Solution Manual For Introduction To Mechatronics And Measurement Systems 5th Edition By David G. Alciatore latest Edition

SolutionsdManual



INTRODUCTION TO M d d




ECHATRONICSAND M d d




EASUREMENT
SYSTEMS

5th edition 20
d d




18




SOLUTIONSMANUAL d




David G. Alciatore, PhD, PE
d d d d




Department of Mechanical Engineering Colorad
d d d d

o State University
d d

Fort Collins, CO 80523
d d d




IntroductiondtodMechatronicsdanddMea surementdSystems 1

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Thisdmanualdcontainsdsolutionsdtodthedend-of-
chapterdproblemsdindthedfifthdeditiondofd"IntroductiondtodMechatronicsdanddMeasurementdSystems."
d Onlydadfewdofdthedopen-

endeddproblemsdthatddodnotdhavedaduniquedanswerdaredleftdfordyourdcreativedsolutions.d Moredinforma
tion,dincludingdandexampledcoursedoutline,dadsuggesteddlaboratorydsyllabus,dMathcad/Matlabdfilesdf
ordexamplesdindthedbook,danddotherdsupplementaldmaterialdaredprovideddondthedbookdwebsitedat:

mechatronics.colostate.edu

Wedhavedclass-
testeddthedtextbookdfordmanydyears,dandditdshoulddbedrelativelydfreedfromderrors.d However,difdyoudno
ticedanyderrorsdordhavedsuggestionsdordadvicedconcerningdthedtextbook'sdcontentdordapproach,dpleased
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.d Idwilldpostdcorrectionsdfordreportedderrorsdondthedbookdweb
site.

Thankdyoudfordchoosingdmydbook.d Idhopeditdhelpsdyoudprovidedyourdstudentsdwithdandenjoya
bledanddfruitfuldlearningdexperiencedindthedexcitingdcross-disciplinarydsubjectdofdmechatronics.




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2 IntroductiondtodMechatronicsdanddMeasurementdSystems




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2.1 Dd=d0.06408dind=d0.001628dm.
dDd -2 –6
Ad =d --------- =d 2.082dd10
4
d=d1.7dxd10-8d m, Ld=d1000dm
L
Rd =d ------ =d 8.2
A

2.2
4d
(a) R1d = 21dd10 d20% sod 168kddR1dd252kd

3d
(b) R2d = 07dd10 d20% so 5.6kddR2dd8.4k

(c) Rsd = R1d +dR2 =d 217kdd 20% sod 174kddRsd d260k

R1d dRd 2
(d) Rpd = ------------------
dR
1d +dR2
R R2M I N
R =d ----------1M--IN---------------- =d 5.43k
pMIN R1 +dR
MIN 2 MIN

R1 R2
R =d ----------M--A--X---------M--A--X---- =d 8.14k
pMAX R1
MAX
+dR2 MAX

2d 1
2.3 R1d = 10dd10 , R2d = 25dd10
2d 1d
R1R2 (10d d 10 )(25d d 10 ) 1
Rd = = = 20dd10
2d 1
R1d +dR2 10dd10 +d25dd10
ad=d2d=dred,d bd=d0d=dblack,d cd=d1d=dbrown,d dd=dgold

2.4 Indseries,dthedtrimdpotdwilldadddandadjustabledvaluedrangingdfromd0dtoditsdmaximumdvaluedtodt
hedoriginaldresistordvalueddependingdondthedtrimdsetting.d Whendindparallel,dthedtrimdpotdcoul
ddbed0dperhapsdcausingdadshort.d Furthermore,dthedtrimdvaluedwilldnotdbedadditivedwithdthedfi
xeddresistor.

2.5 Whendthedlastdconnectiondisdmade,dadsparkdoccursdatdthedpointdofdconnectiondasdthedcomplete
ddcircuitdisdformed.d Thisdsparkdcoulddignitedgasesdproduceddindthedbattery.d Thednegativedter
minaldofdthedbatterydisdconnecteddtodthedframedofdthedcar,dwhichdservesdasdadgrounddreference
dthroughoutdthedvehicle.




IntroductiondtodMechatronicsdanddMea surementdSystems 3

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