Download Problem Sheet 1

DesignEngineering–EA1.3Electronics
SolutionstoProblemSheet1(Lectures1,2&3)
1. Circuit(a)isaparallelcircuit:thereareonlytwonodesandallfourcomponentsare
connectedbetweenthem.
Circuit(b)isaseriescircuit:eachnodeisconnectedtoexactlytwocomponentsandthe
samecurrentmustflowthrougheach.
2. For-Bthevoltageandcurrentcorrespondtothepassivesignconvention(i.e.thecurrent
arrowintheoppositedirectiontothevoltagearrow)andsothepowerabsorbedbyBis
givenbyVxI=20W.
FordeviceAweneedtoreversethedirectionofthecurrenttoconformtothepassivesign
convention.ThereforethepowerabsorbedbyAisVxI=-20W.
Asmustalwaysbetrue,thetotalpowerabsorbedbyallcomponentsiszero.
3. Thepowerabsorbedispositiveifthevoltageandcurrentarrowsgoinoppositedirections
andnegativeiftheygointhesamedirection.Soweget:(a)PV=+4,PI=-4,(b)PV=+4,PI=
-4,(c)PV=-4,PI=+4,(a)PV=-4,PI=+4.Inallcases,thetotalpowerabsorbedisPV+PI=0.
4. Wecanfindapath(shownhighlightedbelow)fromthebottomtothetopoftheVXarrow
thatpassesonlythroughvoltagesourcesandsowejustaddtheseuptogetthetotal
potentialdifference:VX=(-3)+(+2)+(+9)=+8V.
5. Ifweaddupthecurrentsflowingoutoftheregionshownhighlightedbelow,weobtain
IX–5–1+2=0.HenceIX=4A.
6. Thethreeseriesresistorsareequivalenttoasingleresistorwithavalueof1+5+2=8kΩ.
7. Thethreeseriesresistorsareequivalenttoasingleresistorwithavalueof
=0.588kΩ.
8. Wecanfirstcombinetheparallel2kΩand3kΩresistorstogive(2x3)/(2+3)=1.2kΩ. This
istheninserieswiththe4kΩresistorwhichmakes5.2kΩinall.Nowwejusthavethree
resistorsinparalleltogiveatotalof1/(1/1+1/5+1/5.2)=1/1.39=0.718kΩ.
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9. Theresistanceis1/8x10 =125kΩ.
EA1.3ProblemSheet1-Solutions(v1.0)
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10. [Method1]:Theresistorsareinseriesandsoformapotentialdivider.Thetotalseries
resistanceis7kΩ,sothevoltagesacrossthethreeresistorsare14x1/7=2V,14x2/7=4V
2
and14x4/7=8V.ThepowerdissipatedinaresistorisV /R,soforthethreeresistors,this
2
2
2
gives2 /1=4mW,4 /2=8mWand8 /4=16mW.
[Method2]:Thetotalresistanceisis7kΩsothecurrentflowinginthecircuitis14/7=2mA.
ThevoltageacrossaresistorisIxRwhich,infortheseresistors,gives2x1=2V,2x2=4V
and2x4=8V.ThepowerdissipatedisVxIwhichgives2x2=4mW,4x2=8mWand8x2
=16mW.Thecurrentthroughthevoltagesourceis2mA,sothepoweritissupplyingisVI=
14x2=28mW.Thisis,inevitably,equaltothesumofthepowerdissipatedbythethree
resistors:4+8+16=28.
11. [Method1]:Theresistorsareinparallelandsoformacurrentdivider:the21mAwilldivide
inproportiontotheconductances:1mS,0.5mSand0.25mS.Thetotalconductanceis1:75
mS,sothethreeresistorcurrentsare21x1/1.75=12mA,21x0.5/1.75=6mAand21x
2
2
2
0.25/1.75=3mA.ThepowerdissipatedinaresistorisI Rwhichgives12 x1=144mW,6 2
x2=72mWand3 x4=36mW.
[Method2]:Theequivalentresistanceofthethreeresistorsis1/(1/1+1/2+1/4)=4/7kΩ.
Thereforethevoltageacrossallcomponentsintheparallelcircuitis21x4/7=12V.The
currentthrougharesistorisV/Rwhichgives12/1=12mA,12/2=6mAand12/4=3mA.
ThepowerdissipatedinaresistorisVxIwhichgives12x12=144mW,12x6=72mWand
12x3=36mW.Thepowersuppliedbythecurrentsourceis12x21=252mWwhichas
expectedequals144+72+36.
12. Theresistorsformapotentialdivider,soY/X=4/(R1+4).Sowewant4/(R1+4)=1/4=>R1+
4=16=>R1=12kΩ.
