Solutions Manual for Engineering Circuit Analysis 10th Edition by Hayt, Kemmerly, and Durbin

Solutions Manual for Engineering Circuit
Analysis 10th Edition Hayt Kemmerly Durbin
Phillips Complete Latest Updated


written by

LearnWithLeo




The Marketplace to Buy and Sell your Study Material

At Stuvia, you will find the best notes, summaries, flashcards & other study material. Search for your
school or uni and find the study material you need.




www.stuvia.com




Downloaded by: LearnWithLeo | Want to earn $1.236
Distribution of this document is illegal extra per year?

, Stuvia.com - The Marketplace to Buy and Sell your Study Material

Engineering Circuit Analysis 10th Edition by William H. Hayt Complete Chapter Solutions Manual are included
2/16/2026,
(Ch
6:17:20
1 to 18)_remo
AM




Engineering Circuit Analysis 10th Edition Chapter One Exercise Solutions

3  3 sin
1. We need to solve  100  1 which is a transcendental equation. Let’s solve it
3 sin
graphically. This can be done on a graphing calculator, plotting points by hand (with a
little iteration), or using MATLAB script similar to
 q  linspace(0, 0.5 *pi/ 2,1000);
 rel_err  100 *abs(3 *q-3 *sin(q))./sin(q)/ 3;
 plot(q,rel_err,'r.')

Expanding the plot and looking for a point close to 1%, we find a value of q  0.245 radians
is about the limit for the linear approximation if 1% or better accuracy is required.




ering Circuit Analysis 10th Edition by William H. Hayt Complete Chapter
Page Solutions
1 Manual are included (Ch 1 to 18)_removed (1).pdf




Downloaded by: LearnWithLeo | Want to earn $1.236
Distribution of this document is illegal extra per year?

, Stuvia.com - The Marketplace to Buy and Sell your Study Material

Engineering Circuit Analysis 10th Edition by William H. Hayt
th
Complete Chapter Solutions Manual are included
2/16/2026,
(Ch
6:17:20
1 to 18)_remo
AM
Engineering Circuit Analysis 10 Edition Chapter One Exercise Solutions


2. We start by expressing the relative error for the first function in the form

1  x   
1 

100  1 x  1
1
1 x

Which can be simplified to

1  x 1  x   1  0.01
1

or x 2  0.01 which has solutions x  0.1.




ering Circuit Analysis 10th Edition by William H. Hayt Complete Chapter
Page Solutions
2 Manual are included (Ch 1 to 18)_removed (1).pdf




Downloaded by: LearnWithLeo | Want to earn $1.236
Distribution of this document is illegal extra per year?

, Stuvia.com - The Marketplace to Buy and Sell your Study Material

Engineering Circuit Analysis 10th Edition by William H. Hayt
th
Complete Chapter Solutions Manual are included
2/16/2026,
(Ch
6:17:20
1 to 18)_remo
AM
Engineering Circuit Analysis 10 Edition Chapter One Exercise Solutions


 V 
3. We begin by rearranging VC  V0 (1  e t / ) to yield t   ln  1  C  where VC/V0 is
 V0 
specified but t is not. We proceed to construct the expression for relative error, using
 V  V
the approximation that ln 1  C    C :
 V0  V0

 V   VC 
   C    ln  1  
Relative Error  100   0   V0 
V
 V 
 ln  1  C 
 V0 

Mercifully, the time constant (τ) cancels in the numerator and denominator. Thus,

(0.1)  ln(0.1)
(a) Relative Error  100   5.1%
ln(0.1)
 (0.5)   ln(1  0.5)
(b) Relative Error  100   28%
 ln(1  0.5)




ering Circuit Analysis 10th Edition by William H. Hayt Complete Chapter
Page Solutions
3 Manual are included (Ch 1 to 18)_removed (1).pdf




Downloaded by: LearnWithLeo | Want to earn $1.236
Distribution of this document is illegal extra per year?