed To calculate the equivalent resistance of RRed, Fig. 2.57.5 is drawn. As can be seen, the relation below is held between RRed and RGreen.

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

 R  ðR þ RGreen Þ R2 þ RRGreen  RRed ¼ RðR þ RGreen Þ ¼ ¼ 2R þ RGreen R þ ðR þ RGreen Þ

97

ð3Þ

However, since the ladder circuit has an infinite size, RGreen will be approximately equal to RRed because removing one series R and one parallel R will not significantly affect RRed. RGreen ffi RRed

ð4Þ

Solving (3) and (4): R2 þ RRRed ) 2RRRed þ R2Red ffi R2 þ RRRed 2R þ RRed pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 
R  R2 þ 4R2
1  5 2 2 ) RRed þ RRRed
R ffi 0 ) RRed ffi ¼R 2 2 ) RRed ffi

RRed ffi
1:618R, 0:618R However,
1.618R is not an acceptable value for the equivalent resistance of that part of the circuit, since it must be non-negative. Now, the input resistance of the circuit can be determined by using equation (2). Rin ¼

RRRed 0:618R2 ¼ ¼ 0:382R R þ RRed 1:618R

Choice (4) is the answer.

Fig. 2.57 The circuits of solution of problem 2.57

98

2 Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

2.58. To adjust the circuit as an amplifier, the ratio of VV os must be greater than one. To find this ratio, we need to analyze the circuit. Herein, nodal analysis is the best approach. Just two nodes out of four nodes are independent with unknown voltages (e1, Vo), as can be seen in the figure. In this question, note that the quantity shown for each resistor is its conductance and not resistance. KCL in the node with the node voltage of e1: 5ðe1
V s Þ þ 2e1 þ αI x ¼ 0 ) 7e1
5V s þ αI x ¼ 0

ð1Þ

KCL in the node with the node voltage of Vo: 4
αI x þ 3V o þ V o ¼ 0 ) I x ¼ V o α

ð2Þ

On the other hand, we can define Ix based on the node voltage. Using Ohm’s law: I x ¼ 2e1

ð3Þ

4 2 2e1 ¼ V o ) e1 ¼ V o α α

ð4Þ

Solving (2) and (3):

Put (2) and (4) in (1):
7


 2 4 14 þ 4α V 5α V o
5V s þ α V o ¼ 0 ) V o ¼ 5V s ) o ¼ α α α V s 14 þ 4α

The value of VV os must be greater than one to adjust the circuit as an amplifier. Thus: 5α > 1 ) 5α > 14 þ 4α ) α > 14 14 þ 4α Choice (3) is the answer.

Fig. 2.58 The circuit of solution of problem 2.58

2.59. First, we need to calculate the parametric value of input resistance of the circuit. As can be seen in Fig. 2.59.2, a test voltage source is connected to the circuit, since the circuit includes a dependent power source. If the circuit had independent sources, we needed to turn them off. Nodal analysis is the best method to apply in this problem, since the circuit includes just two independent nodes. KCL in node 1: e1 e1
V t e
Vt þ þ αI þ 1 ¼ 0 ) 3e1
2V t þ αI ¼ 0 1 1 1

ð1Þ

2

Solutions of Problems: Circuit Components, Methods of Circuit Analysis, and Circuit Theorems

99

KCL in the supernode:
I t þ

V t
e1 V
e1 V t
αI þ t þ ¼ 0 )
I t þ 2:5V t
2e1
αI ¼ 0 1 1 2

ð2Þ

Defining I based on the node voltages: I¼

V t
e1 ¼ V t
e1 1

ð3Þ

Solving (3) and (1): α
2 V α
3 t

ð4Þ


I t þ 2:5V t
2e1
αðV t
e1 Þ ¼ 0 )
I t þ V t ð2:5
αÞ þ e1 ðα
2Þ ¼ 0

ð5Þ

3e1
2V t þ αðV t
e1 Þ ¼ 0 ) ð3
αÞe1 þ ðα
2ÞV t ¼ 0 ) e1 ¼ Solving (3) and (2):

Solving (4) and (5):    α
2 α2
4α þ 4 V t ðα
2Þ ¼ 0 )
I t þ V t 2:5
α þ ¼0 α
3 α
3  
1  
1 Vt α2
4α þ 4 2:5α
α2
7:5 þ 3α þ α2
4α þ 4 ) ¼ 2:5
α þ ¼ α
3 α
3 It


I t þ V t ð2:5
αÞ þ




) ) Rin ¼

Vt α
3 α
3  ¼ ¼ 1:5α
3:5 1:5 α
73 It Vt α
3 7  0. Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ● Normal ○ Large 1) 7.5e3t A 2) 5e2t A 3) 7.5et A 4) 5e0.5t A

128

3

Problems: First-Order Circuits

Fig. 3.31 The circuit of problem 3.31

3.32. In the circuit of Fig. 3.32, VC(0) ¼ 4 V. For what size of the resistor R will the energy of the capacitor be always constant? Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ● Normal ○ Large 1) 34 Ω 2) 32 Ω 3) 23 Ω 4) 43 Ω

Fig. 3.32 The circuit of problem 3.32

3.33. In the circuit of Fig. 3.33, there is no energy stored in the inductor for t < 0. The switch is closed at t ¼ 1 s. Determine the current flowing through the short-circuit branch at t ¼ 4 s (Isc(t ¼ 4 s)). Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ○ Normal ● Large 1) 0.45 A 2) 2 A 3) 1.995 A 4) None of the above

Fig. 3.33 The circuit of problem 3.33

3

Problems: First–Order Circuits

129

3.34. In the circuit of Fig. 3.34, if the initial voltage of the capacitor is zero, V ðt Þ ¼ 14 ð1  e3t Þuðt Þ is achieved. Now, if the capacitor is replaced by an inductor with the size of 2 H, determine its voltage. Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ○ Normal ● Large 1) 14 e6 uðt Þ V 2) 14 e3t uðt Þ V   2t 3) 14 1  e 3 uðt Þ V t
4) 14 1  e6 uðt Þ V t

Fig. 3.34 The circuit of problem 3.34

3.35. The short-term voltage signal shown in Fig. 3.35.2 is applied to the circuit illustrated in Fig. 3.35.1. Determine the output voltage at t ¼ 2.2 s, assuming IL(0) ¼ 0. Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ○ Normal ● Large 1) 0 V 2) 0.968 V 3) 0.586 V 4) 1.482 V

Fig. 3.35 The circuit of problem 3.35

3.36. In the circuit of Fig. 3.36, the switch is set in “A” position at t ¼ 0. When V(t) reaches the 50% of its primary value, the switch is moved to “B” position. Determine V(t) during the time that the switch is in “B” position. Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ○ Normal ● Large 1) 50e(t  2 ln 2)/2 V 2) 50e(t  ln 2)/3 V 3) 50e(t  ln 2)/2 V 4) 50e(t  2 ln 2)/3 V

130

3

Problems: First-Order Circuits

Fig. 3.36 The circuit of problem 3.36

3.37. In the circuit of Fig. 3.37, switch S1 is closed at t ¼ 0. Additionally, switch S2 is closed when its voltage reaches 9 V. Determine VC(t) after the second switching. Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ○ Normal ● Large 1) 2) 3) 4)

6 þ 1:8e 3 V t1:89 6  2:25e 3 V t2:2 6 þ 1:8e 0:75 V t1:89 6 þ 2:25e 0:75 V t2:2

Fig. 3.37 The circuit of problem 3.37

3.38. The voltage pulse shown in Fig. 3.38.1 is applied to the circuit of Fig. 3.38.2. Determine the time-dependent current equation of the circuit. Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ○ Normal ● Large 1) I(t) ¼ 20etu(t)  10e(t  2)u(t  1) A 2) I(t) ¼ 10etu(t)  10e(t  1)u(t  1) A 3) I(t) ¼ 20etu(t)  20e(t  1)u(t  1) A 4) I(t) ¼ 20etu(t) + 20e(t  1)u(t  1) A

3

Problems: First–Order Circuits

131

Fig. 3.38 The circuit of problem 3.38

3.39. In the circuit of Fig. 3.39, the switch has been in the closed position for a long time, but it is opened at t ¼ 0. Determine V(t) for t > 0. Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ○ Normal ● Large 1) 2) 3) 4)

12e9t V 9 6e4t V 4 12e9t V 9 7:5e4t V 4

Fig. 3.39 The circuit of problem 3.39

3.40. The circuit of Fig. 3.40 is in zero state at t ¼ 0. Calculate IC(t). Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ○ Normal ● Large 1) 2) 3) 4)

1 RC  4R e uðt Þ V. t 1 2RC  2R e uðt Þ V. t 1 2RC  4R e uðt Þ V. It is impossible to calculate it without knowing Vs(t). t

132

3

Problems: First-Order Circuits

Fig. 3.40 The circuit of problem 3.40

3.41. The network shown in Fig. 3.41 includes a resistor and a capacitor that have been connected in parallel. The response of
the circuit for a unit step function is V ðt Þ ¼ 43 1  e1:5t uðt Þ V . If the 2 Ω resistor is replaced by a 0.5 F capacitor, determine the output voltage. Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ○ Normal ● Large 1) 2(1  e0.25t)u(t) V 2) 2(1  e0.5t)u(t) V 3) 4(1  e0.5t)u(t) V 4) 4(1  e0.25t)u(t) V

Fig. 3.41 The circuit of problem 3.41

4

Solutions of Problems: First-Order Circuits

Abstract

In this chapter, the problems of the third chapter are fully solved, in detail and step-by-step. In solving the problems of the first-order circuits, the subjects discussed in the first chapter, that is, the methods of circuit analysis and circuit theorems, are applied. In this chapter, the concepts and the parameters of equivalent capacitance, equivalent inductance, time constant of a capacitive first-order circuit, time constant of an inductive first-order circuit, transient response, voltage continuity principle of capacitor, current continuity principle of inductor, primary value, final value, steady state condition, stored energy of circuit, switching operation, and current-voltage and voltage-current relations of capacitor and inductor are explained. 4.1. First, we need to calculate the Thevenin resistance of the circuit. As can be seen in Fig. 4.1.2, the independent voltage and current sources need to be turned off.

64

RTh ¼ Rin ¼ ð1 þ 2 þ 3Þ
4 ¼ 6
4 ¼ ¼ 2:4 Ω 6þ4 The time constant of the RC circuit: τ ¼ RTh C ¼ 2:4  0:5 ¼ 1:2 sec Choice (1) is the answer.

Fig. 4.1 The circuits of solution of problem 4.1

4.2. Although the circuit includes three energy-saving components, the circuit is a first-order circuit. To calculate the time constant of an RC circuit, we need to determine the Thevenin resistance seen by the equivalent capacitor. Herein, we must turn off the independent power source. In this problem, the capacitors have a parallel connection, as can be seen in Fig. 4.2.2. Thus: # Springer Nature Switzerland AG 2020 M. Rahmani-Andebili, DC Electrical Circuit Analysis, https://doi.org/10.1007/978-3-030-50711-4_4

133

134

4 Solutions of Problems: First-Order Circuits

Ceq ¼ 2 þ 2 þ 3 ¼ 7 F Calculating the Thevenin resistance seen by the capacitor in the circuit of Fig. 4.2.3:





RTh ¼ 6
3 þ 1
3 þ 3 ¼ ð2 þ 1Þ
3 þ 3 ¼ 1:5 þ 3 ¼ 4:5 Ω The time constant of the RC circuit: τ ¼ RTh C eq ¼ 7  4:5 ¼ 31:5 sec Choice (3) is the answer.

Fig. 4.2 The circuits of solution of problem 4.2

ð1Þ

4

Solutions of Problems: First‐Order Circuits

135

4.3. The Thevenin resistance seen by the capacitor is:
R R
RTh ¼ R1
R2 ¼ 1 2 R1 þ R2 The time constant of the circuit: τ ¼ RTh C ¼

R1 R2 R R C C ¼ 1 2 R1 þ R2 R1 þ R2

Choice (1) is the answer.

Fig. 4.3 The circuits of solution of problem 4.3.

4.4. To determine the time constant of the first-order RL circuit, we need to calculate the Thevenin resistance of the circuit seen by the inductor. Since the circuit includes at least one independent source (independent current source), we must apply a test source (test voltage source) to determine the value of VI tt, as can be seen in Fig. 4.4.2. In this problem, nodal analysis is applied. Defining i based on the node voltage, by using Ohm’s law: i¼

Vt 6

ð1Þ

KCL in the supernode: I t  2i þ i þ

Using ð1Þ Vt V V V V ¼ 0 )  It  i þ t ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)  I t  t þ t ¼ 0 )  I t þ t ¼ 0 3 3 6 3 6 )

Vt ¼ 6 ) RTh ¼ 6 Ω It

Time constant of the circuit: τ¼ Choice (4) is the answer.

Leq 0:06 ¼ 0:01 sec ¼ 6 RTh

ð2Þ

136

4 Solutions of Problems: First-Order Circuits

Fig. 4.4 The circuits of solution of problem 4.4

4.5. First, we need to determine the primary voltage of the capacitor before the switching operation. Figure 4.5.2 shows the status of the circuit before the switch is closed. As can be seen, the independent current source is replaced by an open circuit branch because I ¼ 0 before closing the switch. Moreover, since the switch has been open for a long time, the capacitor has reached its steady state condition (open circuit). Therefore, the primary voltage of the capacitor is 6 V (Vc(0) ¼ 6 V ), since there is no voltage drop across the 6 Ω resistor. After closing the switch, the configuration of the circuit is updated, as is illustrated in Fig. 4.5.3. Due to the voltage continuity of capacitor, the instant voltage of the capacitor will be 6 V even after the switching operation. V c ð 0þ Þ ¼ V c ð 0 Þ ¼ 6 V

ð1Þ

Now, by applying KCL in the indicated supernode, we have: 66 6  10 þ 5I ð0þ Þ þ I c ð0þ Þ þ ¼ 0 ) 5I ð0þ Þ þ I c ð0þ Þ  1 ¼ 0 6 4

ð2Þ

Defining I(0+) based on the node voltages: I ð 0þ Þ ¼

6  10 ¼ 1 A 4

ð2Þ,ð3Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 5ð1Þ þ I c ð0þ Þ  1 ¼ 0 ) I c ð0þ Þ ¼ 6 A Choice (2) is the answer.

ð3Þ

4

Solutions of Problems: First‐Order Circuits

137

Fig. 4.5 The circuits of solution of problem 4.5

4.6. The circuit is in the steady state condition. Therefore, the capacitor can be modeled by an open circuit branch, as is shown in Fig. 4.6.2. Using the voltage division formula: 4 5 V C ð1Þ ¼ V a  V b ¼  24   24 ¼ 4 þ 26 5 þ 10



 4 5 3   24 ¼   24 30 15 15

) V C ð1Þ ¼ 4:8 V

ð1Þ

Using the voltage-charge relation of the capacitor: QC ð1Þ ¼ V C ð1ÞC ¼ 4:8  100  106 ¼ 0:48 mC The absolute value of the charge stored in the capacitor is: jQC ð1Þj ¼ j0:48 mC j ¼ 0:48 mC Choice (4) is the answer.

ð2Þ

138

4 Solutions of Problems: First-Order Circuits

Fig. 4.6 The circuits of solution of problem 4.6

4.7. To calculate the time constant of an RC circuit, we need to determine the Thevenin resistance of the circuit seen by the capacitor. As is shown in Fig. 4.7.2, all the independent sources are turned off, and a test source with the voltage and current of Vt and It is connected to the terminal. In this problem, nodal analysis is applied. Defining I1 based on the node voltage: I1 ¼ 

e2 3

ð1Þ

KCL in the supernode: Using ð1Þ e2 e2  V t 4 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 2e2  V t þ e2 ¼ 0 þ  3I 1  I 1 ¼ 0 ) 2e2  V t  4I 1 ¼ 0 ¼ 3 1 1 )  Vt þ

10 3 e ¼ 0 ) e2 ¼ V t 3 2 10

KCL in node 1: I t þ

Using ð1Þ V t  e2 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)  I t þ V t  e2  e2 ¼ 0 )  I t þ V t  2e2 ¼ 0 þ 3I 1 ¼ 0 ¼ 1

ð2Þ

4

Solutions of Problems: First‐Order Circuits

139

  Using ð2Þ 3 6 4 V t ¼ 0 )  It þ V t  V t ¼ 0 )  It þ V t ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)  I t þ V t  2 10 10 10 )

V t 10 V ¼ ¼ 2:5 ) RTh ¼ t ¼ 2:5 Ω It 4 It

ð3Þ

The time constant of the RC circuit: τ ¼ RTh C ¼ 2:5  2 ¼ 5 sec Choice (4) is the answer.

