Solutions for Introduction to Electric Circuits, 9th Edition by Svoboda (All Chapters included)
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Module
Introduction to Electric Circuits 9e Svoboda
Institution
Introduction To Electric Circuits 9e Svoboda
Complete Solutions Manual for Introduction to Electric Circuits, 9th Edition by James A. Svoboda, Richard C. Dorf ; ISBN13: 9781118560600.....(Full Chapters are included and organized in reverse order from Chapter 17 to 1)...Chapter 1 Electric Circuit Variables
Chapter 2 Circuit Elements
Chapter ...
Introduction to Electric Circuits, \9th
Edition by James A. Svoboda
Complete Chapter Solutions Manual
are included (Ch 1 to 17)
** Immediate Download
** Swift Response
** All Chapters included
,Table of Contents are given below
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
,Solutions Manual organized in reverse order, with the last chapter displayed
first, to ensure that all chapters are included in this document.
(Complete Chapters included Ch17-1)
Chapter 17- Two-Port and Three Port Networks
Exercises
25 Ω
Exercise 17.2-1 Find the T circuit equivalent to the Π
circuit shown in Figure E 17.2-1.
100 Ω 125 Ω
Answers: R1 = 10 Ω R2 = 12.5 Ω, and R3 = 50 Ω
Figure E 17.2-1
Solution:
R a Rc 100(25)
R1 = = = 10 Ω
R a + Rb + Rc 250
R b Rc 125(25)
R2 = = = 12.5 Ω
R a + Rb + Rc 250
R a Rb 100(125)
R3 = = = 50 Ω
R a + Rb + Rc 250
1
, Exercise 17.3-1 Find the Z and Y parameters of the 21 Ω
circuit of Figure E 17.3-1. + +
⎡ 1 1⎤
⎢ − ⎥ V1 42 Ω 10.5 Ω V2
⎡18 6 ⎤ 14 21
Answers: Z = ⎢ ⎥ , Y=⎢ ⎥ – –
⎣ 6 9⎦ ⎢− 1 1 ⎥
⎣⎢ 21 7 ⎥⎦ Figure E 17.3-1
Solution:
1
−Y12 = −Y21 = S
21
Y11 + Y12 =
1
42
⇒ Y11 =
1
42
− −1 =( ) 3
=
21 42 14
1
S
10.5 42(10.5)
Since I = I 2 , then V1 = I2 = 6 I2
73.5 73.5
V ⎡18 6⎤
= Z 21 = 1 I1 = 0 = 6 Ω Z= ⎢ Ω
9 ⎥⎦
Z 12
I2 ⎣6
2
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