Exam (elaborations)
Solutions Manual INTRODUCTION TO MECHATRONICS AND MEASUREMENT SYSTEMS 4th Edition
- Course
- Institution
Solutions Manual INTRODUCTION TO MECHATRONICS AND MEASUREMENT SYSTEMS 4th Edition.
[Show more]
Seller
Follow
QuizMerchant
Reviews received
Content preview
Solutions ManualINTRODUCTION TO
MECHATRONICS AND
MEASUREMENT
SYSTEMS
4th edition
2012(c)
SOLUTIONS MANUAL
David G. Alciatore
and
Michael B. Histand
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 third 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 files for examples
in the book, and other supplemental material are provided on the Internet at:
mechatronics.colostate.edu
We have class-tested the textbook for several 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 us via e-mail at David.Alciatore@colostate.edu.
We will post corrections for reported errors on our Web site.
Thank you for choosing our book. We hope it helps you provide your students with an
enjoyable and fruitful learning experience in the cross-disciplinary subject of mechatronics.
2 Introduction to Mechatronics and Measurement Systems
,Solutions Manual
2.1 D = 0.06408 in = 0.001628 m.
2
A = D
–6
---------- = 2.082 10
4
= 1.7 x 10-8 m, L = 1000 m
R = L
------- = 8.2
A
2.2
4
(a) R 1 = 21 10 20% so 168k R 1 252k
3
(b) R 2 = 07 10 20% so 5.6k R 2 8.4k
(c) R s = R 1 + R 2 = 217k 20% so 174k R s 260k
R1 R2
(d) R p = ------------------
-
R1 + R2
R 1MIN R 2MIN
R pMIN = -------------------------------
- = 5.43k
R 1MIN + R 2 MIN
R 1MAX R 2MAX
R pMAX = ---------------------------------
- = 8.14k
R 1 MAX + R 2 MAX
2 1
2.3 R 1 = 10 10 , R 2 = 25 10
2 1
R = ------------------- = --------------------------------------------------- = 20 10
R1 R2 10 10 25 10 1
R1 + R2 2 1
10 10 + 25 10
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 From KCL, I s = I 1 + I 2 + I 3
V V V V
so from Ohm’s Law -------s- = ------s + ------s + ------s
R eq R1 R2 R3
1 - = -----
1- + -----
1- + ----- R1 R2 R3
1- so R = ---------------------------------------------------
Therefore, ------- eq -
R eq R1 R2 R3 R2 R3 + R1 R3 + R1 R2
Is Is
2.9 From Ohm’s Law and Question 2.8, V = -------
- = ----------------------------------------------------
R eq R2 R3 + R1 R3 + R1 R2
----------------------------------------------------
R1 R2 R3
and for one resistor, V = I 1 R 1
R2 R3
Therefore, I 1 = ---------------------------------------------------- I s
R 2 R 3 + R 1 R 3 + R 1 R 2
R1 R2 R1 R2
2.10 lim ------------------- = ------------
- = R2
R1 R 1 + R 2 R1
dV dV 1 dV 2
2.11 I = C eq ------- = C 1 ---------- = C 2 ----------
dt dt dt
From KVL,
V = V1 + V2
so
dV dV dV
------- = ---------1- + ---------2-
dt dt dt
Therefore,
I - = -----
I - + -----
I- 1 1 1 C1 C2
------- so -------- = ------ + ------ or C eq = ------------------
-
C eq C1 C2 C eq C1 C2 C1 + C2
4 Introduction to Mechatronics and Measurement Systems
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller QuizMerchant. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $17.99. You're not tied to anything after your purchase.
Can Stuvia be trusted?
4.6 stars on Google & Trustpilot (+1000 reviews)
77254 documents were sold in the last 30 days
Founded in 2010, the go-to place to buy study notes for 14 years now