Solutions Manual Introduction to Mechatronics and Measurement Systems 1 INTRODUCTION TO MECHATRONICS AND MEASUREMENT SYSTEMS 5th edition 2025

Solutions Manual



INTRODUCTION TO
MECHATRONICS AND
MEASUREMENT
SYSTEMS

5th edition
2018




SOLUTIONS MANUAL

David G. Alciatore, PhD, PE




Department of Mechanical Engineering
Colorado State University
Fort Collins, CO 80523



Introduction to Mechatronics and Measurement Systems 1

,Solutions Manual


This manual contains solutions to the end-of-chapter problems in the fifth edition of
"Introduction to Mechatronics and Measurement Systems." Only a few of the open-ended
problems that do not have a unique answer are left for your creative solutions. More information,
including an example course outline, a suggested laboratory syllabus, Mathcad/Matlab files for
examples in the book, and other supplemental material are provided on the book website at:

mechatronics.colostate.edu

We have class-tested the textbook for many years, and it should be relatively free from
errors. However, if you notice any errors or have suggestions or advice concerning the textbook's
content or approach, please feel free to contact me via e-mail at . I
will post corrections for reported errors on the book website.

Thank you for choosing my book. I hope it helps you provide your students with an
enjoyable and fruitful learning experience in the exciting cross-disciplinary subject of
mechatronics.




2 Introduction to Mechatronics and Measurement Systems

, Solutions Manual

2.1 D = 0.06408 in = 0.001628 m.
2
D -
A = --------- = 2.082  10–6
4
 = 1.7 x 10-8 m, L = 1000 m
L
R = -- --- = 8.2
A

2.2

(a) R1 = 21  104  20% so 168k  R1  252k

(b) R2 = 07  103  20% so 5.6k  R2  8.4k


(c) Rs = R1 + R2 = 217k  20% so 174k  Rs  260k

R1R2
(d) Rp = ------------------
R1 + R2
R1 R2
Rp = ----------M--I-N--------M--I--N--- = 5.43k
MIN R1 + R
MIN 2 MIN

R1 R2
Rp = ----------M--A---X---------M--A--X---- = 8.14k
MAX R1 +R
MAX 2 MAX



2.3 R1 = 10  102 , R2 = 25  101
R1R2 10  10225  101 1
R = = = 20  10
R1 + R2 10  102 + 25  101
a = 2 = red, b = 0 = black, c = 1 = brown, d = gold

2.4 In series, the trim pot will add an adjustable value ranging from 0 to its maximum value to
the original resistor value depending on the trim setting. When in parallel, the trim pot
could be 0 perhaps causing a short. Furthermore, the trim value will not be additive with
the fixed resistor.

2.5 When the last connection is made, a spark occurs at the point of connection as the
completed circuit is formed. This spark could ignite gases produced in the battery. The
negative terminal of the battery is connected to the frame of the car, which serves as a
ground reference throughout the vehicle.




Introduction to Mechatronics and Measurement Systems 3

, Solutions Manual

2.6 No, as long as you are consistent in your application, you will obtain correct answers. If
you assume the wrong current direction, the result will be negative.

2.7 Place two 100 resistors in parallel and you immediately have a 50 resistance.

2.8 Put two 50 resistors in series: 50  50  

2.9 Put a 100 resistor in series with the parallel combination of two 100 resistors:
100  100100100  100  


2.10 From KCL, Is = I1 + I2 + I3
Vs V V V
so from Ohm’s Law -------- = -----s- + -----s- + ------s
Req R1 R2 R3
1 1 1 1
Therefore, --- ---- = ---- + ---- + ---- so R = R1R2R3
----------------------------------------------------
Req R1 R2 R3 eq
R2R3 + R1R3 + R1R2

Is Is
2.11 From Ohm’s Law and Question 2.10, V = - ---- = ----------------------------------------------------
Req R2R3 + R1R3 + R1R2
R1R2R3

and for one resistor, V = I1R1
 R2R3 
Therefore, I1 =  I

R 2R 3 + R 1R 3 + R 1R 2 s


 R1R2  = ------------
R1R2
2.12 lim ------------------ = R2
R1   R1 + R2  R1



dV dV dV
2.13 I = C eq ------ = C 1---------1 = C 2---------2
dt dt dt
From KVL,
V = V1 + V2
so
d-- -V
--- = dV1 dV2
--------- + ---------
dt dt dt




4 Introduction to Mechatronics and Measurement Systems