Fig. 4.7 The circuits of solution of problem 4.7

4.8. Based on the given information: V C ð 0 Þ ¼ 0 V

ð1Þ

Due to the voltage continuity of the capacitor in lack of a power source with an infinite size (e.g., impulse power source): V C ð 0þ Þ ¼ V C ð 0 Þ ¼ 0 V

ð2Þ

To calculate the Thevenin resistance of the circuit, the independent current source is turned off, as is shown in Fig. 4.8.2. As can be noticed from Fig. 4.8.2, the circuit is symmetric if we calculate the Thevenin resistance seen from terminal A-B. Thus, as is shown in Fig. 4.8.3, we can fold the circuit around the A-B axis. By doing this, the two ends of the 3 Ω resistors are short-circuited; therefore, they are eliminated. However, the other resistors are paralleled two by two. Figure 4.8.4 illustrates the simplified circuit of Fig. 4.8.3. RTh ¼ Rin ¼ 0:5 þ 1 þ 0:5 ¼ 2 Ω

ð3Þ

τ ¼ RTh C ¼ 2  2 ¼ 4 sec

ð4Þ

The time constant of the RC circuit:

140

4 Solutions of Problems: First-Order Circuits

The final voltage of the capacitor can be calculated by assuming the capacitor as the open circuit branch, as can be seen in Fig. 4.8.5: V C ð1Þ ¼ I s  ð0:5 þ 1 þ 0:5Þ ¼ 1  2 ¼ 2 V Now, by using the general form of the capacitor voltage in a first-order circuit with DC power supply, we have:   t t t V C ðt Þ ¼ V C ð1Þ þ ðV C ð0þ Þ  V C ð1ÞÞeτ ¼ 2 þ ð0  2Þe4 ¼ 2 1  e4 V Choice (4) is the answer.

Fig. 4.8 The circuits of solution of problem 4.8

ð5Þ

4

Solutions of Problems: First‐Order Circuits

141

Fig. 4.8 (continued)

4.9. Based on the given information: V C ð 0 Þ ¼ 0 V

ð1Þ

Due to the voltage continuity of the capacitor in lack of a power source with an infinite size (e.g., impulse power source): V C ð 0þ Þ ¼ V C ð 0 Þ ¼ 0 V

ð2Þ

The final voltage of the capacitor can be calculated by using the current division formula and Ohm’s law in the circuit of Fig. 4.9.2. I 1,3 ¼

2þ4 5¼3A 2þ4þ1þ3

ð3Þ

I 2,4 ¼

1þ3 5¼2A 2þ4þ1þ3

ð4Þ

142

4 Solutions of Problems: First-Order Circuits

V C ð1Þ ¼ V C þ  V C  ¼ 3I 1,3  4I 2,4 ¼ 3  3  4  2 ¼ 9  8 ¼ 1 V

ð5Þ

To determine the Thevenin resistance of the circuit, we need to turn off the independent current source, as is shown in Fig. 4.9.3.

21

RTh ¼ ð1 þ 2Þ
ð3 þ 4Þ ¼ 3
7 ¼ Ω ð6Þ 10 The time constant of the circuit is: τ ¼ RTh C ¼

21 21 2¼ sec 10 5

ð7Þ

Using the general form of the capacitor voltage in a first-order circuit with DC power supply: V C ðt Þ ¼ 1 þ ð0  1Þe

21t 5

¼ 1  e21t V 5

Choice (1) is the answer.

Fig. 4.9 The circuits of solution of problem 4.9

4

Solutions of Problems: First‐Order Circuits

143

4.10. First, to simplify the problem, we should determine the Thevenin equivalent circuit seen by the capacitor. As is shown in Fig. 4.10.2, the input resistance of the circuit can be calculated by using the series-parallel rule.







RTh ¼ Rin ¼ 1k
1k þ 0:5k
1k ¼ ð0:5k þ 0:5k Þ
1k ¼ 1k
1k ¼ 0:5 kΩ

ð1Þ

As can be seen in Fig. 4.10.3, we can use the voltage division formula, twice, to calculate the open circuit voltage.

1k1:5k 1:5k 1k
ð0:5k þ 1k Þ 1kþ1:5k 2:5k
 8 ¼  8 ¼ 8¼3V e1 ¼ 4k 1k1:5k
1k þ 1k þ 1k
ð0:5k þ 1kÞ 2:5k 1kþ1:5k V Th ¼ V oc ¼

1k 1k e ¼ 3¼2V 1k þ 0:5k 1 1:5k

ð2Þ

ð3Þ

The Thevenin equivalent circuit is illustrated in Fig. 4.10.4. Based on the given information: V C ð 0 Þ ¼ 0 V

ð4Þ

Due to the voltage continuity of the capacitor in lack of a power source with an infinite size (e.g., impulse power source): V C ð 0þ Þ ¼ V C ð 0 Þ ¼ 0 V

ð5Þ

The final voltage of the capacitor will be 2 V, since the capacitor will behave like an open circuit. Thus: V C ð 1Þ ¼ 2 V

ð6Þ

τ ¼ RTh C ¼ 0:5 kΩ  2 μF ¼ 1 ms

ð7Þ

The time constant of the RC circuit:

Using the general form of the capacitor voltage in a first-order circuit with DC power supply:   t 3 t V C ðt Þ ¼ V C ð1Þ þ ðV C ð0þ Þ  V C ð1ÞÞeτ ¼ 2 þ ð0  2Þe1103 ¼ 2 1  e10 t       3 3 V C ðt ¼ 1msÞ ¼ V C t ¼ 103 s ¼ 2 1  e10 10 ¼ 2 1  e1 V Choice (2) is the answer.

ð8Þ

144

4 Solutions of Problems: First-Order Circuits

Fig. 4.10 The circuits of solution of problem 4.10

4.11. To simplify the problem, we can determine the Thevenin equivalent circuit seen by the inductor. The input resistance of the circuit can be calculated by using the series-parallel rule, as is shown in Fig. 4.11.2.
46
RTh ¼ Rin ¼ 1 þ ð1 þ 3Þ
ð2 þ 4Þ ¼ 1 þ ¼ 3:4 Ω 4þ6

ð1Þ

The Thevenin voltage of the circuit will be zero, since there is no power source in the circuit. Therefore, we have the Thevenin equivalent circuit shown in Fig. 4.11.3. Based on the given information: I L ð0Þ ¼ 1 A

ð2Þ

Due to the current continuity of the inductor in lack of a power source with an infinite size (e.g., impulse power source): I L ð 0þ Þ ¼ I L ð 0Þ ¼ 1 A The final current of the inductor will be 0 A, since there is no power source in the circuit. Thus:

ð3Þ

4

Solutions of Problems: First‐Order Circuits

145

I L ð1Þ ¼ 0 A

ð4Þ

The time constant of the RC circuit: τ¼

L 1 ¼ sec RTh 3:4

ð5Þ

Using the general form of the inductor current in a first-order circuit with DC power supply: I L ðt Þ ¼ I L ð1Þ þ ðI L ð0þ Þ  I L ð1ÞÞeτ ¼ 0 þ ð1  0Þe3:4t ¼ e3:4t t

ð6Þ

Now, by using the current division formula and Ohm’s law, we can calculate Vab(t) in the circuit of Fig. 4.11.4. I 1,3 ¼

2þ4  e3:4t ¼ 0:6e3:4t A 2þ4þ1þ3

ð7Þ

I 2,4 ¼

1þ3  e3:4t ¼ 0:4e3:4t A 2þ4þ1þ3

ð8Þ

V ab ðt Þ ¼ V a ðt Þ  V b ðt Þ ¼ 3I 1,3  4I 2,4 ¼ 3  0:6e3:4t  4  0:4e3:4t ¼ ð1:8  1:6Þe3:4t ) V ab ðt Þ ¼ 0:2e3:4t V Choice (2) is the answer.

Fig. 4.11 The circuits of solution of problem 4.11

146

4 Solutions of Problems: First-Order Circuits

4.12. First, we should determine the Thevenin equivalent circuit seen by the capacitor. Using the current division formula and Ohm’s law to calculate the open circuit voltage in Fig. 4.12.2:  V Th ¼ V oc ¼ V þ oc  V oc ¼

Is I 1 s1¼0V 2 2

ð1Þ

Calculating the Thevenin resistance of the circuit seen by the capacitor (see Fig. 4.12.3):



RTh ¼ Rin ¼ ð1 þ 1Þ
ð1 þ 1Þ ¼ 2
2 ¼ 1 Ω

ð2Þ

Now, we have the Thevenin equivalent circuit shown in Fig. 4.12.4. KCL in the indicated node: I C ðt Þ þ I R ðt Þ ¼ 0 I C ðt Þ ¼ C

d d d V ðt Þ ¼ 1  V C ðt Þ ¼ V C ðt Þ dt C dt dt I R ðt Þ ¼

V C ðt Þ ¼ V C ðt Þ 1

Solving (3), (4), and (5): d V ðt Þ þ V C ðt Þ ¼ 0 dt C Choice (3) is the answer.

Fig. 4.12 The circuits of solution of problem 4.12

ð3Þ ð4Þ ð5Þ

4

Solutions of Problems: First‐Order Circuits

147

4.13. Based on the given information: V C ð0 Þ ¼ 2 V

ð1Þ

V C ð0þ Þ ¼ V C ð0 Þ ¼ 2 V

ð2Þ

Due to the voltage continuity of capacitor:

To determine the Thevenin resistance of the circuit, we need to turn off the voltage source, as can be seen in Fig. 4.13.2.

RTh ¼ 1
1 ¼ 0:5 Ω

ð3Þ

τ ¼ RTh C ¼ 0:5  1 ¼ 0:5 sec

ð4Þ

The time constant of the circuit:

The final voltage of the capacitor can be calculated by using the voltage division formula in Fig. 4.13.3. V C ð 1Þ ¼

1  5 ¼ 2:5 V 1þ1

ð5Þ

Using the general form of the capacitor voltage in a first-order circuit with DC power supply: V C ðt Þ ¼ V C ð1Þ þ ðV C ð0þ Þ  V C ð1ÞÞeτ ¼ 2:5 þ ð2  2:5Þe0:5 ¼ 2:5  4:5e2t t

t

ð6Þ

Now, by applying KCL in the indicated supernode of the circuit of Fig. 4.13.4, we have: V C ðt Þ  5 þ I C ðt Þ þ I ðt Þ ¼ 0 ) I ðt Þ ¼ V C ðt Þ þ 5  I C ðt Þ 1

ð7Þ

The voltage-current relation of the capacitor: I C ðt Þ ¼ C

Using ð6Þ    d d V ðt Þ ¼ 2:5  4:5e2t ¼ ð4:5Þ  2e2t ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) I C ðt Þ ¼ 1  dt C dt ) I C ðt Þ ¼ 9e2t

Solving (6), (7), and (8):   I ðt Þ ¼ V C ðt Þ þ 5  I C ðt Þ ¼  2:5  4:5e2t þ 5  9e2t ¼ 2:5  4:5e2t Choice (2) is the answer.

ð8Þ

148

4 Solutions of Problems: First-Order Circuits

Fig. 4.13 The circuits of solution of problem 4.13

4.14. To calculate the time constant of the circuit, first, we need to determine the Thevenin resistance seen by the capacitor, as is shown in Fig. 4.14.2. Herein, since the circuit includes a dependent source, we must apply a test source. Moreover, the independent current source must be turned off. Using Ohm’s law for the 4 Ω resistor: Ix ¼  KCL in the supernode:

Vt 4

ð1Þ

4

Solutions of Problems: First‐Order Circuits

149

I x  I t þ kI x þ

Vt V ¼ 0 )  I t þ ðk  1ÞI x þ t ¼ 0 3 3

    Using ð1Þ Vt Vt 1k 1 þ ¼ 0 )  It þ þ Vt ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)  I t þ ðk  1Þ  4 3 4 3 )  It þ

3  3k þ 4 V 12 12 Vt ¼ 0 ) t ¼ ) RTh ¼ 12 7  3k 7  3k It

ð2Þ

As we know, the time constant of an RC circuit is: τ ¼ RTh C ¼

12 24 2¼ 7  3k 7  3k

ð3Þ

Based on the given information: τ ¼ 8 sec Solving (3) and (4): 24 4 ¼ 8 ) 3 ¼ 7  3k ) 3k ¼ 4 ) k ¼ 7  3k 3 Choice (1) is the answer.

Fig. 4.14 The circuits of solution of problem 4.14

ð4Þ

150

4 Solutions of Problems: First-Order Circuits

4.15. As we know, the time constant of first-order RC and RL circuits can be calculated by using (1) and (2), respectively: τRC ¼ RTh C

ð1Þ

L RTh

ð2Þ

τRC ¼ τRL

ð3Þ

and L ¼ C

ð4Þ

τRL ¼ Based on the given information:

Solving (1), (2), and (3): RTh C ¼

Using ð4Þ L ) RTh 2 C ¼ L ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) RTh 2 ¼ 1 RTh

ð5Þ

Therefore, we just need to calculate the Thevenin resistance seen by the capacitor. The circuit includes a dependent source; thus we need to apply a test source. Additionally, the independent sources must be turned off, as can be seen in Fig. 4.15.2. Defining Ix based on the node voltages: Ix ¼ 

e1 ¼ e1 1

ð6Þ

KCL in the top node of the test source: I t þ

V t  e1 ¼ 0 ) e1 ¼ V t  RI t R

ð7Þ

KCL in the supernode: I t þ

Using ð6Þ e1 e1 e1  2I x ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)  I t þ 3e1  2ðe1 Þ ¼ 0 þ þ ¼ 0 )  I t þ 3e1  2I x ¼ 0 ¼ 1 1 1

Using ð7Þ V 1 )  I t þ 5e1 ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)  I t þ 5ðV t  RI t Þ ¼ 0 ) 5V t  I t ð1 þ 5RÞ ¼ 0 ) t ¼ R þ 5 It ) RTh ¼ R þ

1 5

Solving (5) and (8):   1 2 1 4 6 ¼ 1 ) R þ ¼ 1 ) R ¼ Ω,  Ω Rþ 5 5 5 5 Negative value of R is not acceptable. Thus: R¼ Choice (1) is the answer.

4 ¼ 0:8 Ω 5

ð8Þ

4

Solutions of Problems: First‐Order Circuits

151

Fig. 4.15 The circuits of solution of problem 4.15

4.16. Based on the given information: V C ð0þ Þ ¼ 10 V

ð1Þ

The Thevenin resistance seen by the capacitor (see Fig. 4.16.2):
12  4
RTh ¼ Rin ¼ 9 þ ð4 þ 8Þ
ð3 þ 1Þ ¼ 9 þ ¼ 9 þ 3 ¼ 12 Ω 12 þ 4

ð2Þ

The time constant of the circuit: τ ¼ RTh C ¼ 12  3 ¼ 36 sec

ð3Þ

The final voltage of the capacitor will be zero, since there is no power source in the circuit. V C ð 1Þ ¼ 0 V

ð4Þ

Using the general form of the capacitor voltage in a first-order circuit with DC power supply: V C ðt Þ ¼ V C ð1Þ þ ðV C ð0þ Þ  V C ð1ÞÞeτ ¼ 0 þ ð10  0Þe36 ¼ 10e36 V t

Using the voltage division formula in the circuit of Fig. 4.16.1:

t

t

ð5Þ

152

4 Solutions of Problems: First-Order Circuits

V ab ðt Þ ¼ V a ðt Þ  V b ðt Þ ¼

8 1 5 V ðt Þ   V x ðt Þ ¼ V x ðt Þ 8þ4 x 1þ3 12



ð4 þ 8Þ
ð3 þ 1Þ t t 3 30 t
 10e36 ¼ V x ðt Þ ¼  10e36 ¼ e36
3 þ 9 12 ð4 þ 8Þ
ð3 þ 1Þ þ 9

ð6Þ

ð7Þ

Solving (6) and (7): V ab ðt Þ ¼

t 5 30 36t 150 36t  e ¼ e ¼ 1:04e36 V 12 12 144

Choice (2) is the answer.

Fig. 4.16 The circuits of solution of problem 4.16

4.17. Before closing the switch at t ¼ 0, the circuit is like the one shown in Fig. 4.17.2. As can be seen, the inductor is modeled by an open circuit branch, since the circuit is in its steady state condition. I L ð 0 Þ ¼ 5 A

ð1Þ

After the switching operation, the circuit is changed to the one illustrated in Fig. 4.17.3. Due to current continuity of the inductor: I L ð 0þ Þ ¼ I L ð 0 Þ ¼ 5 A

ð2Þ

To calculate the Thevenin resistance seen by the inductor, we need to turn off the independent current source, as is shown in Fig. 4.17.4. RTh ¼ Rin ¼ 100 Ω

ð3Þ

The time constant of the circuit: 1

τ¼

L 1 ¼ 10 ¼ sec RTh 100 1000

ð4Þ

Figure 4.17.5 shows how to calculate the final current of the inductor. As can be seen, the final current will be zero, since the whole current will pass through the short-circuit branch. I L ð1Þ ¼ 0 A

ð5Þ

4

Solutions of Problems: First‐Order Circuits

153

Using the general form of the current of an inductive first-order circuit with DC power supply: I L ðt Þ ¼ I L ð1Þ þ ðI L ð0þ Þ  I L ð1ÞÞeτ ¼ 0 þ ð5  0Þe t



t 1 1000

¼ 5e1000t

Now, by applying KCL in the indicated node of the circuit of Fig. 4.17.3, we have:   5 þ I ðt Þ þ I L ðt Þ ¼ 0 ) I ðt Þ ¼ 5  I L ðt Þ ¼ 5  5e1000t ¼ 5 1  e1000t A Choice (3) is the answer.

