资源描述：
《Solution.Manual.for.Electric.Circuits.9th.Edition 外文学习材料》由会员上传分享，免费在线阅读，更多相关内容在学术论文-天天文库。
1、1CircuitVariablesAssessmentProblemsAP1.1Useaproductofratiostoconverttwo-thirdsthespeedoflightfrommeterspersecondtomilespersecond:823×10m100cm1in1ft1mile124,274.24miles····=31s1m2.54cm12in5280feet1sNowsetupaproportiontodeterminehowlongittakesthissignaltotravel1100miles:124,274.2
2、4miles1100miles=1sxsTherefore,1100−3x==0.00885=8.85×10s=8.85ms124,274.24AP1.2Tosolvethisproblemweuseaproductofratiostochangeunitsfromdollars/yeartodollars/millisecond.Webeginbyexpressing$10billioninscientificnotation:9$100billion=$100×10Nowwedeterminethenumberofmillisecondsinoneye
3、ar,againusingaproductofratios:1year1day1hour1min1sec1year····=365.25days24hours60mins60secs1000ms31.5576×109msNowwecanconvertfromdollars/yeartodollars/millisecond,againwithaproductofratios:$100×1091year100·==$3.17/ms1year31.5576×109ms31.5576©2010PearsonEducation,Inc.,UpperSaddleR
4、iver,NJ.Allrightsreserved.ThispublicationisprotectedbyCopyrightandwrittenpermissionshouldbeobtainedfromthepublisherpriortoanyprohibitedreproduction,storageinaretrieval1–1system,ortransmissioninanyformorbyanymeans,electronic,mechanical,photocopying,recording,orlikewise.Forinformat
5、ionregardingpermission(s),writeto:RightsandPermissionsDepartment,PearsonEducation,Inc.,UpperSaddleRiver,NJ07458.1–2CHAPTER1.CircuitVariablesAP1.3RememberfromEq.(1.2),currentisthetimerateofchangeofcharge,ordqi=Inthisproblem,wearegiventhecurrentandaskedtofindthetotaldtcharge.Todothi
6、s,wemustintegrateEq.(1.2)tofindanexpressionforchargeintermsofcurrent:Ztq(t)=i(x)dx0Wearegiventheexpressionforcurrent,i,whichcanbesubstitutedintotheaboveexpression.Tofindthetotalcharge,welett→∞intheintegral.ThuswehaveZ∞20∞20q=20e−5000xdx=e−5000x=(e−∞−e0)total−5000−5000002020=(0−1)==
7、0.004C=4000µC−50005000AP1.4RecallfromEq.(1.2)thatcurrentisthetimerateofchangeofcharge,ordqi=.Inthisproblemwearegivenanexpressionforthecharge,andaskedtodtfindthemaximumcurrent.FirstwewillfindanexpressionforthecurrentusingEq.(1.2):dqd1t1i==−+e−αtdtdtα2αα2d1dtd1=−e−αt−e−αtdt
8、α2dtαdtα21t1=0−e−αt−αe−αt−−αe−αtααα2
