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Engineering Circuit Analysis,
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S
10th Edition
SOLUTIONS
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MANUAL
AAP
William H. Hayt, Steven M. Durbin, Jamie Phillips
PPPR
Comprehensive Solutions Manual for Instructors
ROOV
and Students
||| ||| || ||| || | || ||| | || ||| |
VE D
9781265845292
E
© William H. Hayt, Steven M. Durbin & Jamie Phillips. All rights
reserved. Reproduction or distribution without permission is
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prohibited.
?
© MEDGEEK
,ST TU
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
UV V
S
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;
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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.
?_?_
AAP
PPPR
ROOV
VE D
E
D? ?
?
, Engineering Circuit Analysis 10th Edition Chapter One Exercise Solutions
ST TU
2. We start by expressing the relative error for the first function in the form
1 x
1
100 1 x 1
UV V
S
1
1 x
Which can be simplified to
1 x 1 x 1 0.01
IAIA
1
or x 2 0.01 which has solutions x 0.1.
?_?_
AAP
PPPR
ROOV
VE D
E
D? ?
?
, Engineering Circuit Analysis 10th Edition Chapter One Exercise Solutions
ST TU
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
UV V
S
V V
the approximation that ln 1 C C :
V0 V0
V VC
C ln 1
Relative Error 100 0 V0
V
V
IAIA
ln 1 C
V0
Mercifully, the time constant (τ) cancels in the numerator and denominator. Thus,
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(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)
AAP
PPPR
ROOV
VE D
E
D? ?
?
ST TU
Engineering Circuit Analysis,
UV V
S
10th Edition
SOLUTIONS
IAIA
?_?_
MANUAL
AAP
William H. Hayt, Steven M. Durbin, Jamie Phillips
PPPR
Comprehensive Solutions Manual for Instructors
ROOV
and Students
||| ||| || ||| || | || ||| | || ||| |
VE D
9781265845292
E
© William H. Hayt, Steven M. Durbin & Jamie Phillips. All rights
reserved. Reproduction or distribution without permission is
D? ?
prohibited.
?
© MEDGEEK
,ST TU
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
UV V
S
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;
IAIA
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.
?_?_
AAP
PPPR
ROOV
VE D
E
D? ?
?
, Engineering Circuit Analysis 10th Edition Chapter One Exercise Solutions
ST TU
2. We start by expressing the relative error for the first function in the form
1 x
1
100 1 x 1
UV V
S
1
1 x
Which can be simplified to
1 x 1 x 1 0.01
IAIA
1
or x 2 0.01 which has solutions x 0.1.
?_?_
AAP
PPPR
ROOV
VE D
E
D? ?
?
, Engineering Circuit Analysis 10th Edition Chapter One Exercise Solutions
ST TU
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
UV V
S
V V
the approximation that ln 1 C C :
V0 V0
V VC
C ln 1
Relative Error 100 0 V0
V
V
IAIA
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)
AAP
PPPR
ROOV
VE D
E
D? ?
?