Fig. 4.17 The circuits of solution of problem 4.17

ð6Þ

154

4 Solutions of Problems: First-Order Circuits

4.18. Based on the given information: V C ð 0 Þ ¼ 2 V

ð1Þ

Due to the voltage continuity of the capacitor in lack of a power source with an infinite size (e.g., impulse power source): V C ð 0þ Þ ¼ V C ð 0 Þ ¼ 2 V

ð2Þ

To calculate the Thevenin resistance seen by the capacitor, we need to apply a test source, since the circuit includes a dependent power source, as can be seen in Fig. 4.18.2. Defining i based on the node voltages: i¼

V t  e1 ¼ V t  e1 1

ð3Þ

KCL in the right-side supernode: i þ

Using ð3Þ 1 e1 1 10 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) e1  3ðV t  e1 Þ ¼ 0 ) e  3V t ¼ 0  2i ¼ 0 ) e1  3i ¼ 0 ¼ 3 3 3 1 3 ) e1 ¼

9 V 10 t

ð4Þ

KCL in the left-side supernode: I t þ

Using ð4Þ V t V t  e1 3 3 9 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)  I t þ V t  V t ¼ 0 þ ¼ 0 )  I t þ V t  e1 ¼ 0 ¼ 2 2 10 2 1 )  It þ

6 V 5 5 V ¼ 0 ) t ¼ ) RTh ¼ Ω 10 t 3 3 It

ð5Þ

5  3 ¼ 5 sec 3

ð6Þ

The time constant of the circuit: τ ¼ RTh C ¼

The final voltage of the capacitor will be zero, since there is no power source in the circuit, as can be seen in 3.38.1. V C ð 1Þ ¼ 0 V

ð7Þ

Using the general form of the capacitor voltage in a first-order circuit with DC power supply: V C ðt Þ ¼ V C ð1Þ þ ðV C ð0þ Þ  V C ð1ÞÞeτ ¼ 0 þ ð2  0Þe5 ¼ 2e5 V t

t

t

ð8Þ

Based on the given information: V C ðt 0 Þ ¼ 0:5  2 ¼ 1 V Solving (8) and (9): t0

t0

2e 5 ¼ 1 ) e 5 ¼ 0:5 )  Choice (1) is the answer.

t0 ¼ ln ð0:5Þ ) t 0 ¼ 5 ln ð2Þ sec 5

ð9Þ

4

Solutions of Problems: First‐Order Circuits

155

Fig. 4.18 The circuits of solution of problem 4.18

4.19. As can be seen in the circuit of Fig. 4.19.2, the equivalent capacitance of the circuit seen by terminal a0  b0 (C 0eq ) is approximately equal to the one seen by terminal a  b (Ceq), since the size of the circuit is infinite. Therefore, we can write: C 0eq ffi Ceq

ð1Þ

Now, we can calculate the total capacitance of the circuit by using the circuit of Fig. 4.19.3: 2C0eq þ 3 C0eq þ 1 1 1 1 1 1 þ ¼2þ 0 ¼ 0 ) C eq ffi 0 ffi þ 0 Ceq 1 Ceq þ 1 1 Ceq þ 1 2Ceq þ 3 Ceq þ 1 Solving (1) and (2): C eq ffi

Ceq þ 1 ) 2C eq 2 þ 3Ceq ffi Ceq þ 1 ) 2C eq 2 þ 2Ceq  1 ffi 0 2Ceq þ 3

) Ceq ffi

2 

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 22  4  2  ð1Þ 22

pffiffiffiffiffi 2  12 ¼ ¼ 1:36 F, 0:36 F 4

Ceq ffi  1.36 F is not acceptable. Therefore: C eq ffi 0:36 F Choice (2) is the answer.

ð2Þ

156

4 Solutions of Problems: First-Order Circuits

Fig. 4.19 The circuits of solution of problem 4.19

4.20. First, we need to find the voltage of the capacitor before the switching operation by using the circuit of Fig. 4.20.2. Based on the given information: V C ð 0 Þ ¼ 0

ð1Þ

Because of the voltage continuity of the capacitor in lack of a power source with an infinite size (e.g., impulse power source): V C ð 0þ Þ ¼ V C ð 0 Þ ¼ 0

ð2Þ

After a long time, the capacitor will be an open circuit, and its voltage can be calculated by using Ohm’s law: V C ð1Þ ¼ 1  R1 ¼ R1

ð3Þ

τ ¼ RTh C ¼ R1 C

ð4Þ

The time constant of the RC circuit:

Using the general form of the capacitor voltage in a first-order circuit with DC power supply:

4

Solutions of Problems: First‐Order Circuits

157

V C ðt Þ ¼ V C ð1Þ þ ðV C ð0þ Þ  V C ð1ÞÞeτ ¼ R1 þ ð0  R1 Þe t

R t C 1

   t ¼ R1 1  e R1 C

ð5Þ

After closing the switch at t ¼ t1 ¼ R1C, the circuit is updated and converted to the one shown in Fig. 4.20.3. The primary voltage of the capacitor for the circuit of Fig. 4.20.3 can be calculated as follows:   t  R C    1  1 V C ðt 1 þ Þ ¼ V C ðt 1  Þ ¼ R1 1  e R1 C ¼ R1 1  e R1 C ¼ R1 1  e1

ð6Þ

The final voltage of the capacitor can be determined by using Ohm’s law and assuming the capacitor as the open circuit branch after a long time: V C ð 1Þ ¼ 1 

R1 R2 RR ¼ 1 2 R1 þ R2 R1 þ R2

ð7Þ

By studying the general form of the capacitor voltage in a first-order circuit with DC power supply (see equation (8)), it is concluded that, to maintain the voltage of the capacitor constant, the transient part of the voltage must be eliminated, as is presented in (9): V C ðt Þ ¼ V C ð1Þ þ ðV C ðt 1 þ Þ  V C ð1ÞÞe ) ðV C ðt 1 þ Þ  V C ð1ÞÞe

tt1 τ

tt1 τ

¼0

In other words, the final voltage of the capacitor must be equal to its primary voltage. Using ð7Þ, ð6Þ ) V C ð 1Þ ¼ V C ð t 1 þ Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)

    R1 R2 R2 ¼ R1 1  e1 ) ¼ 1  e1 R1 þ R2 R1 þ R2

      ) R2 ¼ R1 1  e1 þ R2 1  e1 ) R2 e1 ¼ R1 1  e1 ) R2 ¼ R1 ðe  1Þ Choice (4) is the answer.

Fig. 4.20 The circuits of solution of problem 4.20

ð8Þ ð9Þ

158

4 Solutions of Problems: First-Order Circuits

4.21. Figure 4.21.2 shows the circuit before closing the switch. Since the switch was open for a long time, the circuit has reached its steady state condition, and the capacitor is like an open circuit. The voltage of the capacitor (before the switching) can be calculated by using Ohm’s law in Fig. 4.21.2. V C ð0 Þ ¼ 5  2 ¼ 10 V

ð1Þ

The voltage of the capacitor, after the switching, can be determined based on the capacitor voltage continuity law (in lack of a power source with an infinite size (e.g., impulse power source)): V C ð0þ Þ ¼ V C ð0 Þ ¼ 10 V

ð2Þ

Figure 4.21.3 illustrates the circuit after closing the switch. In this condition, the Thevenin resistance of the circuit is 1 Ω, since the 2 Ω resistor is short-circuited by the closed switch and the current source is turned off (see Fig. 4.21.4). RTh ¼ Rin ¼ 1 Ω

ð3Þ

τ ¼ RTh C ¼ 1  1 ¼ 1 sec

ð4Þ

The time constant of the RC circuit:

The final voltage of the capacitor will be zero as can be noticed from Fig. 4.21.5, since the whole current flows through the short-circuit branch. Or, by applying KVL in the right mesh, we have: V C ð1Þ þ 0 ¼ 0 V ) V C ð1Þ ¼ 0

ð5Þ

Using the general form of the capacitor voltage in a first-order circuit with DC power supply: V C ðt Þ ¼ V C ð1Þ þ ðV C ð0þ Þ  V C ð1ÞÞeτ ¼ 0 þ ð10  0Þet ¼ 10et t

ð6Þ

However, we are interested in the current of the switch at t ¼ 1 sec: KCL in the supernode in the circuit of Fig. 4.21.6: 5 þ I þ I C ðt Þ þ

00 ¼0 2

ð7Þ

Using the voltage-current relation of the capacitor: I C ðt Þ ¼ C

d d V ðt Þ ¼ 1  ð10et Þ ¼ 10et dt C dt

Solving (7) and (8): I ðt Þ ¼ 5  I C ðt Þ ¼ 5 þ 10et ) I ðt ¼ 1Þ ¼ 5 þ 10e1 ¼ 5ð1 þ 2Þe1 A Choice (4) is the answer.

ð8Þ

4

Solutions of Problems: First‐Order Circuits

Fig. 4.21 The circuits of solution of problem 4.21

159

160

4 Solutions of Problems: First-Order Circuits

4.22. The circuit seems to be a second-order circuit, but it is the series connection of two first-order circuits. In other words, we can individually analyze the first-order circuits. Using KVL in the indicated mesh in the circuit of Fig. 4.22.2: V o ðt Þ  V C ðt Þ þ V L ðt Þ ¼ 0 ) V o ðt Þ ¼ V C ðt Þ þ V L ðt Þ

ð1Þ

Based on the given information: V C ð 0 Þ ¼ 1 V

ð2Þ

Due to the voltage continuity of the capacitor in lack of a power source with an infinite size (e.g., impulse power source) in the RC circuit: V C ð 0þ Þ ¼ V C ð 0 Þ ¼ 1 V

ð3Þ

The final voltage of the capacitor will be zero, since the RC circuit is not connected to any power source. Thus: V C ð 1Þ ¼ 0

ð4Þ

τ ¼ RTh C ¼ 1  0:5 ¼ 0:5 sec

ð5Þ

The time constant of the RC circuit:

Using the general form of the capacitor voltage in a first-order circuit with DC power supply: V C ðt Þ ¼ V C ðt Þ ¼ V C ð1Þ þ ðV C ð0þ Þ  V C ð1ÞÞeτ ¼ 0 þ ð1  0Þe0:5 ¼ e2t t

t

ð6Þ

Using the current-voltage relation of the inductor: V L ðt Þ ¼ L

d d I ðt Þ ¼ 1 r ðt Þ ¼ uðt Þ dt L dt

ð7Þ

In (7), r(t) is ramp function with the following definition:

r ðt Þ ¼

0 t 0, unit step function (u(t)) is applied in (12). V L ðt Þ ¼ 2e  3t uðt Þ V 2

Choice (2) is the answer.

Fig. 4.28 The circuits of solution of problem 4.28

ð12Þ

170

4 Solutions of Problems: First-Order Circuits

4.29. The circuit for t ¼ 0 is shown in Fig. 4.29.2. The circuit is in its steady state condition; thus, the capacitor is open circuit. By applying KVL in the left-side mesh: 4 þ V 1 ð0 Þ ¼ 0 ) V 1 ð0 Þ ¼ 4 V

ð1Þ

By applying KVL in the right-side mesh: Using ð1Þ V 1 ð0 Þ þ V C ð0 Þ þ 2V 1 ð0 Þ ¼ 0 ) V C ð0 Þ ¼ V 1 ð0 Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) V C ð0 Þ ¼ 4 V

ð2Þ

Due to the voltage continuity of the capacitor in lack of a power source with an infinite size (e.g., impulse power source): V C ð0þ Þ ¼ V C ð0 Þ ¼ 4 V

ð3Þ

The circuit for t ¼ 0+ is illustrated in Fig. 4.29.3 while modeling the initial voltage of the capacitor by the 4 V independent voltage source. Applying KVL in the mesh: V 1 ð0þ Þ þ ð4Þ þ 2V 1 ð0þ Þ ¼ 0 ) V 1 ð0þ Þ ¼ 4 V

ð4Þ

Applying Ohm’s law for the 2 Ω resistor: Using ð4Þ 1 1 V 1 ð0þ Þ ¼ 2I C ð0þ Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) I C ð0þ Þ ¼   V 1 ð0þ Þ ¼   4 ¼ 2 A 2 2 Applying the voltage-current relation of the capacitor: I C ð0þ Þ ¼ C

d d d d V ð0þ Þ ¼ 1  V C ð0þ Þ ) V C ð0þ Þ ¼ I C ð0þ Þ ) V C ð0þ Þ ¼  2 V=sec dt C dt dt dt

Choice (4) is the answer.

Fig. 4.29 The circuits of solution of problem 4.29

ð5Þ

4

Solutions of Problems: First‐Order Circuits

171

4.30. Figure 4.30.2 shows the circuit for t ¼ 0, where the capacitor is open circuit, since the circuit is in the steady state condition. Based on the given information: V C ð 0 Þ ¼ 1 V

ð1Þ

Due to the presence of the impulse voltage source (δ(t)) in the circuit, the voltage of the capacitor will not be continuous. To calculate the value of VC(0+), we need to use the current-voltage relation of the capacitor, as follows: 1 V C ð0þ Þ ¼ V C ð0 Þ þ C

Z0þ 0

1 I C ðt Þ:dt ¼ 1 þ 0:5

Z0þ

Z0þ I C ðt Þ:dt ¼ 1 þ 2

0

I C ðt Þ:dt

ð2Þ

0

As we know, in the presence of the impulse voltage source, capacitors and inductors behave like short-circuit and open circuit branches, respectively. Therefore, we have the circuit of Fig. 4.30.3 for t ¼ 0+, where the whole current will flow through the capacitor. Using Ohm’s law: I C ðt Þ ¼

3δðt Þ 2

ð3Þ

Solving (2) and (3): V C ð 0þ Þ ¼ 1 þ 2

Z0þ 

3δðt Þ :dt ¼ 1 þ 3 2

0

Z0þ δðt Þ:dt

ð4Þ



0

As we know: Z0þ δðt Þ:dt ¼ 1 

0

Solving (4) and (5): V C ð 0þ Þ ¼ 1 þ 3  1 ¼ 4 V Choice (2) is the answer.

Fig. 4.30 The circuits of solution of problem 4.30

ð5Þ

172

4 Solutions of Problems: First-Order Circuits

4.31. The circuit seems to be a third-order circuit; however, it is a first-order RL circuit that needs to be simplified, as is shown in Fig. 4.31.2.
36
Leq ¼ 3
6 þ 3 ¼ þ3¼2þ3¼5H 3þ6

ð1Þ

The circuit for t ¼ 0 is illustrated in Fig. 4.31.3, where the inductor has been replaced by a short-circuit branch, since the circuit is in the steady state condition. Using Ohm’s law for the 3 Ω resistor: I L ð 0 Þ ¼

15 ¼5A 3

ð2Þ

Due to the current continuity of the inductor in lack of a power source with an infinite size (e.g., impulse power source): I L ð 0þ Þ ¼ I L ð 0 Þ ¼ 5 A

ð3Þ

The final current of the inductor is zero because the circuit lacks a power source, as can be seen in Fig. 4.31.4. Therefore: I L ð1Þ ¼ 0 A

ð4Þ

Moreover, the Thevenin resistance of the circuit is: RTh ¼ 3 þ 7 ¼ 10 Ω

ð5Þ

The time constant of the circuit: τ¼

Leq 5 ¼ ¼ 0:5 sec RTh 10

ð6Þ

Using the general form of the current of an inductive first-order circuit with DC power supply: I L ðt Þ ¼ I L ð1Þ þ ðI L ð0þ Þ  I L ð1ÞÞeτ ¼ 0 þ ð5  0Þe0:5 ¼ 5e2t t

From the circuit: I ðt Þ ¼ I L ðt Þ ¼ 5e2t Choice (2) is the answer.

t

ð7Þ

4

Solutions of Problems: First‐Order Circuits

173

Fig. 4.31 The circuits of solution of problem 4.31

4.32. The energy of a capacitor will not change if its voltage remains constant, as can be noticed from (1): 1 W C ðt Þ ¼ C ðV C ðt ÞÞ2 2

ð1Þ

V C ð 0 Þ ¼ 4 V

ð2Þ

Based on the given information:

Due to the voltage continuity of the capacitor in lack of a power source with an infinite size (e.g., impulse power source): V C ð 0þ Þ ¼ V C ð 0 Þ ¼ 4 V

ð3Þ

To have a constant energy in the capacitor: Using ð3Þ V C ð 1 Þ ¼ V C ð 0þ Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) V C ð1Þ ¼ 4 V

ð4Þ

To calculate the final voltage of the capacitor, we need to assume that the capacitor is an open circuit branch, as can be seen in Fig. 4.32.2. This circuit can be analyzed by using the superposition theorem.

174

4 Solutions of Problems: First-Order Circuits

Using the voltage division formula in Fig. 4.32.3: V C1 ð1Þ ¼

4þ2 12 2¼ 4þ2þR 6þR

ð5Þ

Using the current division formula and Ohm’s law in Fig. 4.32.4: V C2 ð1Þ ¼



 4 12R 3 R¼ 4þ2þR 6þR

V C ð1Þ ¼ V C1 ð1Þ þ V C2 ð1Þ ¼

12 12R 12 þ 12R þ ¼ 6þR 6þR 6þR

Solving (4) and (7): 12 þ 12R 3 ¼ 4 ) 12 þ 12R ¼ 24 þ 4R ) 8R ¼ 12 ) R ¼ Ω 6þR 2 Choice (2) is the answer.

