Solution Manual For Introduction To Electric Circuits 6th Edition

Solution Manual For Introduction To Electric Circuits 6th Edition

Info icon This preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Solution Manual to accompany Introduction to Electric Circuits, 6e By R. C. Dorf and J. A. Svoboda 1
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Table of Contents Chapter 1 Electric Circuit Variables Chapter 2 Circuit Elements Chapter 3 Resistive Circuits Chapter 4 Methods of Analysis of Resistive Circuits Chapter 5 Circuit Theorems Chapter 6 The Operational Amplifier Chapter 7 Energy Storage Elements Chapter 8 The Complete Response of RL and RC Circuits Chapter 9 The Complete Response of Circuits with Two Energy Storage Elements Chapter 10 Sinusoidal Steady-State Analysis Chapter 11 AC Steady-State Power Chapter 12 Three-Phase Circuits Chapter 13 Frequency Response Chapter 14 The Laplace Transform Chapter 15 Fourier Series and Fourier Transform Chapter 16 Filter Circuits Chapter 17 Two-Port and Three-Port Networks 2
Image of page 2
Errata for Introduction to Electric Circuits, 6th Edition Errata for Introduction to Electric Circuits, 6th Edition Page 18 , voltage reference direction should be + on the right in part B: Page 28 , caption for Figure 2.3-1: "current" instead of "cuurent" Page 41, line 2: "voltage or current" instead of "voltage or circuit" Page 41, Figure 2.8-1 b: the short circuit is drawn as an open circuit. Page 42, line 11: "Each dependent source ..." instead of "Each dependent sources..." Page 164, Table 5.5-1: method 2, part c, one should insert the phrase "Zero all independent sources, then" between the "(c)" and "Connect a 1-A source. . ." The edited phrase will read: "Zero all independent sources, then connect a 1-A source from terminal b to terminal a. Determine Vab. Then Rt = Vab/1." Page 340, Problem P8.3-5: The answer should be . Page 340, Problem P8.3-6: The answer should be . Page 341, Problem P.8.4-1: The answer should be Page 546, line 4: The angle is instead of . Page 554, Problem 12.4.1 Missing parenthesis: Page 687, Equation 15.5-2: Partial t in exponent: http://www.clarkson.edu/~svoboda/errata/6th.html (1 of 2)5/10/2004 7:41:43 PM
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Errata for Introduction to Electric Circuits, 6th Edition Page 757, Problem 16.5-7: H b (s) = V 2 (s) / V 1 (s) and H c (s) = V 2 (s) / V s (s) instead of H b (s) = V 1 (s) / V 2 (s) and H c (s) = V 1 (s) / V s (s). http://www.clarkson.edu/~svoboda/errata/6th.html (2 of 2)5/10/2004 7:41:43 PM
Image of page 4
Chapter 1 – Electric Circuit Variables Exercises Ex. 1.3-1 ( ) 2 2 3 2 3 0 0 0 8 4 A 8 8 ( ) (0) (8 4 ) 0 2 2 C 3 3 t t t i t t t q t i d q d t t τ τ τ τ τ τ = = + = + = = 2 Ex. 1.3-3 ( ) ( ) ( ) 0 0 0 4 4 0 4sin3 0 cos3 cos3 C 3 3 t t t q t i d q d t τ τ τ τ τ = + = + = − = − + 4 3 Ex. 1.3-4 ( ) ( ) ( ) 2 2 0 0 ( ) 2 0 2 2 2 t t dq t i t i t t dt e t < = = > < < Ex. 1.4-1 i 1 = 45 µ A = 45 × 10 -6 A < i 2 = 0.03 mA = .03 × 10 -3 A = 3 × 10 -5 A < i 3 = 25 × 10 -4 A Ex. 1.4-2 ( )( ) = 4000 A 0.001 s 4 C q i t = = Ex. 1.4-3 9 6 3 45 10 9 10 5 10 q i t × = = = × × = 9 µ A Ex. 1.4-4 19 9 19 10 19 9 electron C electron C = 10billion 1.602 10 = 10 10 1.602 10 s electron s e electron C = 10 1.602 10 electron s C 1.602 10 1.602 nA s i     × × ×         × × = × = lectron 1-1
Image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Ex. 1.6-1 (a) The element voltage and current do not adhere to the passive convention in Figures 1.6-1 B and 1.6-1 C so the product of the element voltage and current is the power supplied by these elements.
Image of page 6
Image of page 7
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern