# Mecatronics

INTRODUCTION TO

MECHATRONICS AND

MEASUREMENT

SYSTEMS

3rd edition

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: www.engr.colostate.edu/mechatronics

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

–6

πD

A = --------- = 2.082 × 10

4

ρ = 1.7 x 10-8 Ωm, L = 1000 m

R = ρL = 8.2Ω

-----A

2.2

4

(a)

R 1 = 21 × 10 ± 20% so 168kΩ ≤ R 1 ≤ 252kΩ

(b)

R 2 = 07 × 10 ± 20% so 5.6kΩ ≤ R 2 ≤ 8.4kΩ

(c)

R s = R 1 + R 2 = 217kΩ ± 20% so 174kΩ ≤ R s ≤ 260kΩ

(d)

R1 R2

R p = -----------------R1 + R2

3

R 1MIN R 2MIN

R pMIN = ------------------------------- = 5.43kΩ

R 1MIN + R 2 MIN

R 1MAX R 2MAX

R pMAX = --------------------------------- = 8.14kΩ

R 1 MAX + R 2 MAX

2.3

2

R 1 = 10 × 10 , R 2 = 25 × 10

1

2

1

R1 R2

( 10 × 10 ) ( 25 × 10 - = 20 × 10 1

)

R = ------------------ = -------------------------------------------------2 1

R1 + R2

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 0Ω perhaps causing a short. Furthermore, the trim value will not be additive with the fixed resistor.

2.5

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.

2.6

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

Introduction to Mechatronics and Measurement Systems

3

Solutions Manual

2.7

From KCL, I s = I 1 + I 2 + I 3

Vs

Vs Vs Vs

so from Ohm’s Law ------- = ----- + ----- + ----R eq

R1 R2 R3

R1 R2 R3

11- 1- 1Therefore, ------- = ----- + ----- + ----- so R eq = --------------------------------------------------R2 R3 + R1 R3 + R1 R2 R eq

R1 R2 R3

Is

Is

From Ohm’s Law and Question 2.7, V = ------- = ---------------------------------------------------R eq R2 R3 + R1 R3 + R1 R2

--------------------------------------------------R1 R2 R3

2.8

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

2.9

R1 R2

R1 R2

lim ------------------ = ------------ = R 2

R1

R 1 → ∞ R 1 + R 2

2.10

dV 1

dV 2

dV

I = C eq ------ = C 1 --------- = C 2 --------dt

dt

dt

From KVL,

V = V1 + V2

so

dV 1

dV = --------- + dV 2

- -------------dt

dt

dt

Therefore,

C1 C2

I- I11- 1I------- = ----- + ----- so ------- = ----- + ----- or C eq = -----------------C1 C2 C eq

C1 C2

C1 + C2

C eq

2.11

V = V1 = V2

dV 1

dV 2

dV

dV

I 1 = C 1 --------- = C 1 ------ and I 2 = C 2...

Continue Reading

Please join StudyMode to read the full document