Engineering Circuit Analysis 10th Edition Solutions Manual William H. Hayt Circuit Analysis Solutions Engineering Circuit Analysis Chapters 1–18 Solutions

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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.




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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.




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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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Engineering Circuit Analysis 10th Edition Chapter Two Exercise Solutions

1. Convert the following to engineering notation:

(a) 0.045 W  45 103 W  45 mW

(b) 2000 pJ  2000 1012  2 109 J  2 nJ

(c) 0.1 ns  0.1109  100 1012 s  100 ps

(d) 39, 212 as  3.9212 104 1018  39.212 1015 s  39.212 fs

(e) 3 

(f) 18,000 m  10 103 m  18 km

(g) 2,500,000,000,000 bits  2.5 1012 bits  2.5 terabits

3
 1015 atoms  102 cm 
  10 atoms/m  it’s unclear what a “zeta atom” is 
21 3
(h)  3 
 cm  1 m 




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