Fig. 4.32 The circuits of solution of problem 4.32

ð6Þ ð7Þ

4

Solutions of Problems: First‐Order Circuits

175

4.33. As can be seen in Fig. 4.33.2, since there is no energy stored in the inductor and no power source exists in the circuit for t < 0, we have: I L ð 0 Þ ¼ 0

ð1Þ

Between 0 < t < 1, we have the circuit of Fig. 4.33.3, which is a first-order circuit. Due to the current continuity of the inductor in lack of a power source with an infinite size (e.g., impulse power source): I L ð0þ Þ ¼ I L ð0 Þ ¼ 0

ð2Þ

To determine the time constant of the circuit, we need to calculate the equivalent resistance of the circuit seen by the inductor. Thus, we need to turn off the independent current source, as is shown in Fig. 4.33.4. Req ¼ 0:25 þ 0:5 ¼ 0:75 Ω

ð3Þ

The time constant of an inductive first-order circuit is: τ¼

L 0:5 2 ¼ sec ¼ Req 0:75 3

ð4Þ

Now, we need to determine the final current of the inductor (see Fig. 4.33.5), assuming that the configuration of the circuit will remain the same. As we know, the inductor will behave like a short-circuit branch in its steady state condition, as can be seen in Fig. 4.33.5. Therefore, by using the current division formula, we have: I L ð 1Þ ¼

0:25 6¼2A 0:25 þ 0:5

ð5Þ

Next, by using the general form of the inductor current in a first-order circuit with DC power supply, we can determine the time-dependent current of the inductor. I L ðt Þ ¼ I L ð1Þ þ ðI L ð0þ Þ  I L ð1ÞÞeτ ) I L ðt Þ ¼ 2 þ ð0  2Þe t

2t 3

  3 ¼ 2 1  e  2t A

ð6Þ

Since the switch is closed at t ¼ 1 sec, we need to determine the current of the inductor after the switching operation. Using the current equation of the inductor:   3 I L ð1 Þ ¼ 2 1  e2 ¼ 1:55 A

ð7Þ

The current of the inductor is contiguous during any short interval, therefore: I L ð1þ Þ ¼ I L ð1 Þ ¼ 1:55 A

ð8Þ

After the switching operation at t ¼ 1, the inductor is parallel to a short-circuit branch, as is shown in Fig. 4.33.6. By using the voltage-current relation of the inductor, we can find out that the current of the inductor will remain constant for t > 1.   1 I L ðt Þ ¼ I L t  0 þL

Zt V L ðt Þ:dt ) I L ðt Þ ¼ 1:55 þ t 0

1 0:5

Zt 0:dt ) I L ðt Þ ¼ 1:55 A for t > 1 1

ð9Þ

176

4 Solutions of Problems: First-Order Circuits

Now, by applying KCL in the indicated supernode at t ¼ 4: 

0:25  6 þ I L ðt ¼ 4Þ þ I sc ðt ¼ 4Þ ¼ 0 )  2 þ 1:55 þ I sc ðt ¼ 4Þ ¼ 0 0:25 þ 0:5 ) I sc ðt ¼ 4Þ ¼ 0:45 A

Choice (1) is the answer.

Fig. 4.33 The circuits of solution of problem 4.33

4.34. The voltage equation of the capacitor is as follows: V ðt Þ ¼

 1 1  e3t uðt Þ 4

ð1Þ

By comparing (1) and the general form of the capacitor voltage in a first-order circuit with DC power supply (see (2)), we can extract the time constant of the network, as is presented in (3): V C ðt Þ ¼ V C ð1Þ þ ðV C ð0þ Þ  V C ð1ÞÞeτ t

ð2Þ

4

Solutions of Problems: First‐Order Circuits

177

)τ¼

1 sec 3

ð3Þ

On the other hand, we know that the general form of time constant of a capacitive first-order circuit is: τ ¼ RTh C

ð4Þ

Solving (3) and (4) and using C ¼ 1 F: RTh  1 ¼

1 1 ) RTh ¼ Ω 3 3

ð5Þ

Moreover, since the capacitor will behave like an open circuit after a long time, the final voltage of the capacitor will be equal to the Thevenin voltage of the circuit, as can be noticed from Fig. 4.34.3.  1 1 1  e3t uðt Þ ¼ V t!1 4 4

V Th ¼ V C ð1Þ ¼ lim

ð6Þ

Now, we know the parameters of the Thevenin equivalent circuit of the network (RTh and VTh) connected to the 2 H inductor. This circuit is an inductive first-order circuit, which is shown in Fig. 4.34.4. Calculating the time constant of the circuit: τ¼

L 2 ¼ ¼ 6 sec RTh 13

ð7Þ

The general form of the current of an inductive first-order circuit with DC power supply is: I L ðt Þ ¼ I L ð1Þ þ ðI L ð0þ Þ  I L ð1ÞÞeτ t

ð8Þ

The current of the inductor before connecting to the circuit was zero (IL(0) ¼ 0 A). Since the current of the inductor must be continuous in lack of a power source with an infinite size (e.g., impulse power source), its current, after the connection, will not change. Hence: I L ð 0þ Þ ¼ I L ð 0 Þ ¼ 0 A

ð9Þ

Moreover, after a long time, the inductor will behave like a short circuit. Therefore, its final current can be calculated as follows: I L ð 1Þ ¼

V Th 14 3 ¼ ¼ A RTh 13 4

ð10Þ

Using (7), (8), (9), and (10):   t   t 3 3 3 1  e 6 I L ð t Þ ¼ þ 0  e6 ¼ 4 4 4

ð11Þ

We know that the voltage of an inductor can be calculated by using V L ðt Þ ¼ L dIdtL ðtÞ. Thus: V L ðt Þ ¼ 2 

     t d 3 3 1 t 1 t 1  e 6 ¼ 2   e 6 ¼ e 6 V dt 4 4 6 4

Since (12) is credible for t > 0, we need to use a unit step function (u(t)) in (12), as follows: 1 t V L ð t Þ ¼ e  6 uð t Þ V 4 Choice (1) is the answer.

ð12Þ

178

4 Solutions of Problems: First-Order Circuits

Fig. 4.34 The circuits of solution of problem 4.34

4.35. To solve this problem, we need to write the differential equation of the circuit. As is given in Fig. 4.35.2, the input voltage is a sign wave. V s ðt Þ ¼ 10 sin ð2t Þ

ð1Þ

To formulate the differential equation of the problem, we can write KVL in the mesh of the circuit of Fig. 4.35.3. V s ðt Þ þ L

d I ðt Þ þ RI R ðt Þ ¼ 0 dt L

ð2Þ

From the circuit, it is clear that: I R ðt Þ ¼ I L ðt Þ

ð3Þ

Solving (1), (2), (3), and the given information in the problem: 10 sin ð2t Þ þ 2:5

d d I ðt Þ þ 5I L ðt Þ ¼ 0 ) I L ðt Þ þ 2I L ðt Þ ¼ 4 sin ð2t Þ dt L dt

ð4Þ

Equation (4) is a first-order differential equation that has the solution of: I L ðt Þ ¼ sin ð2t Þ  cos ð2t Þ þ e2t

ð5Þ

Equation (5) is valid just for the period of 0, π2 . After t ¼ π2 sec , the voltage source is turned off, and the circuit is converted to the one shown in Fig. 4.35.4. Now, we need to determine the primary current of the inductor.   π ¼ sin ðπ Þ  cos ðπ Þ þ eπ ¼ 0  ð1Þ þ 0:0432 ¼ 1:0432 A IL t ¼ 2

ð6Þ

Due to the current continuity of the inductor in lack of a power source with an infinite size (e.g., impulse power source):     πþ π ¼ 1:0432 A IL t ¼ ¼ IL t ¼ 2 2

ð7Þ

Moreover, the final current of the inductor will be zero, since there is no power source in the circuit. In other words, the whole energy of the inductor will be wasted in the resistor.

4

Solutions of Problems: First‐Order Circuits

179

I L ð 1Þ ¼ 0

ð8Þ

In addition, the time constant of the circuit is: τ¼

L 2:5 ¼ ¼ 0:5 sec R 5

ð9Þ

Using the general form of the inductor current in a first-order circuit with DC power supply for the circuit of Fig. 4.35.4 (π2 < t < 1):   þ  tπ2 tπ 2 π I L ð t Þ ¼ I L ð 1Þ þ I L  I L ð1Þ e τ ¼ 0 þ ð1:0432  0Þe 0:5 2 tπ 2

) I L ðt Þ ¼ 1:0432e 0:5 for

π 0. Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ○ Normal ● Large 1) 2 + 6et + 4e2t 2) 2 + 8et  6e2t 3) 2  6et + 6e2t 4) 2  6et + 8e2t

208

5 Problems: Second-Order and Higher-Order Circuits

Fig. 5.28 The circuit of problem 5.28

5.29. In the circuit of Fig. 5.29, after closing the switch at t ¼ 0, a critically damped general response is achieved for the circuit. Determine the value of IL(t) after the switching operation. Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ○ Normal ● Large 1) (0.4 + 2.25t)e500tu(t) A 2) (0.4 + 0.225t)e200tu(t) A 3) (0.4 + 225t)e1000tu(t) A 4) (0.4 + 225t)e500tu(t) A

Fig. 5.29 The circuit of problem 5.29

5.30. In the circuit of Fig. 5.30, the switch has been open for a long time. Calculate the time-dependent equation of the current of the switch (I ), if it is closed at t ¼ 0. Difficulty level ○ Easy ○ Normal ● Hard Computation amount ○ Small ○ Normal ● Large 1) 2) 3) 4)

I ðt Þ ¼ 10  2e4  4 sin ð2t Þ A I(t) ¼ 8 + 2e4t + 4 sin (2t) A I(t) ¼ 10  2e4t  4 sin (2t) A I(t) ¼ 10  2e4t + 4 sin (2t) A t

Fig. 5.30 The circuit of problem 5.30

6

Solutions of Problems: Second-Order and Higher-Order Circuits

Abstract

In this chapter, the problems of the fifth chapter are fully solved, in detail and step-by-step. In solving the problems of this chapter, all the subjects of the previous chapters are applied. Moreover, the concepts of quality factor, circuit responses, impulse function, time-dependent second-order homogeneous and nonhomogeneous differential equations, characteristic equation, and general solution of a second-order differential equations are explained. 6.1. To analyze a circuit in the steady state condition, we need to replace the capacitors and inductors by the open circuit and short-circuit branches, respectively, as can be seen in Fig. 6.1.2. By simplifying the circuit of Fig. 6.1.2, we can get the circuit shown in Fig. 6.1.3. To determine the output voltage, we can use the superposition theorem. Vo ¼

1 1  10 þ 41¼5þ2¼7V 1þ1 1þ1

Choice (2) is the answer.

Fig. 6.1 The circuits of solution of problem 6.1 # Springer Nature Switzerland AG 2020 M. Rahmani-Andebili, DC Electrical Circuit Analysis, https://doi.org/10.1007/978-3-030-50711-4_6

209

210

6

Solutions of Problems: Second-Order and Higher-Order Circuits

6.2. To calculate the energy stored in the energy-saving components of the circuit, we need to analyze the circuit in the steady state condition. As can be seen in Fig. 6.2.2, in the steady state condition, the capacitor and inductor must be replaced with the open circuit and short-circuit branches, respectively. Since the capacitor is an open circuit branch, the whole current of the current source flows through the inductor. Therefore: IL ¼ 2 A

ð1Þ

V C ¼ 10  2 ¼ 20 V

ð2Þ

1 1 W L ¼ LI L 2 ¼  5  103  22 ¼ 102 J ¼ 10 mJ 2 2

ð3Þ

Using Ohm’s law:

Energy stored in the inductor:

Energy stored in the capacitor: 1 1 W C ¼ CV C 2 ¼  25  106  202 ¼ 5  103 J ¼ 5 mJ 2 2

ð4Þ

Total energy stored in the circuit: W ¼ W L þ W C ¼ 10 mJ þ 5 mJ ¼ 15 mJ Choice (3) is the answer.

Fig. 6.2 The circuits of solution of problem 6.2

6.3. Figure 6.3.2 shows the circuit for t ¼ 0+. As can be seen, the capacitor has been replaced by a short-circuit branch, since it has the primary voltage of zero. Moreover, the inductor has been replaced by an open circuit branch because it has the primary current of zero. Based on the given information: d V ð0þ Þ ¼ 10 V= sec dt C Using the voltage-current relation of the capacitor:

ð1Þ

6

Solutions of Problems: Second‐Order and Higher‐Order Circuits

I C ð 0þ Þ ¼ C

211

Using ð1Þ d V C ð 0þ Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) I C ð0þ Þ ¼ C  10 ¼ 10C dt

ð2Þ

Applying KVL in the indicated loop: 10 þ 2I C ð0þ Þ ¼ 0 ) I C ð0þ Þ ¼ 5 A

ð3Þ

Solving (2) and (3): 10C ¼ 5 ) C ¼

1 F 2

Choice (2) is the answer.

Fig. 6.3 The circuits of solution of problem 6.3

6.4. Since the voltage source is an impulse function, the inductor and capacitor will behave like an open circuit and short circuit, respectively, as is shown in Fig. 6.4.2. Based on the given information: V s ðt Þ ¼ δðt Þ

ð1Þ

2 I R ðt Þ ¼ δ ðt Þ 3

ð2Þ

212

6

Solutions of Problems: Second-Order and Higher-Order Circuits

Applying KVL in the circuit: V s ðt Þ þ RI R ðt Þ þ 1  I R ðt Þ ¼ 0 ) I R ðt Þ ¼ )

V s ðt Þ Using ð1Þ, ð2Þ 2 δ ðt Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) δðt Þ ¼ Rþ1 Rþ1 3

2 1 1 ¼ ) 2R þ 2 ¼ 3 ) R ¼ Ω 3 Rþ1 2

Choice (3) is correct.

Fig. 6.4 The circuits of solution of problem 6.4

6.5. First, we need to analyze the circuit for t < 0, shown in the circuit of Fig. 6.5.2, where the inductor has been replaced by a short-circuit branch. Using Ohm’s law: I L ð 0 Þ ¼

32 ¼4A 8

ð1Þ

Moreover, since the capacitor is not connected to a power source, its primary voltage is zero. V C ð 0 Þ ¼ 0

ð2Þ

Due to the continuity of the capacitor voltage and inductor current in lack of power source with an infinite size (e.g., impulse power source), we can conclude that: I L ð 0þ Þ ¼ I L ð 0 Þ ¼ 4 A

ð3Þ

V C ð 0þ Þ ¼ V C ð 0 Þ ¼ 0 V

ð4Þ

To analyze the circuit at t ¼ 0+, we can model the capacitor and inductor by using a voltage source (with the size of 0 V) and a current source (with the size of 4 A), respectively, as is shown in Fig. 6.5.3. Now, we can calculate the voltage of the inductor by applying KVL in the loop: V L ð0þ Þ  8  4  12  4 ¼ 0 ) V L ð0þ Þ ¼ 80 V Choice (3) is the answer.

6

Solutions of Problems: Second‐Order and Higher‐Order Circuits

213

Fig. 6.5 The circuits of solution of problem 6.5

6.6. First, we need to analyze the circuit for t ¼ 0. Since the circuit has reached its steady state, the capacitor and inductor are open circuit and short circuit, respectively, as can be seen in Fig. 6.6.3. Herein, the 6 Ω resistor is eliminated because its two ends are short-circuited. By using Ohm’s law for the whole circuit, we have: I L ð 0 Þ ¼

12 ¼ 3 A 4þ0

ð1Þ

Using Ohm’s law for the 4 Ω resistor: V C ð0 Þ ¼ 4  ð3Þ ¼ 12 V

ð2Þ

Due to the current continuity of the inductor: I L ð0þ Þ ¼ I L ð0 Þ ¼ 3 A

ð3Þ

Due to the voltage continuity of the capacitor: V C ð0þ Þ ¼ V C ð0 Þ ¼ 12 V

ð4Þ

The circuit for t ¼ 0+ is shown in Fig. 6.6.4, where the primary voltage of the capacitor and the initial current of the inductor have been modelled by using 12 V voltage source and 3 A current source, respectively. By applying KVL in the indicated mesh, we have: 12 þ ð12Þ þ V L ð0þ Þ ¼ 0 ) V L ð0þ Þ ¼ 24 V

ð5Þ

214

6

Solutions of Problems: Second-Order and Higher-Order Circuits

KCL in the indicated supernode: I C ð0þ Þ 

Using ð5Þ V ð0þ Þ 12 24 ¼0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)  I C ð0þ Þ þ 3  3 þ þ ð3Þ þ L ¼0 6 4 6 )  I C ð 0þ Þ þ 4 ¼ 0 ) I C ð 0þ Þ ¼ 4 A

Choice (2) is the answer.

