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Solution Manual
to accompany

Introduction to Electric Circuits, 6e
By R. C. Dorf and J. A. Svoboda

1

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 CompleteResponse 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 Networks2

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 anopen 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

Errata forIntroduction to Electric Circuits, 6th Edition

Page 757, Problem 16.5-7: Hb(s) = V2(s) / V1(s) and Hc(s) = V2(s) / Vs(s) instead of Hb(s) = V1(s) / V2 (s) and Hc(s) = V1(s) / Vs(s).

http://www.clarkson.edu/~svoboda/errata/6th.html (2 of 2)5/10/2004 7:41:43 PM

Chapter 1 – Electric Circuit Variables
Exercises
Ex. 1.3-1 i (t ) = 8 t 2 − 4 t A q(t ) = Ex. 1.3-3



t 0

i dτ + q(0) =∫

t 0

t 8 8 (8τ 2 − 4τ ) dτ + 0 = τ 3 −2τ 2 = t 3 − 2 t 2 C 0 3 3

q ( t ) = ∫ i (τ ) dτ + q ( 0 ) = ∫ 4sin 3τ dτ + 0 = −
0 0

t

t

4 4 4 t cos 3τ 0 = − cos 3 t + C 3 3 3

Ex. 1.3-4

dq ( t ) i (t ) = dt

0  i (t ) = 2  −2( t − 2 ) −2e

t 2

Ex. 1.4-1

i1 = 45 µA = 45 × 10-6 A < i2 = 0.03 mA = .03 × 10-3 A = 3 × 10-5 A < i3 = 25 × 10-4 A
Ex. 1.4-2
∆ q = i∆ t=

( 4000 A )( 0.001 s )

= 4 C

Ex. 1.4-3

i=

∆ q 45 × 10−9 = = 9 × 10−6 = 9 µA −3 ∆t 5 × 10

Ex. 1.4-4
electron   C   −19 i = 10 billion  1.602 ×10 electron  = s    C   9 electron   −19 10×10  1.602 × 10 electron  s    electron C = 1010 × 1.602 ×10−19 electron s C = 1.602 × 10−9 = 1.602 nA s

1-1

Ex. 1.6-1

(a) The element voltage and current do notadhere to the passive convention in Figures 1.6-1B and 1.6-1C so the product of the element voltage and current is the power supplied by these elements. (b) The element voltage and current adhere to the passive convention in Figures 1.6-1A and 1.6-1D so the product of the element voltage and current is the power delivered to, or absorbed by these elements. (c) The element voltage and current donot adhere to the passive convention in Figure 1.6-1B, so the product of the element voltage and current is the power delivered by this element: (2 V)(6 A) = 12 W. The power received by the element is the negative of the power delivered by the element, -12 W. (d) The element voltage and current do not adhere to the passive convention in Figure 1.6-1B, so the product of the element voltage and...
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