,ACCESS Test Bank for Engineering Circuit Analysis 10th Edition Hayt
Engineering Circuit Analysis 10th Edition Chapter One Exercise Solutions
3 − 3sin
1. We need to solve 100 = 1 which is a transcendental equation. Let’s solve it
3sin
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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,ACCESS Test Bank for Engineering Circuit Analysis 10th Edition Hayt
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 x2 = 0.01 which has solutions x = 0.1.
© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior
written consent of McGraw Hill LLC.
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, ACCESS Test Bank for Engineering Circuit Analysis 10th Edition Hayt
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 the
V V
approximation that ln 1 − C − C :
V0 V0
V VC
− − C + ln 1 −
V V0
Relative Error = 100 0
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)
© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior
written consent of McGraw Hill LLC.
mynursytest.store
Engineering Circuit Analysis 10th Edition Chapter One Exercise Solutions
3 − 3sin
1. We need to solve 100 = 1 which is a transcendental equation. Let’s solve it
3sin
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.
mynursytest.store
,ACCESS Test Bank for Engineering Circuit Analysis 10th Edition Hayt
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 x2 = 0.01 which has solutions x = 0.1.
© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior
written consent of McGraw Hill LLC.
mynursytest.store
, ACCESS Test Bank for Engineering Circuit Analysis 10th Edition Hayt
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 the
V V
approximation that ln 1 − C − C :
V0 V0
V VC
− − C + ln 1 −
V V0
Relative Error = 100 0
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)
© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior
written consent of McGraw Hill LLC.
mynursytest.store