Fig. 6.6 The circuits of solution of problem 6.6

ð6Þ

6

Solutions of Problems: Second‐Order and Higher‐Order Circuits

215

6.7. As can be seen in Fig. 6.7.2 that shows the circuit for t ¼ 0, the whole current of the current source passes through the inductor; therefore: I L ð 0 Þ ¼ 4 A

ð1Þ

V C ð0 Þ þ 1  4 ¼ 0 ) V C ð0 Þ ¼ 4 V

ð2Þ

KVL in the left-side mesh:

Due to the current continuity of the inductor in lack of a power source with an infinite size (e.g., impulse power source): I L ð 0þ Þ ¼ I L ð 0 Þ ¼ 4 A

ð3Þ

Due to the voltage continuity of the capacitor in lack of a power source with an infinite size (e.g., impulse power source): V C ð 0þ Þ ¼ V C ð 0 Þ ¼ 4 V

ð4Þ

Figure 6.7.3 shows the circuit for t ¼ 0+, where the 4 V voltage source and the 4 A current source are applied to show the primary voltage of the capacitor and the initial current of the inductor, respectively. KCL in the indicated supernode: 4  12 þ I C ð0þ Þ  4 ¼ 0 ) I C ð0þ Þ ¼ 12

ð5Þ

Using ð5Þ V ðt Þ þ I C ð0þ Þ  1 þ 4 ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)  V ðt Þ þ 12  1 þ 4 ¼ 0 ) V ðt Þ ¼ 16 V

ð6Þ

KVL in loop 1:

KVL in loop 2: Using ð3Þ, ð6Þ V ðt Þ þ V L ð0þ Þ þ I L ð0þ Þ  1 ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)  16 þ V L ð0þ Þ þ 4  1 ¼ 0 ) V L ð0þ Þ ¼ 12 V Using the current-voltage relation of the inductor: V L ð 0þ Þ ¼ L

V ð0þ Þ d d d I L ð 0þ Þ ¼ 3 I L ð 0þ Þ ) I L ð 0þ Þ ¼ L 3 dt dt dt

Using ð7Þ d 12 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) I L ð0þ Þ ¼ ¼ 4 A= sec dt 3 Choice (3) is the answer.

ð7Þ

216

6

Solutions of Problems: Second-Order and Higher-Order Circuits

Fig. 6.7 The circuits of solution of problem 6.7

6

Solutions of Problems: Second‐Order and Higher‐Order Circuits

217

6.8. Figure 6.8.2 illustrates the circuit for t ¼ 0. As can be seen, since the power source of the circuit is short-circuited, the initial voltage of the capacitor and the primary current of the inductor are zero. In other words: I L ð 0 Þ ¼ 0 A

ð1Þ

V C ð 0 Þ ¼ 0 V

ð2Þ

Due to the current continuity of the inductor in lack of a power source with an infinite size (e.g., impulse power source): I L ð 0þ Þ ¼ I L ð 0 Þ ¼ 0 A

ð3Þ

Due to the voltage continuity of the capacitor in lack of a power source with an infinite size (e.g., impulse power source): V C ð 0þ Þ ¼ V C ð 0 Þ ¼ 0 V

ð4Þ

Therefore, the capacitor and the inductor need to be modeled by a short-circuit branch and an open circuit branch for t ¼ 0+, respectively, as is shown in Fig. 6.8.3. In Fig. 6.8.3, the whole current of the current source flows through the left-side 1 Ω resistor. Therefore: V 1 ð 0þ Þ ¼ 3  1 ¼ 3 V

ð5Þ

V 1 ð0þ Þ þ V L ð0þ Þ þ 0 ¼ 0 ) V L ð0þ Þ ¼ V 1 ð0þ Þ ¼ 3 V

ð6Þ

KVL in the indicated loop of Fig. 6.8.3:

The current-voltage relation of the inductor: V L ð0þ Þ ¼ L

Using ð6Þ d d d I L ð0þ Þ ¼ 1  I L ð0þ Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) I L ð0þ Þ ¼ 3 A= sec dt dt dt

The circuit for t > 0 is shown in Fig. 6.8.4. By applying KCL in node “A,” we have: 3 þ

d V 1 ðt Þ dt d d d d þ I L ðt Þ ¼ 0 ) V ðt Þ þ I L ðt Þ ¼ 0 ) V 1 ðt Þ ¼  I L ðt Þ ¼  3 V=sec 1 dt 1 dt dt dt

Choice (4) is the answer.

ð7Þ

218

6

Solutions of Problems: Second-Order and Higher-Order Circuits

Fig. 6.8 The circuits of solution of problem 6.8

6

Solutions of Problems: Second‐Order and Higher‐Order Circuits

219

6.9. Based on the given information: V C ð0Þ ¼ 6 V

ð1Þ

I L ð0Þ ¼ 5 A

ð2Þ

Due to the continuity of the voltage of the capacitor and the current of the inductor: V C ð0þ Þ ¼ V C ð0Þ ¼ 6 V

ð3Þ

I L ð 0þ Þ ¼ I L ð 0Þ ¼ 5 A

ð4Þ

KVL in the indicated mesh of the circuit of Fig. 6.9.2: I L ðt Þ  1 þ V L ðt Þ þ V C ðt Þ ¼ 0

ð5Þ

The current-voltage relation of the inductor: V L ðt Þ ¼ L

d d I ðt Þ ¼ 3 I L ðt Þ dt L dt

Solving (5) and (6) for t ¼ 0+: I L ð 0þ Þ þ 3

Using ð3Þ, ð4Þ d d I L ð 0þ Þ þ V C ð 0þ Þ ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 5 þ 3 I L ð0þ Þ þ ð6Þ ¼ 0 dt dt )

d 1 I ð0þ Þ ¼ A= sec dt L 3

Choice (1) is the answer.

Fig. 6.9 The circuits of solution of problem 6.9

ð6Þ

220

6

Solutions of Problems: Second-Order and Higher-Order Circuits

6.10. To determine the behavior of a second-order circuit, its characteristic equation must be known. Therefore, first, we need to formulate the differential equation of the circuit. Herein, we can write the differential equation based on the voltage of the capacitor or the current of the inductor. Applying KVL in the indicated mesh of the circuit of Fig. 6.10.2: V L ðt Þ þ 1:5I L ðt Þ þ V C ðt Þ ¼ 0 ) V C ðt Þ ¼ V L ðt Þ  1:5I L ðt Þ

ð1Þ

Applying KCL in the supernode: I L ðt Þ þ

V L ðt Þ V ðt Þ þ I C ðt Þ þ C ¼ 0 1 2

ð2Þ

The voltage-current relation of the capacitor:

I C ðt Þ ¼ C

d d d V ðt Þ ¼ 1  V C ðt Þ ¼ V C ðt Þ dt C dt dt

ð3Þ

Applying (3) in (2): I L ðt Þ þ V L ðt Þ þ

V ðt Þ d V ðt Þ þ C ¼ 0 2 dt C

ð4Þ

Using ð1Þ d 1 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) I L ðt Þ þ V L ðt Þ þ ðV L ðt Þ  1:5I L ðt ÞÞ þ ðV L ðt Þ  1:5I L ðt ÞÞ ¼ 0 dt 2 )

1 3 d 3 d I ðt Þ þ V L ðt Þ þ V L ðt Þ  I ðt Þ ¼ 0 4 L 2 dt 2 dt L

ð5Þ

The current-voltage relation of the inductor: V L ðt Þ ¼ L

d d d I ðt Þ ¼ 1  I L ðt Þ ¼ I L ðt Þ dt L dt dt

ð6Þ

Solving (5) and (6):

 1 3 d d d 3 d d2 1 I L ðt Þ þ I L ðt Þ þ I L ðt Þ  I L ðt Þ ¼ 0 ) 2 I L ðt Þ þ I L ðt Þ ¼ 0 4 2 dt dt dt 2 dt 4 dt

ð7Þ

The characteristic equation of (7) and its roots are: 1 1 s2 þ ¼ 0 ) s1,2 ¼  j 4 2

ð8Þ

Since the characteristic equation has pure imaginary roots, the circuit has undamped or oscillating behavior. Choice (2) is correct.

6

Solutions of Problems: Second‐Order and Higher‐Order Circuits

221

Fig. 6.10 The circuits of solution of problem 6.10

6.11. First, we need to analyze the circuit for t ¼ 0, as is shown in Fig. 6.11.2. Since the circuit has reached its steady state condition, the capacitor is open circuit and the inductor is short circuit. V C ð0 Þ þ 1  1  V C ð0 Þ ¼ 0 ) V C ð0 Þ ¼

1 V 2

ð1Þ

As can be seen in the circuit, the whole current of the current source flows through the inductor. I L ð 0 Þ ¼ 1 A

ð2Þ

Due to the continuity of voltage of the capacitor and the current of the inductor: V C ð 0þ Þ ¼ V C ð 0 Þ ¼

1 V 2

I L ð 0þ Þ ¼ I L ð 0 Þ ¼ 1 A

ð3Þ ð4Þ

Figure 6.11.3 illustrates the circuit for t ¼ 0+, where the primary voltage of the capacitor and the primary current of the inductor are modeled by the voltage and current sources, respectively. Now, by applying KVL in the indicate mesh, we have: V C ð0þ Þ þ ð1  I C ð0þ ÞÞ  1  V C ð0þ Þ þ V L ð0þ Þ ¼ 0   Using ð3Þ 1 )  2V C ð0þ Þ þ 1  I C ð0þ Þ þ V L ð0þ Þ ¼ 0 ¼ þ 1  I C ð 0þ Þ þ V L ð 0þ Þ ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)  2 2 ) V L ð 0þ Þ ¼ I C ð 0þ Þ

ð5Þ

222

6

Solutions of Problems: Second-Order and Higher-Order Circuits

KCL in the supernode: 1 þ I C ð0þ Þ þ 1 þ

V L ð0þ Þ  10 ¼ 0 ) I C ð0þ Þ þ V L ð0þ Þ  10 ¼ 0 1

ð6Þ

Solving (5) and (6): I C ð0þ Þ þ I C ð0þ Þ  10 ¼ 0 ) I C ð0þ Þ ¼ 5 V The voltage-current relation of the capacitor: I C ðt Þ ¼ C

Using ð7Þ d d d d V C ðt Þ ¼ 1  V C ðt Þ ) V C ðt Þ ¼ I C ð0þ Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) V C ðt Þ ¼ 5 V= sec dt dt dt dt

Choice (4) is correct.

Fig. 6.11 The circuits of solution of problem 6.11

ð7Þ

6

Solutions of Problems: Second‐Order and Higher‐Order Circuits

223

6.12. We need to define all the variables of the differential equations of the circuit based on IR(t). KVL in the left-side mesh of the circuit of Fig. 6.12.2: V S ðt Þ þ RI R ðt Þ þ V L ðt Þ ¼ 0

ð1Þ

Using the current-voltage relation of the inductor: V L ðt Þ ¼ L

d I ðt Þ dt L

ð2Þ

KCL in the supernode: I R ðt Þ þ I L ðt Þ þ 2I R ðt Þ  I S ðt Þ ¼ 0 ) I L ðt Þ ¼ I R ðt Þ þ I S ðt Þ Solving (1) and (2): V S ðt Þ þ RI R ðt Þ þ L

Using ð3Þ d d ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)  V S ðt Þ þ RI R ðt Þ þ L ðI R ðt Þ þ I S ðt ÞÞ ¼ 0 I L ðt Þ ¼ 0 ¼ dt dt )  V S ðt Þ þ RI R ðt Þ  L )L

d d I ðt Þ þ L I S ðt Þ ¼ 0 dt R dt

d d I ðt Þ  RI R ðt Þ ¼ L I S ðt Þ  V S ðt Þ dt R dt

Choice (1) is correct.

Fig. 6.12 The circuits of solution of problem 6.12

ð3Þ

224

6

Solutions of Problems: Second-Order and Higher-Order Circuits

6.13. To calculate the energy of a current source, its time-dependent voltage equation must be known. KVL in the loop of the circuit of Fig. 6.13.2: V S ðt Þ þ V R ðt Þ þ V L ðt Þ þ V C ðt Þ ¼ 0

ð1Þ

The current-voltage relation of the resistor: V R ðt Þ ¼ RI R ðt Þ ¼ RI S ðt Þ ¼ 1:5  10t ¼ 15t V

ð2Þ

The current-voltage relation of the inductor: V L ðt Þ ¼ L

d d d I ðt Þ ¼ L I S ðt Þ ¼ 1  ð10t Þ ¼ 10 V dt L dt dt

ð3Þ

The current-voltage relation of the capacitor:

V C ðt Þ ¼ V C ð0Þ þ

1 C

ðt I C ðt Þ:dt ¼ V C ð0Þ þ 0




V C ðt Þ ¼ 20 þ 10 10

1 C

 2

t 2

ðt I S ðt Þ:dt ¼ 20 þ

1 0:1

0

ðt 10t:dt ¼ 0

¼ 20 þ 50t 2 V

ð4Þ

Applying (2), (3), and (4) in (1): V S ðt Þ þ 15t þ 10 þ 20 þ 50t 2 ¼ 0 ) V S ðt Þ ¼ 15t þ 30 þ 50t 2 Now, we can calculate the energy of the current source for the given interval, that is, [0, 1]: ðt

ð1

W S ¼ V S ðt ÞI S ðt Þ:dt ¼ 0

  15t þ 30 þ 50t 2  10t:dt

0

ð1

 1 W S ¼ 150t 2 þ 300t þ 500t 3 :dt ¼ 50t 3 þ 150t 2 þ 125t 4 0 0

W S ¼ ð50 þ 150 þ 125Þ  ð0 þ 0 þ 0Þ ¼ 325 J Choice (2) is correct.

Fig. 6.13 The circuits of solution of problem 6.13

ð5Þ

6

Solutions of Problems: Second‐Order and Higher‐Order Circuits

225

6.14. Before the switching operation, the circuit is in the steady state situation. Therefore, the inductors can be replaced by the short-circuit branches, as can be seen in Fig. 6.14.2. To calculate the inductors’ currents before opening the switch, we can use the superposition theorem in Fig. 6.14.2. I L1 ðt ¼ 0 Þ ¼

R 12 R V 12 V 0 12 þ V 0   0 ¼ þ þ ¼ R þ R R þ R2 R þ R R þ R2 3R 3R 3R

ð1Þ

R 12 V 12 V 0 12  2V 0    0 ¼ ¼ R þ R R þ R2 R þ R2 3R 32 R 3R

ð2Þ

I L2 ðt ¼ 0 Þ ¼

From the circuit of Fig. 6.14.3, it is clear that right after the switching (t ¼ 0+), the inductors’ currents are equal in magnitude but opposite in direction. I L1 ðt ¼ 0þ Þ ¼ I L2 ðt ¼ 0þ Þ

ð3Þ

To avoid having any impulse voltage across the switch, the abovementioned relation between them must be held before the switching (t ¼ 0) as well. Therefore: I L1 ðt ¼ 0 Þ ¼ I L2 ðt ¼ 0 Þ Solving (1), (2), and (4): 12 þ V 0 12  2V 0 ¼ ¼ 0 ) 12 þ V 0 ¼ 12 þ 2V 0 ) V 0 ¼ 24 V 3R 3R Choice (4) is the answer.

Fig. 6.14 The circuits of solution of problem 6.14

ð4Þ

226

6

Solutions of Problems: Second-Order and Higher-Order Circuits

6.15. Before the switching operation, since the circuit is in the steady state condition, the capacitor and inductors behave like an open circuit branch and short-circuit branches, respectively, as is shown in Fig. 6.15.2. By analyzing the circuit, we can calculate the voltage of the capacitor and the current of each inductor. By simplifying the circuit of Fig. 6.15.2 and showing in Fig. 6.15.3, the total current of the circuit is: I ð 0 Þ ¼

18 ¼6A 1þ2

ð1Þ

The current division for the inductors (see Fig. 6.15.2): I L1 ð0 Þ ¼

2 6¼4A 1þ2

ð2Þ

I L2 ð0 Þ ¼

1 6¼2A 1þ2

ð3Þ

Using (1) and Ohm’s law for the 2 Ω resistor in the circuit of Fig. 6.15.2: V C ð0 Þ ¼ 2I ð0 Þ ¼ 2  6 ¼ 12 V

ð4Þ

Since there is no power source with an infinite size (e.g., impulse power source) in the circuit, the voltage of the capacitor and the current of each inductor will continue during the switching operation. Hence: I L1 ð0þ Þ ¼ I L1 ð0 Þ ¼ 4 A

ð5Þ

I L2 ð0þ Þ ¼ I L2 ð0 Þ ¼ 2 A

ð6Þ

V C ð0þ Þ ¼ V C ð0 Þ ¼ 12 V

ð7Þ

For a very short interval, that is, [0, 0+], we can model the capacitor and the inductors by using a voltage source and current sources, as is illustrated in Fig. 6.15.4. Now, by applying KVL in the indicated mesh, we can calculate the voltage of the switch right after the switching operation. V S ð0þ Þ ¼ 2  4 þ 12 ¼ 4 V Choice (3) is the answer.