13. Theresistorsformapotentialdivider,soY/X=R2/(R1+R2)=1/10.
SowewantR2/(R1+R2)=1/10andR1+R2=10MΩ.Substitutingoneintotheotherand
cross-multiplyinggives10xR2=10MΩ =>R2=1MΩ.Substitutingthisintothesimplerof
thetwoinitialequationsgivesR1=10-1=9MΩ.
14. (a)3k=1.5k+1.5k=3.3k||33k,(b)4k=3.9k+100,(c)3.488k=3:9k||33k.
TomakeanexhaustivesearchforcreatingaresistanceofR,youneedtoconsidertwo
possibilities:
(i)
(ii)
fortworesistorsinseries,thelargestofthetworesistorsmustbeintherange
[1/2R,R]or
fortworesistorsinparallel,thesmallestresistormustbeintherange[R,2R].
Inbothcasesthereareatmostfourpossibilities,soyouneedtoconsideruptoeight
possibilitiesinall.So,forexample,forR=3:5k,wewouldconsiderthefollowing
possibilities:(i)1.5k+1.8k=3.3k,1.8k+1.8k=3.6k,2.2k+1.2k=3.4k,2.7k+0.82k=
3.52kand(ii)3.9k||33k=3.488k,4.7k||15k=3.579k,5.6k||10k=3.59k,6.8k||6.8k
=3.4k.Thechoicewithleasterroristheonegivenabove.
Sinceweareinterestedinpercentageerrors,weneedtoconsidertheratiobetween
resistorvalues.Thelargestratiobetweensuccessiveresistorsistheseriesis15/12=1.25
(thisincludesthewraparoundratioof100/82=1.22).Theworst-casepercentageerrorwill
ariseifourtargetresistanceisthemeanofthesetwovalues,13.5.Thepercentageerrorin
choosingeitheroneisthen1.5/13.5=11:1%.
EA1.3ProblemSheet1-Solutions(v1.0)
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15. Whenbothswitchesareclosed,Voutis5Volts.OtherwiseVoutis0Volts.Ifa“TRUE”input
meansclosingaswitchthena“TRUE”outputisrepresentedby5Volts.
n
16.Using2 wheren=4gives16differentpossiblesymbols.
17. a)(5)10≡(101)2
b)(99)10≡(1100011)2 c)(1024)10≡(10000000000)2
18. a)(1010)2≡(10)10
b)(10000000)2≡(128)10
c)(11111111)2≡(255)10
19. a)(64)10≡(40)16
b)(98)10≡(62)16
20. Convertingbetweenbinaryandoctalorhexisquitestraight-forwardbecausethebasesare
relatedbyintegerpowers.Thismeansthateachgroupofbinarydigitsisdirectlyrelatedtoa
hexoroctaldigit.Theoctalbaseisthe3rdpowerofthebinarybasehenceagroupof3
binarydigitsisequivalentto1octaldigit.Thehexbaseisthe4thpowerofthebinarybase
andhenceagroupof4binarydigitsisequivalenttoonehexdigit.Itiseasytoseewhy
writinginbinaryisnotpopularwhenconsideringproblem4c)forexample.Hexandoctal
aremuchmorecompactand,hence,morepopular.
a)(F8)16≡(11111000)2
b)(144)16≡(101000100)2
Notethateachhexdigitisequivalenttofourbinarydigitsandthatthesegroupsoffourbits
havebeenseparatedbyspaces.Theuseofsuchspacesaidsclaritybutisoptional.
21. a)01101001 b)0000withacarry-outof1 c)00101010 d)1011
Inb),theadditionhasoverflowedthesizeoftheinputnumbers(givenas4bits).Theextra1
generatedfromaddingtheMSBsiscalledacarry.Ifthiscarryisnothandledcorrectly,itwill
causeanerrorinthecalculation.
Ind),theanswerwantstobenegative-exceptyoumaynotyethavestudiedhowto
representnegativenumbers.Theanswergivenhereturnsouttobecorrectin2’s
complementform-moreonthatlaterinthecourse.
22. Totalnumberofdifferentsymbols=26+26+3=55.Ifwechoosen=5wecanrepresent
32symbolswhichistoofew.Ifwechoosen=6wecanrepresent64symbolswhichisOK.
Wecanonlychooseintegersforn.Thusweneed6bits.
23. Totalnumberofsymbolsusedinthemessageis17.(Don'tforgettoincludethespace
betweenthewordsandthefull-stopattheend).FromQuestion2aboveweknowthatwe
need6bitspersymbol.Therefore,thetotalnumberofbitsinthemessageis6x17=102.
Atarateof9600bitspersecond,102bitswouldtake102/9600=10.625ms.
24. LookuptheASCIItable(L2S26).“ImperialCollege”inASCIIis(hexadecimal):
EA1.3ProblemSheet1-Solutions(v1.0)
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49,6D,70,65,72,69,61,6C,20,43,6F,6C,6C,65,67,65.Ittakes16bytes.
EA1.3ProblemSheet1-Solutions(v1.0)
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