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Solutions for Engineering Circuit Analysis, 10th Edition Hayt (All Chapters included)

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Complete Solutions Manual for Engineering Circuit Analysis, 10th Edition by William H. Hayt, Steven M. Durbin, Jamie Phillips ; ISBN13: 9781264149919.....(Full Chapters included Chapter 1 to 18)...Chapter 1: Introduction Chapter 2: Basic Components and Electric Circuits Chapter 3: Voltage and Cur...

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  • 31 de mayo de 2024
  • 1291
  • 2023/2024
  • Examen
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  • solutions
  • manual
  • engineering
  • circuit
  • analysis
  • 10th edition
  • 10e
  • hayt
  • 9781264149919
  • solutions manual

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  • Electronics
  • Electronics

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Por: mizhouubcca • 1 mes hace

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Engineering Circuit Analysis
10th Edition by William H. Hayt



Complete Chapter Solutions Manual
are included (Ch 1 to 18)




** Immediate Download
** Swift Response
** All Chapters included

,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.

, Engineering Circuit Analysis 10th 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.

, Engineering Circuit Analysis 10th 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)

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