6

Solutions of Problems: Second‐Order and Higher‐Order Circuits

227

Fig. 6.15 The circuits of solution of problem 6.15

6.16. Due to the presence of the voltage source with an infinite size (Vs(t) ¼ δ(t)) in the circuit, for the interval of [0, 0+], the capacitors and inductors will behave like the short-circuit and open circuit branches, respectively, as can be seen in Fig. 6.16.2. The requested output is the current of 2 H inductor. To calculate its current, we need to use the voltagecurrent relation of the inductor for that inductor as follows:

I L ðt Þ ¼

1 L

ðt 1

V L ðt Þ:dt ) I 2H ð0þ Þ ¼

8 0 0. The components are in parallel; thus: V C ðt Þ ¼ V L ðt Þ

ð1Þ

The voltage-current relation of the capacitor: I C ðt Þ ¼ C

Using ð1Þ d d d V C ðt Þ ¼ 200  106 V C ðt Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) I C ðt Þ ¼ 2  104 V L ðt Þ dt dt dt

ð2Þ

The current-voltage relation of the inductor: V L ðt Þ ¼ L

d I ðt Þ dt L

ð3Þ

1 V ðt Þ ¼ 0 20 L

ð4Þ

KCL in the supernode: I C ðt Þ þ I L ðt Þ þ Applying (2) in (4): 2  104

d 1 V ðt Þ þ I L ðt Þ þ V L ðt Þ ¼ 0 dt L 20



 Using ð3Þ d d 1 d ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 2  104 L I L ðt Þ þ I L ðt Þ þ L I L ðt Þ ¼ 0 dt dt 20 dt ) 2  104  L  )

d2 L d I ðt Þ ¼ 0 I L ðt Þ þ I L ðt Þ þ 2 20 dt L dt

d2 d 5000 I ðt Þ ¼ 0 I ðt Þ þ 250 I L ðt Þ þ dt L L dt 2 L

ð5Þ

Hence, the characteristic equation of the circuit is: s2 þ 250s þ

5000 ¼0 L

ð6Þ

Based on the given information, the response of the circuit is critically damped; thus, the delta (discriminant) of the quadratic characteristic equation is zero. Δ¼0

ð7Þ

Solving (7) and (6): ð250Þ2  4ð1Þ Choice (4) is correct.



 5000 4  5000 ¼0)L¼ ¼ 0:32 H L 2502

ð8Þ

6

Solutions of Problems: Second‐Order and Higher‐Order Circuits

235

Fig. 6.20 The circuits of solution of problem 6.20

6.21. First, we need to analyze the circuit for t ¼ 0. Since the circuit is in the steady state condition, the capacitor and the inductors are open circuit branch and short-circuit branches, respectively, as is shown in Fig. 6.21.2. Using Ohm’s law for the whole circuit: I L1 ð0 Þ ¼ I L2 ð0 Þ ¼

12 12 ¼ ¼1A 3 þ 2 þ 6 þ 1 12

ð1Þ

Using Ohm’s law for the 3 Ω resistor: V C ð0 Þ ¼ 3I L1 ð0 Þ ¼ 3  1 ¼ 3 V

ð2Þ

Due to the current continuity of the inductor in lack of a power source with an infinite size (e.g., impulse power source): I L1 ð0þ Þ ¼ I L1 ð0 Þ ¼ 1 A

ð3Þ

Due to the voltage continuity of the capacitor in lack of a power source with an infinite size (e.g., impulse power source):

236

6

Solutions of Problems: Second-Order and Higher-Order Circuits

V C ð 0þ Þ ¼ V C ð 0 Þ ¼ 3 V

ð4Þ

For t ¼ 0+, the switch is closed and the circuit is updated, as is illustrated in Fig. 6.21.3. Applying KVL in the indicated mesh: eð0þ Þ þ 1  1  12 þ 3 ¼ 0 ) eð0þ Þ ¼ 8 V

ð5Þ

To determine the value of dtd eð0þ Þ, we need to analyze the circuit for t > 0, as can be seen in Fig. 6.21.4. Applying KVL in the left-side mesh: eðt Þ þ 1  I L1 ðt Þ  12 þ V C ðt Þ ¼ 0

ð6Þ

for t ¼ 0þ d d d d d d eðt Þ þ I L1 ðt Þ þ V C ðt Þ ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) eð0þ Þ þ I L1 ð0þ Þ þ V C ð0þ Þ ¼ 0 dt dt dt dt dt dt

ð7Þ

Applying derivate operator to (6):

KCL in the supernode for t ¼ 0+: I L1 ð0þ Þ  I C ð0þ Þ 

Using ð3Þ, ð4Þ V C ð 0þ Þ 3 ¼0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 1  I C ð0þ Þ  ¼ 0 ) I C ð0þ Þ ¼ 0 A 3 3

ð8Þ

Using the voltage-current relation of the capacitor: I C ðt Þ ¼ C

for t ¼ 0þ d d d 1 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) I C ð0þ Þ ¼ 2 V C ð0þ Þ ) V C ð0þ Þ ¼ I C ð0þ Þ V C ðt Þ ¼ dt dt dt 2 )

d 1 V ð 0þ Þ ¼  0 ¼ 0 dt C 2

ð9Þ

Applying KVL in the right-side mesh: for t ¼ 0þ eðt Þ  2  I L1 ðt Þ  V L1 ðt Þ ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) eð0þ Þ  2I L1 ð0þ Þ  V L1 ð0þ Þ ¼ 0 Using ð3Þ, ð5Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 8  2ð1Þ  V L ð0þ Þ ¼ 0 ) V L1 ð0þ Þ ¼ 6 V

ð10Þ

Using the voltage-current relation of the capacitor: V L1 ðt Þ ¼ L1

for t ¼ 0þ d d d I L1 ðt Þ ) V L1 ðt Þ ¼ 1  I L1 ðt Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) V L1 ð0þ Þ ¼ I L1 ð0þ Þ dt dt dt

ð11Þ

Solving (10) and (11): d I ð0þ Þ ¼ 6 A= sec dt L1 Solving (7), (9), and (12): d d eð0þ Þ þ 6 þ 0 ¼ 0 ) eð0þ Þ ¼ 6 V= sec dt dt Choice (3) is the answer.

ð12Þ

6

Solutions of Problems: Second‐Order and Higher‐Order Circuits

Fig. 6.21 The circuits of solution of problem 6.21

237

238

6

Solutions of Problems: Second-Order and Higher-Order Circuits

Fig. 6.21 (continued)

6.22. First, we need to calculate the characteristic equation of the circuit. Then, the quality factor of the circuit can be calculated by using (1): Q¼

ω0 2α

ð1Þ

Applying KVL in the left-side mesh of the circuit in Fig. 6.22.2: dt d d d2 d I1 þ V C ¼ 0 ) 2 I1 þ 3 2 I1 þ V C ¼ 0 dt dt dt dt d

2I 1 þ 3

ð2Þ

KCL in node “A”: IC ¼ I1  I2

ð3Þ

The voltage-current relation of the capacitor: IC ¼ C

Using ð3Þ d d d d 1 1 V C ¼ 2 V C ) V C ¼ IC ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) V C ¼ ðI 1  I 2 Þ dt dt dt 2 dt 2

ð4Þ

Solving (2) and (4): 2

d d2 1 d d2 I 1 þ 3 2 I 1 þ ðI 1  I 2 Þ ¼ 0 ) 4 I 1 þ 6 2 I 1 þ I 1  I 2 ¼ 0 dt 2 dt dt dt

ð5Þ

KVL in the right-side mesh of the circuit: d dt

4I 2  V C ¼ 0 ) 4

Using ð4Þ d d d 1 I2  V C ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 4 I 2  ðI 1  I 2 Þ ¼ 0 dt dt dt 2 ) I1 ¼ 8

d I þ I2 dt 2

ð6Þ

6

Solutions of Problems: Second‐Order and Higher‐Order Circuits

239

Solving (5) and (6): 4




 d d d2 d d 8 I2 þ I2 þ 6 2 8 I2 þ I2 þ 8 I2 þ I2  I2 ¼ 0 dt dt dt dt dt ) 32

d2 d d3 d2 d I I þ 4 þ 48 I þ 6 I þ 8 I2 ¼ 0 2 2 2 dt dt dt 2 dt 3 dt 2 2

Ð dt d3 d2 d d2 d ) 48 3 I 2 þ 38 2 I 2 þ 12 I 2 ¼ 0 ) 48 2 I 2 þ 38 I 2 þ 12I 2 ¼ 0 dt dt dt dt dt )

d2 19 d 1 I þ I ¼0 I þ dt 2 2 24 dt 2 4 2

ð7Þ

Therefore, the characteristic equation of the circuit is: s2 þ

19 1 sþ ¼0 24 4

ð8Þ

As we know, the general form of the characteristic equation of a circuit is: s2 þ 2αs þ ω20 ¼ 0

ð9Þ

α¼

19 1 , ω0 ¼ 48 2

ð10Þ

Q¼

1 ω0 12 2 ¼ ¼ 19 2α 24 19

Now, by comparing (8) and (9):

Therefore, by using (1), we have:

Choice (4) is the answer.

Fig. 6.22 The circuits of solution of problem 6.22

240

6

Solutions of Problems: Second-Order and Higher-Order Circuits

6.23. Figure 6.23.2 illustrates the circuit for t > 0. Due to the current continuity of the inductor in lack of a power source with an infinite size (e.g., impulse power source): I L ð 0þ Þ ¼ I L ð 0 Þ ¼ 0 A

ð1Þ

Due to the voltage continuity of the capacitor in lack of a power source with an infinite size (e.g., impulse power source): V C ð 0þ Þ ¼ V C ð 0 Þ ¼ 0 V

ð2Þ

Applying KCL in the indicated node of the circuit of Fig. 6.23.2: I C ðt Þ  I L ðt Þ  2I L ðt Þ ¼ 0 ) I C ðt Þ ¼ 3I L ðt Þ

ð3Þ

Applying KVL in the indicated loop: Using ð3Þ 10 þ V L ðt Þ þ V C ðt Þ þ I C ðt Þ  1 ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)  10 þ V L ðt Þ þ V C ðt Þ þ 3I L ðt Þ ¼ 0 d dt

)

d d d V ðt Þ þ V C ðt Þ þ 3 I L ðt Þ ¼ 0 dt L dt dt

ð4Þ ð5Þ

The current-voltage relation of the inductor: V L ðt Þ ¼ L

d d d I ðt Þ ¼ 1  I L ðt Þ ) V L ðt Þ ¼ I L ðt Þ dt L dt dt

ð6Þ

The voltage-current relation of the capacitor: I C ðt Þ ¼ C

Using ð3Þ d d d d 2 V C ðt Þ ¼ 1:5  V C ðt Þ ) V C ðt Þ ¼ I C ðt Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) V C ðt Þ ¼ 2I L ðt Þ dt dt dt 3 dt

ð7Þ

Solving (5), (6), and (7):
 d d d d2 d I L ðt Þ þ 2I L ðt Þ þ 3 I L ðt Þ ¼ 0 ) 2 I L ðt Þ þ 3 I L ðt Þ þ 2I L ðt Þ ¼ 0 dt dt dt dt dt

ð8Þ

The characteristic equation of the circuit is: s2 þ 3s þ 2 ¼ 0 ) s1,2 ¼ 1,  2

ð9Þ

The general form of the solution of equation (8) is: I L ðt Þ ¼ Aet þ Be2t

ð10Þ

Ae0 þ Be20 ¼ 0 ) A þ B ¼ 0

ð11Þ

Applying (1) in (10):

Solving (4) for t ¼ 0+: Using ð1Þ, ð2Þ 10 þ V L ð0þ Þ þ V C ð0þ Þ þ 3I L ð0þ Þ ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)  10 þ V L ð0þ Þ þ 0 þ 3  0 ¼ 0

6

Solutions of Problems: Second‐Order and Higher‐Order Circuits

) V L ð0þ Þ ¼ 10 V

241

ð12Þ

Solving (6) for t ¼ 0+: V L ð 0þ Þ ¼

Using ð12Þ d d I ð 0þ Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) I L ð0þ Þ ¼ 10 A= sec dt L dt

ð13Þ

Calculating the first derivate of (10): d I ðt Þ ¼ Aet  2Be2t dt L

ð14Þ

10 ¼ Ae0  2Be20 ) 10 ¼ A  2B

ð15Þ

10 ¼ A  2ðAÞ ) A ¼ 10, B ¼ 10

ð16Þ

Applying (13) in (14):

Solving (11) and (15):

Solving (16) and (10):   I L ðt Þ ¼ 10et  10e2t uðt Þ Choice (3) is the answer.

Fig. 6.23 The circuits of solution of problem 6.23

242

6

Solutions of Problems: Second-Order and Higher-Order Circuits

6.24. We need to write the differential equation of the circuit based on the voltage of the capacitor. Figure 6.24.2 illustrates the circuit for t > 0. Applying KCL in node “A”: V C ðt Þ ¼ 0 )  I L ðt Þ þ I C ðt Þ þ 2:5V C ðt Þ ¼ 0 0:4

ð1Þ

d d d I ðt Þ þ I C ðt Þ þ 2:5 V C ðt Þ ¼ 0 dt L dt dt

ð2Þ

I L ðt Þ þ I C ðt Þ þ d dt

) 

The voltage-current relation of the capacitor: I C ðt Þ ¼ C

d d d V ðt Þ ¼ 1  V C ðt Þ ¼ V C ðt Þ dt C dt dt

ð3Þ

Based on the given information: V s ðt Þ ¼ 1 for t > 0

ð4Þ

1 þ V L ðt Þ þ V C ðt Þ ¼ 0

ð5Þ

KVL in the indicated mesh:

The current-voltage relation of the inductor: V L ðt Þ ¼ L

d d d I ðt Þ ¼ 1  I L ðt Þ ¼ I L ðt Þ dt L dt dt

ð6Þ

d I ðt Þ ¼ 1  V C ðt Þ dt L

ð7Þ

Solving (5) and (6):

Solving (7), (3), and (2):  ð1  V C ðt ÞÞ þ

)


 d d d V C ðt Þ þ 2:5 V C ðt Þ ¼ 0 dt dt dt

d2 d V ðt Þ þ 2:5 V C ðt Þ þ V C ðt Þ ¼ 1 dt dt 2 C

ð8Þ

Equation (8) is a second-order nonhomogeneous differential equation; thus, it has both common and particular solutions. Its characteristic equation is: s2 þ 2:5s þ 1 ¼ 0 ) s1,2 ¼ 0:5,  2

ð9Þ

Therefore, the general solution of (8) is: V C,c ðt Þ ¼ Ae0:5t þ Be2t

ð10Þ

Additionally, a constant quantity is chosen as the particular solution of (8), since “1” is a constant value, as follows: V C,p ðt Þ ¼ k

ð11Þ

6

Solutions of Problems: Second‐Order and Higher‐Order Circuits

243

Now, by putting (11) in (8), we have: d2 d k þ 2:5 k þ k ¼ 1 ) k ¼ 1 dt dt 2

ð12Þ

The total solution is the sum of the common solution and particular solution: ) V C ðt Þ ¼ V C,c ðt Þ þ V C,p ðt Þ ¼ Ae0:5t þ Be2t þ 1

ð13Þ

Based on the given information: V C ð 0 Þ ¼ 1 V

ð14Þ

I L ð 0 Þ ¼ 2 A

ð15Þ

Due to the continuity of the voltage of the capacitor and the current of the inductor: V C ð 0þ Þ ¼ V C ð 0 Þ ¼ 1 V

ð16Þ

I L ð 0þ Þ ¼ I L ð 0 Þ ¼ 2 A

ð17Þ

Applying (16) in (13): þ

þ

1 ¼ Ae0:50 þ Be20 þ 1 ) A þ B ¼ 0

ð18Þ

I L ð0þ Þ þ I C ð0þ Þ þ 2:5V C ð0þ Þ ¼ 0 ) I C ð0þ Þ ¼ I L ð0þ Þ  2:5V C ð0þ Þ

ð19Þ

Solving (1) for t ¼ 0+:

Solving (16), (17), and (19): I C ð0þ Þ ¼ 2  2:5  1 ¼ 0:5 A

ð20Þ

Solving (3) and (13): I C ðt Þ ¼

 d d  0:5t V ðt Þ ¼ Ae þ Be2t þ 1 ¼ 0:5Ae0:5t  2Be2t dt C dt

for t ¼ 0þ þ þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) I C ð0þ Þ ¼ 0:5Ae0:50  2Be20 ¼ 0:5A  2B

ð21Þ

0:5 ¼ 0:5A  2B

ð22Þ

1 1 0:5 ¼ 0:5A  2ðAÞ )  0:5 ¼ 1:5A ) A ¼  , B ¼ 3 3

ð23Þ

Solving (21) and (20):

Solving (22) and (18):

Solving (23) and (13): ) V C ðt Þ ¼ 1 þ Choice (1) is the answer.

