Solutions for Introduction to Electric Circuits, 9th Edition by Svoboda (All Chapters included)
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Introduction to Electric Circuits 9e Svoboda
Institución
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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