 1  0:5t þ e2t e 3

244

6

Solutions of Problems: Second-Order and Higher-Order Circuits

Fig. 6.24 The circuits of solution of problem 6.24

6.25. We need to determine the characteristic equation of the circuit based on the preferred variable (I(t)). Defining the voltage of the independent voltage source based on the node voltages in the circuit of Fig. 6.25.2: kI ðt Þ ¼ V L ðt Þ  V C ðt Þ ) V L ðt Þ ¼ kI ðt Þ þ V C ðt Þ

ð1Þ

Applying Ohm’s law for the 2 Ω resistor: V C ðt Þ ¼ 2I ðt Þ

ð2Þ

Applying KCL in the supernode: I L ðt Þ þ

V L ðt Þ 4 7

7 þ I C ðt Þ þ I ðt Þ ¼ 0 ) I L ðt Þ þ V L ðt Þ þ I C ðt Þ þ I ðt Þ ¼ 0 4

ð3Þ

The voltage-current relation of the capacitor: I C ðt Þ ¼ C

d d V ðt Þ ¼ 2 V C ðt Þ dt C dt

ð4Þ

The voltage-current relation of the inductor: ðt

1 I L ðt Þ ¼ I L ð0 Þ þ L 

0



1 V L ðt Þ:dt ¼ 0 þ 1

ðt

ðt V L ðt Þ:dt ¼ 0



V L ðt Þ:dt 0

ð5Þ



In (5), IL(0) is assumed to be zero, since the general response of a circuit does not depend on its primary condition. Using (4) and (5) in (3): ðt 0

7 d V L ðt Þ:dt þ V L ðt Þ þ 2 V C ðt Þ þ I ðt Þ ¼ 0 4 dt

d dt

) V L ðt Þ þ

7 d d2 d V L ðt Þ þ 2 2 V C ðt Þ þ I ðt Þ ¼ 0 4 dt dt dt

Using ð1Þ 7 d d2 d ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) kI ðt Þ þ V C ðt Þ þ ðkI ðt Þ þ V C ðt ÞÞ þ 2 2 V C ðt Þ þ I ðt Þ ¼ 0 4 dt dt dt ) kI ðt Þ þ V C ðt Þ þ



 7 d 7 d d2 kþ1 I ðt Þ þ V C ðt Þ þ 2 2 V C ðt Þ ¼ 0 4 dt 4 dt dt

6

Solutions of Problems: Second‐Order and Higher‐Order Circuits

245

  Using ð2Þ 7 d 7 d d2 I ðt Þ þ ð2I ðt ÞÞ þ 2 2 ð2I ðt ÞÞ ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) kI ðt Þ þ 2I ðt Þ þ k þ 1 4 dt 4 dt dt )4 )

  d2 7 9 d I ðt Þ þ ðk þ 2ÞI ðt Þ ¼ 0 I ðt Þ þ k þ 2 4 2 dt dt

  d2 7 9 d kþ2 I ð t Þ þ k þ I ðt Þ þ I ðt Þ ¼ 0 16 8 dt 4 dt 2

ð6Þ

The characteristic equation of (6) is: s2 þ



 7 9 kþ2 kþ sþ ¼0 16 8 4

ð7Þ

To adjust the circuit in the critically damped condition, the delta (discriminant) of the quadratic characteristic equation must be zero (Δ ¼ 0). Therefore: Δ¼



7 9 kþ 16 8

2

 4ð 1Þ


 kþ2 49 2 63 81 ¼0) k þ kþ k2¼0 4 256 64 64

49 2 1 47 64  ) k  k ¼0)k¼ 256 64 64 1

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   49 47 1 2 þ 4 256 64 64 98 256

1  3 4  192 94 ¼  ,2 ) k ¼ 64 98 4 ¼ 98 49 256

Therefore, k ¼ 2 is one of the answers. Choice (2) is correct.

Fig. 6.25 The circuits of solution of problem 6.25

¼

1 64



qffiffiffiffi

98 256

9 16

246

6

Solutions of Problems: Second-Order and Higher-Order Circuits

6.26. Based on the given information: V C ð 0 Þ ¼ 0 V

ð1Þ

I L ð 0 Þ ¼ 0 A

ð2Þ

Due to the voltage continuity of the capacitor in lack of a power source with an infinite size (e.g., impulse power source): V C ð 0þ Þ ¼ V C ð 0 Þ ¼ 0 V

ð3Þ

Due to the current continuity of the inductor in lack of a power source with an infinite size (e.g., impulse power source): I L ð 0þ Þ ¼ I L ð 0 Þ ¼ 0 A

ð4Þ

Figure 6.26.2 shows the circuit for t ¼ 0+. As can be seen, the capacitor and the inductor have been modeled by a shortcircuit branch and an open circuit branch, respectively, due to (3) and (4). Applying KCL in the supernode: 1 þ I R¼1Ω ð0þ Þ þ 0 ¼ 0 ) I R¼1Ω ð0þ Þ ¼ 1

ð5Þ

V o ð0þ Þ þ 0 þ 1  I R¼1Ω ð0þ Þ ¼ 0 )  V o ð0þ Þ þ 1  1 ¼ 0 ) V o ð0þ Þ ¼ 1 V

ð6Þ

Applying KVL in the loop:

Figure 6.26.3 shows the circuit for t > 0. Using the current-voltage relation of inductor: V L ðt Þ ¼ L

d d I ðt Þ ¼ 2 I L ðt Þ dt L dt

ð7Þ

Applying KVL in the right-side mesh: Using ð7Þ d V o ðt Þ þ 4I L ðt Þ þ V L ðt Þ ¼ 0 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)  V o ðt Þ þ 4I L ðt Þ þ 2 I L ðt Þ ¼ 0 dt for t ¼ 0þ d 1 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) I L ð0þ Þ ¼ V o ð0þ Þ  2I L ð0þ Þ 2 dt Using ð4Þ, ð6Þ d 1 1 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) I L ð0þ Þ ¼  1  2  0 ¼ A= sec dt 2 2

ð8Þ

Applying KCL in the supernode 1: 1 þ I R¼1Ω ðt Þ þ I L ðt Þ ¼ 0 ) I R¼1Ω ðt Þ ¼ 1  I L ðt Þ

ð9Þ

Applying KCL in supernode 2: 10V o ðt Þ  I C ðt Þ þ I L ðt Þ ¼ 0 ) I C ðt Þ ¼ I L ðt Þ  10V o ðt Þ

ð10Þ

6

Solutions of Problems: Second‐Order and Higher‐Order Circuits

247

Applying KVL in the middle mesh: I R¼1Ω ðt Þ  1 þ V C ðt Þ þ 4I L ðt Þ þ V L ðt Þ ¼ 0 Using ð9Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)  ð1  I L ðt ÞÞ þ V C ðt Þ þ 4I L ðt Þ þ V L ðt Þ ¼ 0 )  1 þ V C ðt Þ þ 5I L ðt Þ þ V L ðt Þ ¼ 0

ð11Þ

Using the current-voltage relation of the capacitor and (1): 1 V C ðt Þ ¼ V C ð0 Þ þ C 

ðt 

1 I C ðt Þ:dt ¼ 0 þ 0:5

0

Using ð10Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) V C ðt Þ ¼ 2

ðt

ðt I C ðt Þ:dt ¼ 2

0



I C ðt Þ:dt 

0

ðt ð10V o ðt Þ þ I L ðt ÞÞ:dt 0



Applying (12) and (7) in (11): ðt 1 þ 2

ð10V o ðt Þ þ I L ðt ÞÞ:dt þ 5I L ðt Þ þ 2 0

d dt

) 20V o ðt Þ þ 2I L ðt Þ þ 5

d I ðt Þ ¼ 0 dt L

d d2 I L ðt Þ þ 2 2 I L ðt Þ ¼ 0 dt dt

for t ¼ 0þ d d2 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)20V o ð0þ Þ þ 2I L ð0þ Þ þ 5 I L ð0þ Þ þ 2 2 I L ð0þ Þ ¼ 0 dt dt Using ð4Þ, ð6Þ, ð8Þ 1 d2 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼)  20  1 þ 2  0 þ 5  þ 2 2 I L ð0þ Þ ¼ 0 2 dt )  Choice (2) is correct.

35 d2 d2 35 þ 2 2 I L ð 0þ Þ ¼ 0 ) 2 I L ð 0þ Þ ¼ ðA= sec Þ2 2 4 dt dt

ð12Þ

248

6

Solutions of Problems: Second-Order and Higher-Order Circuits

Fig. 6.26 The circuits of solution of problem 6.26

6

Solutions of Problems: Second‐Order and Higher‐Order Circuits

249

6.27. First, we should note that the circuit is not in the steady state condition for t ¼ 0. Now, by using the current division formula in node “B” in the circuit of Fig. 6.27.2, we can write: I ð0 Þ ¼

1  I L ð 0 Þ 1þ2

ð1Þ

Based on the given information: I ð 0 Þ ¼ 2 A

ð2Þ

Solving (1) and (2): 2¼

1  I L ð 0 Þ ) I L ð 0 Þ ¼ 6 A 1þ2

ð3Þ

Based on the given information: V C1 ð0 Þ ¼ 3 V

ð4Þ

V C2 ð0 Þ ¼ 1 V

ð5Þ

Due to the voltage continuity of the capacitor in lack of a power source with an infinite size (e.g., impulse power source): V C1 ð0þ Þ ¼ V C1 ð0 Þ ¼ 3 V

ð6Þ

V C2 ð0þ Þ ¼ V C2 ð0 Þ ¼ 1 V

ð7Þ

Due to the current continuity of the inductor in lack of a power source with an infinite size (e.g., impulse power source): I L ð 0þ Þ ¼ I L ð 0 Þ ¼ 6 A

ð8Þ

Figure 6.27.3 shows the circuit for t ¼ 0+. As can be seen, the capacitors have been modeled by the voltage sources with the sizes of 3 V and 1 V. Additionally, the inductor has been modeled by a current source of 6 A. Applying KCL in node “A”: V A ð0þ Þ  ð3Þ V A ð0þ Þ  1 3V A ð0þ Þ þ 9 þ V A ð0þ Þ  1 þ 36 þ þ6¼0) ¼0 2 6 6 ) 4V A ð0þ Þ þ 44 ¼ 0 ) V A ð0þ Þ ¼ 11 V

ð9Þ

Using the current division formula in node “B”: I ð 0þ Þ ¼

Using ð8Þ 1 1  I L ð 0þ Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) I ð0þ Þ ¼  6 ¼ 2 A 1þ2 3

ð10Þ

Using Ohm’s law for the right-side 2 Ω resistor: V B ð0þ Þ ¼ 2I ð0þ Þ ¼ 2  2 ¼ 4 V

ð11Þ

Defining the voltage of the inductor based on the node voltages: Using ð9Þ, ð11Þ V L ð 0þ Þ ¼ V A ð 0þ Þ  V B ð 0þ Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) V L ð0þ Þ ¼ 11  4 ¼ 15 V

ð12Þ

250

6

Solutions of Problems: Second-Order and Higher-Order Circuits

Using the current-voltage relation of the inductor for t ¼ 0+: V L ð 0þ Þ ¼ L

V ð0þ Þ Using ð12Þ d d d d 15 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) I L ð0þ Þ ¼ I L ð 0þ Þ ¼ 4 I L ð 0þ Þ ) I L ð 0þ Þ ¼ L 4 dt dt dt dt 4

Using the current division formula in node “B”: I ðt Þ ¼

d for 0þ d dt 1 d 1 d 1 d  I L ðt Þ ) I ðt Þ ¼  I L ðt Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) I ð0þ Þ ¼  I L ð0þ Þ 1þ2 dt 3 dt dt 3 dt


 Using ð13Þ d þ 1 15 5 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) I ð0 Þ ¼  ¼ A=sec dt 3 4 4 Choice (1) is correct.

Fig. 6.27 The circuits of solution of problem 6.27

ð13Þ

6

Solutions of Problems: Second‐Order and Higher‐Order Circuits

251

6.28. To solve this problem, we need to determine the characteristic equation of the circuit. Based on the given information: V C ð 0 Þ ¼ 4 V

ð1Þ

I L ð 0 Þ ¼ 2 A

ð2Þ

Due to the voltage continuity of the capacitor in lack of a power source with an infinite size (e.g., impulse power source): V C ð 0þ Þ ¼ V C ð 0 Þ ¼ 4 V

ð3Þ

Due to the current continuity of the inductor in lack of a power source with an infinite size (e.g., impulse power source): I L ð 0þ Þ ¼ I L ð 0 Þ ¼ 2 A

ð4Þ

2 þ V L ðt Þ þ V C ðt Þ þ 3I L ðt Þ ¼ 0

ð5Þ

KVL in the loop:

The current-voltage relation of the inductor: V L ðt Þ ¼ L

d d d I ðt Þ ¼ 1  I L ðt Þ ¼ I L ðt Þ dt L dt dt

ð6Þ

The current-voltage relation of the capacitor: 1 V C ðt Þ ¼ V C ð0 Þ þ C 

ðt 0

1 I C ðt Þ:dt ¼ 4 þ 0:5

ðt

ðt I C ðt Þ:dt ¼ 4 þ 2 0

I C ðt Þ:dt

ð7Þ

0

Solving (5), (6), and (7): d 2 þ I L ðt Þ þ 4 þ 2 dt

ðt

d dt

I C ðt Þ:dt þ 3I L ðt Þ ¼ 0 ) 0

d2 d I ðt Þ þ 2I C ðt Þ þ 3 I L ðt Þ ¼ 0 dt dt 2 L

ð8Þ

From the circuit, we see that: I C ðt Þ ¼ I L ðt Þ

ð9Þ

d2 d I ðt Þ þ 3 I L ðt Þ þ 2I L ðt Þ ¼ 0 dt dt 2 L

ð10Þ

Solving (8) and (9):

Thus, the characteristic equation of the circuit and its roots are: s2 þ 3s þ 2 ¼ 0 ) s ¼ 1,  2

ð11Þ

252

6

Solutions of Problems: Second-Order and Higher-Order Circuits

Therefore, the general form of the current of the inductor is: I L ðt Þ ¼ Aet þ Be2t

ð14Þ

Now, we need to determine the value of the parameters (“A,” “B,” and “C”): Applying (4) in (14): þ

þ

2 ¼ Ae0 þ Be20 ) A þ B ¼ 2

ð15Þ

Solving (5) for t ¼ 0+: 2 þ V L ð0þ Þ þ V C ð0þ Þ þ 3I L ð0þ Þ ¼ 0 ) V L ð0þ Þ ¼ 2  V C ð0þ Þ  3I L ð0þ Þ Using ð3Þ, ð4Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) V L ð0þ Þ ¼ 2  4  3  2 ¼ 8

ð16Þ

Solving (6) and (14): V L ðt Þ ¼

 d d  t I L ðt Þ ¼ Ae þ Be2t ¼ Aet  2Be2t dt dt

ð17Þ

Solving (16) and (17): þ

þ

8 ¼ Ae0  2Be20 ) A þ 2B ¼ 8

ð18Þ

A þ 2ð2  AÞ ¼ 8 ) A ¼ 4, B ¼ 6

ð19Þ

I L ðt Þ ¼ 4et þ 6e2t

ð20Þ

Solving (15) and (18):

Applying (19) in (14):

Solving (7) and (9): ðt V C ðt Þ ¼ 4 þ 2 0

Using ð20Þ I L ðt Þ:dt ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 4 þ 2

ðt

  4et þ 6e2t :dt

0

h  it  1 ¼ 4 þ 2 4ðet Þ þ 6  e2t  ¼ 4 þ 2 4et  3e2t  ð4  3Þ 2 0 V C ðt Þ ¼ 2 þ 8et  6e2t Choice (2) is correct.

6

Solutions of Problems: Second‐Order and Higher‐Order Circuits

253

Fig. 6.28 The circuits of solution of problem 6.28

6.29. Based on the given information, the circuit will be in critically damped condition if the switch is closed (t > 0). Therefore, the delta (discriminant) of the characteristic equation of the circuit shown in Fig. 6.29.2 must be zero (Δ ¼ 0). Now, we need to determine the differential equation of the circuit. KVL in the left-side mesh: V C ðt Þ þ V L ðt Þ ¼ 0 ) V C ðt Þ ¼ V L ðt Þ

ð1Þ

The voltage-current relation of the capacitor: I C ðt Þ ¼ C

Using ð1Þ d d d V C ðt Þ ¼ 106 V C ðt Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) I C ðt Þ ¼ 106 V L ðt Þ dt dt dt

ð2Þ

The current-voltage relation of the inductor: V L ðt Þ ¼ L

d d I ðt Þ ¼ 4 I L ðt Þ dt L dt

ð3Þ

KCL in the supernode: 1 I C ðt Þ þ I L ðt Þ þ V L ðt Þ ¼ 0 R

ð4Þ

Applying (2) in (4): 106

d 1 V ðt Þ þ I L ðt Þ þ V L ðt Þ ¼ 0 dt L R



 Using ð3Þ 6 d d 1 d ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) 10 4 I L ðt Þ þ I L ðt Þ þ 4 I L ðt Þ ¼ 0 dt dt R dt ) 4  106  )

ð5Þ

d2 4 d I ðt Þ ¼ 0 I ðt Þ þ I L ðt Þ þ R dt L dt 2 L

d2 106 d 106 I L ðt Þ þ I L ðt Þ þ I ðt Þ ¼ 0 2 R dt 4 L dt

ð6Þ

254

6

Solutions of Problems: Second-Order and Higher-Order Circuits

Therefore, the characteristic equation of the circuit is: s2 þ
Δ¼0)

106 R

2

106 106 sþ ¼0 R 4


6
6 2 10 10 106  4ð 1Þ ¼ 106 ) ¼0) ¼ 103 ) R ¼ 1000 Ω 4 R R

ð7Þ ð8Þ

In (8), just the positive value of the resistor, that is, R ¼ 1000 Ω, is acceptable. Now, we can analyze the circuit for t ¼ 0, shown in Fig. 6.29.3. Using the current division formula: I L ð 0 Þ ¼

1000  0:5 ¼ 0:4 A 1000 þ 250

ð9Þ

Using Ohm’s law for the 250 Ω resistor: V C ð0 Þ ¼ I L ð0 Þ  250 ¼ 0:4  250 ¼ 100 V

ð10Þ

Due to the current continuity of the inductor in lack of a power source with an infinite size (e.g., impulse power source): I L ð0þ Þ ¼ I L ð0 Þ ¼ 0:4 A

ð11Þ

Due to the voltage continuity of the capacitor in lack of a power source with an infinite size (e.g., impulse power source): V C ð0þ Þ ¼ V C ð0 Þ ¼ 100 V

ð12Þ

Now, we need to come back to the main problem (analyzing the circuit for t > 0), shown in Fig. 6.29.4. Solving (7) and (8): s2 þ

106 106 sþ ¼ 0 ) s2 þ 1000s þ 250000 ¼ 0 ) ðs þ 500Þ2 ¼ 0 1000 4 ) s ¼ 500,  500

ð13Þ

Therefore, the general form of the current of the inductor is: I L ðt Þ ¼ ðA þ Bt Þe500t uðt Þ

ð14Þ

Now, we need to determine the value of its parameters (“A” and “B”): Applying (11) in (14): þ

0:4 ¼ ðA þ B  0þ Þe5000 ) A ¼ 0:4

ð15Þ

As can be seen in Fig. 6.29.4, the inductor and the capacitor are in parallel and by using (12): V L ð0þ Þ ¼ V C ð0þ Þ ¼ 100 V

ð16Þ

d d I ð0þ Þ ) I L ð0þ Þ ¼ 25 A= sec dt L dt

ð17Þ

Using (3) and (16) for t ¼ 0+: 100 ¼ 4

6

Solutions of Problems: Second‐Order and Higher‐Order Circuits

255

Calculating the first derivative of (14): d d I ðt Þ ¼ ðA þ Bt Þe500t ¼ Be500t  500e500t ðA þ Bt Þ dt L dt for 0þ d þ þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) I L ð0þ Þ ¼ Be5000  500e5000 ðA þ B  0þ Þ ¼ B  500A dt Using ð15Þ d ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) I L ð0þ Þ ¼ B  500  0:4 ¼ B  200 dt

ð18Þ

Solving (17) and (18): 25 ¼ B  200 ) B ¼ 225 Applying (15) and (19) in (14): I L ðt Þ ¼ ð0:4 þ 225t Þe500t uðt Þ Choice (4) is correct.

Fig. 6.29 The circuits of solution of problem 6.29

ð19Þ

256

6

Solutions of Problems: Second-Order and Higher-Order Circuits

Fig. 6.29 (continued)

6.30. The status of the circuit for t ¼ 0 is illustrated in Fig. 6.30.2. As can be seen, the capacitor and the inductors are open circuit and short circuits, respectively, since the circuit has reached its steady state condition. KVL in the indicated loop: 10 þ I L2 ð0 Þ  1 þ 4  I L2 ð0 Þ ¼ 0 ) I L2 ð0 Þ ¼

10 ¼2A 5

ð1Þ

Using Ohm’s law for the 4 Ω resistor: V C ð0 Þ ¼ 4I L2 ð0 Þ ¼ 4  2 ¼ 8 V

ð2Þ

As can be noticed from the circuit of Fig. 6.30.2, the current of the first inductor is zero. I L1 ð0 Þ ¼ 0 A

ð3Þ

Due to the current continuity of the inductor in lack of a power source with an infinite size (e.g., impulse power source): I L1 ð0þ Þ ¼ I L1 ð0 Þ ¼ 0 A

ð4Þ

I L2 ð0þ Þ ¼ I L2 ð0 Þ ¼ 2 A

ð5Þ

Due to the voltage continuity of the capacitor in lack of a power source with an infinite size (e.g., impulse power source): V C ð 0þ Þ ¼ V C ð 0 Þ ¼ 8 V

ð6Þ

Applying KCL in the indicated supernode of Fig. 6.30.3: I 0 ðt Þ þ I L1 ðt Þ þ I L2 ðt Þ þ I ðt Þ ¼ 0 ) I ðt Þ ¼ I 0 ðt Þ  I L1 ðt Þ  I L2 ðt Þ

ð7Þ

Therefore, we need to calculate the value of I0(t), IL1(t), and IL2(t). As can be noticed from Fig. 6.30.3, due to the existence of the short-circuit branch, the circuit includes three separate circuits including zero-order circuit, first-order circuit, and second-order circuit. Zero-order circuit: The zero-order circuit includes the series combination of the voltage source and the 1 Ω resistor, as can be seen in Fig. 6.30.4.

6

Solutions of Problems: Second‐Order and Higher‐Order Circuits

257

Applying Ohm’s law for the 1 Ω resistor: I 0 ðt Þ ¼

10  0 ¼ 10 A 1

ð8Þ

First-order circuit: The first-order circuit includes the series combination of the 4 Ω resistor and the 1 H inductor, as can be seen in Fig. 6.30.5. The primary current of the inductor: I L2 ð0þ Þ ¼ 2 A

ð9Þ

The final current of the inductor is zero, since its circuit does not include any power source. I L2 ð1Þ ¼ 0 A

ð10Þ

L 1 ¼ Ω RTh 4

ð11Þ

The time constant of the first-order circuit: τ¼

Using the general form of the inductor current in a first-order circuit with DC power supply: I L2 ðt Þ ¼ I L2 ð1Þ þ ðI L2 ð0þ Þ  I L2 ð1ÞÞeτ ) I L2 ðt Þ ¼ 0 þ ð2  0Þe4t ¼ 2e4t t

ð12Þ

Second-order circuit: The second-order circuit includes the series combination of the 1 H inductor and the capacitor, as can be seen in Fig. 6.30.6. Applying KVL in the mesh: V L1 ðt Þ þ V C ðt Þ ¼ 0 ) L1

ð13Þ

d d I ðt Þ þ V C ðt Þ ¼ 0 ) 1  I L1 ðt Þ þ V C ðt Þ ¼ 0 dt L1 dt d dt

)

d2 d I ðt Þ þ V C ðt Þ ¼ 0 dt dt 2 L1

ð14Þ

KCL in the indicated node: I L ðt Þ þ I C ðt Þ ¼ 0 ) I C ðt Þ ¼ I L ðt Þ

ð15Þ

Using the voltage-current relation of the capacitor: I C ðt Þ ¼ C

Using ð15Þ d d d d V ðt Þ ¼ 0:25 V C ðt Þ ) V C ðt Þ ¼ 4I C ðt Þ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) V C ðt Þ ¼ 4I L ðt Þ dt C dt dt dt

ð16Þ

Solving (14) and (16): d2 I ðt Þ þ 4I L ðt Þ ¼ 0 dt 2 L1

ð17Þ

258

6

Solutions of Problems: Second-Order and Higher-Order Circuits

Equation (17) is a second-order homogeneous differential equation; therefore, it only has a common solution. Its characteristic equation and the related roots are: s2 þ 4 ¼ 0 ) s ¼ 2j

ð18Þ

Therefore, the general form of the solution of equation (17) is: I L1 ðt Þ ¼ Acosð2t Þ þ Bsinð2t Þ

ð19Þ

To determine the value of the parameters (A and B), we need to apply the primary value of the variables (IL1(t) and VC(t)), that is, IL1(0+) ¼ 0 A and VC(0+) ¼ 8 V. Solving (4) and (19): Acosð2  0þ Þ þ Bsinð2  0þ Þ ¼ 0 ) A  1 þ B  0 ¼ 0 ) A ¼ 0

ð20Þ

V L1 ð0þ Þ ¼ V C ð0þ Þ ¼ 8 V

ð21Þ

Solving (13) and (6):

Using the current-relation of the inductor: V L1 ðt Þ ¼ L1

d d d I ðt Þ ¼ 1  I L1 ðt Þ ¼ I L1 ðt Þ dt L1 dt dt

Using ð19Þ d ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) V L1 ðt Þ ¼ ðAcosð2t Þ þ Bsinð2t ÞÞ ¼ 2Asinð2t Þ þ 2Bcosð2t Þ dt

ð22Þ

Solving (21) and (22): 2Asinð2  0þ Þ þ 2Bcosð2  0þ Þ ¼ 8 )  2A  0 þ 2B  1 ¼ 8 ) B ¼ 4

ð23Þ

Solving (19), (20), and (23): I L1 ðt Þ ¼ 4 sin ð2t Þ Now, solving (7), (8), (12), and (24):   I ðt Þ ¼ I 0 ðt Þ  I L1 ðt Þ  I L2 ðt Þ ¼ 10  ð4 sin ð2t ÞÞ  2e4t I ðt Þ ¼ 10  2e4t þ 4 sin ð2t Þ Choice (4) is the answer.

ð24Þ

6

Solutions of Problems: Second‐Order and Higher‐Order Circuits

Fig. 6.30 The circuits of solution of problem 6.30

259

260

6

Solutions of Problems: Second-Order and Higher-Order Circuits

Fig. 6.30 (continued)

Index

A Analysis, see Mesh analysis; Nodal analysis C Capacitive first-order circuit, 192 Capacitive network, 123 Capacitor, 117–123, 125–127, 132 Capacitor constant, 157 Capacitor voltage, 140, 154, 156–158 Capacitor voltage continuity rule, 182 Characteristic equation, 207, 251, 252 Circuit components, 5 Constant energy, 173 Constant quantity, 242 Continuity law, 158 Conventional methods, 1, 33, 42 Critically damped condition, 208, 253–255 Critically damped response, 204, 234 Current continuity of inductor, 144, 152, 161, 166, 172, 175, 178, 187, 203, 206, 217, 235, 240, 246, 247 Current division formula, 45, 141, 145, 146, 161, 166, 174, 175, 187, 207, 249 Current-relation of inductor, 258 Current sources, 118, 125, 226 Current-voltage relation, 160, 169, 171, 198, 219, 242 Current-voltage relation of capacitor, 224 Current-voltage relation of inductor, 233, 240, 251, 253 D Dependent current source, 17, 62, 68 Dependent voltage, 113 Dependent voltage source, 67, 71, 76, 88, 103, 108 Derivate operator, 204, 236 Differential equation, 178 Dirac Delta Function, 201 E Energy conservation law, 5, 40 Energy-saving components, 133 Equivalent capacitance, 123, 155 F Final value, 182, 192 First-order circuit, 143, 208, 257 capacitor voltage, 140, 142, 147, 151, 163, 164, 180–183 DC power supply, 179, 181–183, 185, 186, 189, 192 inductive, 153, 168, 172, 177

inductor, 161, 175 inductor current, 145, 166 series connection, 160 time constant, 168, 175–177 First-order RC and RL circuits, 150 First-order RL circuit, 116, 135 H Heuristic method, 7, 15, 42, 44, 59, 68 I Impulse function, 196, 211 Impulse power source, 169, 172, 173, 175, 177, 178, 180, 181 Impulse voltage, 201, 225 Impulse voltage source, 171 Independent current source, 5, 15, 18, 40, 41, 66, 80, 139, 152, 163, 175 Independent voltage source, 19, 41, 56, 170, 182, 206, 229, 244 Inductor, 121, 126, 128, 129 Inductor and voltage continuity, 202, 229 K Kirchhoff current law (KCL) nodal analysis, 11, 26, 30, 52–54, 56, 85, 86, 100, 111 node voltage, 98, 107 supernode, 3, 4, 34, 35, 37, 38, 43, 46, 54, 59, 64, 65, 67–69, 71, 74, 76–78, 88, 94, 102, 103, 108, 113 Kirchhoff voltage law (KVL) bottom mesh, 89 circuit loop, 4, 23, 37, 46, 56, 91 current source vs. mesh currents, 48 mesh, 42 mesh analysis, 12, 24, 36, 49, 54, 55, 57, 62, 64, 92, 93 mesh current, 48 supermesh, 9, 36, 47, 68, 81, 82, 95 top loop, 44 top mesh, 81, 88, 102 L LC circuit, 203, 231–233 Linear time-invariant (LTI), 202, 227 M Magnetic flux conservation law, 187 Mesh analysis, 3, 9, 12, 15, 17, 19–21, 24, 25, 36, 37, 49, 54, 57, 62, 64, 68, 73, 78, 80, 82, 83, 92, 95 Mesh current, 9, 78, 80, 82, 83, 92, 95

# Springer Nature Switzerland AG 2020 M. Rahmani-Andebili, DC Electrical Circuit Analysis, https://doi.org/10.1007/978-3-030-50711-4

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262 N Negative R value, 150 Nodal analysis, 2, 11, 16–19, 25, 26, 30, 34, 35, 37, 40, 43, 46, 52, 54, 56, 58, 60, 65, 66, 69, 72, 74, 76, 79, 85, 86, 94, 98, 111, 138 Node voltage, 18, 35, 43, 53, 60, 65, 69, 71, 74–77, 79, 85, 98–100, 103, 108, 113, 135, 136, 138, 154 Norton equivalent circuit, 14, 27, 29, 62 Norton resistance, 12, 19, 55, 56, 77, 102 O Ohm’s law, 2, 17, 20, 28, 35, 46, 54, 56, 73, 76, 78, 82, 83, 88, 94, 95, 98, 103, 113, 141, 145, 146, 148, 156–158, 162, 164, 165, 169–172, 179, 187, 197, 210, 213, 226, 229, 235 Open circuit branch, 140, 152, 157, 185 P Parallel connection, 188 Parallel current sources, 51 Parallel resistors, 51, 188 Power source, 4, 39 Primary value, 125, 129, 180 Primary voltage, 119, 122, 123, 125, 136, 157, 164, 182, 213 Q Quadratic characteristic equation, 204, 206, 234, 245 Quality factor (Q), 205, 238, 239 R Ramp function, 160 Red-color boundary, 25, 96 Resistive circuit, 126 Resistors, 22, 23, 27, 37, 49, 80, 83, 92, 100, 121, 128, 132 RLC circuit, 200 S Second-order circuit, 160, 199, 220, 257, 258 Second-order homogeneous differential equation, 258 Second-order nonhomogeneous differential equation, 206, 242 Series connection, 188 Series-parallel rule, 10, 30, 49, 110, 143, 144 Short-circuit branch, 25, 96, 152, 158, 161, 172, 175, 187, 197, 212 Short-circuit current, 56, 100, 124 Short-circuited nodes, 23, 90 Short-term voltage signal, 129 Source transformation technique, 188 Source transformation theorem, 11, 13, 28, 50, 51, 61, 104 Steady state condition, 117, 136, 137, 158, 161, 163, 171, 172, 175, 187 current continuity, 198, 215, 217 current-voltage relation, 217 Ohm’s law, 195, 210 superposition theorem, 195, 209 voltage continuity, 198, 215, 217 voltage-current relation, 199, 222 Supermesh, 36 Superposition theorem, 15, 66, 126, 165, 166, 173, 188, 225 Switching, 123, 130, 201, 226, 227

Index Switching operation, 136, 152, 156, 158, 161, 175, 180, 187 Symmetric nodes, 30, 110 T Thevenin circuit, 28, 106 Thevenin equivalent circuit, 9, 14, 21–24, 46, 62, 64, 73, 83–85, 92, 109, 143, 144, 146 Thevenin resistance, 4, 6–8, 10, 12, 13, 15–17, 31, 38, 41–43, 46, 49, 55, 56, 69, 83, 91, 106, 108, 109, 112, 133, 138, 139 capacitor, 116, 134, 135, 138, 146, 151 circuit, 172, 180 circuit constant, 148 dependent source, 150, 154 independent current source, 142 independent voltage, 133 inductor, 152, 161 resistor, 158 time constant, 180 voltage source, 147 Thevenin voltage, 6, 13, 15–17, 22, 31, 42, 58, 64, 68–70, 74, 83, 88, 91, 109, 112, 113, 168, 177 Time constant, 115, 116, 118, 121, 133, 135, 139, 142, 143, 145, 149, 151, 154, 156, 166, 169, 172, 192 Time-dependent current, 175 Time-dependent deferential equation, 200, 223 Time-dependent equation, 205, 208, 240, 241, 256 Time-dependent first-order differential equation, 130, 178 Time-dependent voltage equation, 162, 200, 224 U Unit step function, 177 Unstable response, 199 V Voltage-charge relation, 137 Voltage continuity of capacitor, 119, 127, 129, 136, 139, 141, 143, 147, 154, 156, 160, 162, 163, 165, 170, 173, 184, 186, 189, 192, 203, 206, 217, 235, 240, 246, 247 Voltage-current relation, 147, 158, 164, 170, 182, 185, 189, 196, 210, 242 Voltage-current relation of capacitor, 229, 233, 234, 236, 238, 244 Voltage division, 165 Voltage division formula, 4, 38, 61, 137, 143, 147, 151, 166, 174, 181, 182, 228, 229 Voltage equation, 168, 176 Voltage nodes, 8, 45 Voltage pulse, 130 Voltage source, 6, 24, 43, 60, 91, 92, 98, 178, 212, 226, 227 W Wheatstone bridge, 166, 188 Z Zero-order circuit, 208, 256