Table Of Content1
Circuit Variables
Assessment Problems
AP1.1 Tosolvethisproblemweuseaproductofratiostochangeunitsfromdollars/yearto
dollars/millisecond. Webeginbyexpressing$10billioninscientificnotation:
$100billion = $100×109
Nowwedeterminethenumberofmillisecondsinoneyear,againusingaproductof
ratios:
1year 1day 1hour 1min 1sec 1year
· · · · =
365.25days 24hours 60mins 60secs 1000ms 31.5576×109 ms
Nowwecanconvertfromdollars/yeartodollars/millisecond,againwithaproduct
ofratios:
$100×109 1year 100
· = = $3.17/ms
1year 31.5576×109 ms 31.5576
AP1.2 First,werecognizethat1ns=10−9 s. Thequestionthenaskshowfarasignalwill
travelin10−9 sifitistravelingat80%ofthespeedoflight. Rememberthatthe
speedoflightc = 3×108 m/s. Therefore,80%ofcis(0.8)(3×108) = 2.4×108
m/s. Now,weuseaproductofratiostoconvertfrommeters/secondto
inches/nanosecond:
2.4×108 m 1s 100cm 1in (2.4×108)(100) 9.45in
· · · = =
1s 109 ns 1m 2.54cm (109)(2.54) 1ns
Thus,asignaltravelingat80%ofthespeedoflightwilltravel9.45(cid:1)(cid:1) ina
nanosecond.
1–1
1–2 CHAPTER1. CircuitVariables
AP1.3 RememberfromEq. (1.2),currentisthetimerateofchangeofcharge,ori = dq In
dt
thisproblem,wearegiventhecurrentandaskedtofindthetotalcharge. Todothis,
wemustintegrateEq. (1.2)tofindanexpressionforchargeintermsofcurrent:
(cid:1)
t
q(t) = i(x)dx
0
Wearegiventheexpressionforcurrent,i,whichcanbesubstitutedintotheabove
expression. Tofindthetotalcharge,welett → ∞intheintegral. Thuswehave
(cid:1) (cid:2)
∞ 20 (cid:2)∞ 20
q = 20e−5000xdx = e−5000x(cid:2) = (e∞ −e0)
total −5000 (cid:2) −5000
0 0
20 20
= (0−1) = = 0.004C = 4000µC
−5000 5000
AP1.4 RecallfromEq. (1.2)thatcurrentisthetimerateofchangeofcharge,ori = dq. In
dt
thisproblemwearegivenanexpressionforthecharge,andaskedtofindthe
maximumcurrent. FirstwewillfindanexpressionforthecurrentusingEq. (1.2):
(cid:3) (cid:4) (cid:5) (cid:6)
dq d 1 t 1
i = = − + e−αt
dt dt α2 α α2
(cid:4) (cid:5) (cid:4) (cid:5) (cid:4) (cid:5)
d 1 d t d 1
= − e−αt − e−αt
dt α2 dt α dt α2
(cid:4) (cid:5) (cid:4) (cid:5)
1 t 1
= 0− e−αt −α e−αt − −α e−αt
α α α2
(cid:4) (cid:5)
1 1
= − +t+ e−αt
α α
= te−αt
Nowthatwehaveanexpressionforthecurrent,wecanfindthemaximumvalueof
thecurrentbysettingthefirstderivativeofthecurrenttozeroandsolvingfort:
di d
= (te−αt) = e−αt +t(−α)eαt = (1−αt)e−αt = 0
dt dt
Sincee−αt neverequals0forafinitevalueoft,theexpressionequals0onlywhen
(1−αt) = 0. Thus,t = 1/α willcausethecurrenttobemaximum. Forthisvalue
oft,thecurrentis
1 1
i = e−α/α = e−1
α α
Rememberintheproblemstatement,α = 0.03679. Usingthisvalueforα,
1
i = e−1 ∼= 10A
0.03679
Problems 1–3
AP1.5 Startbydrawingapictureofthecircuitdescribedintheproblemstatement:
AlsosketchthefourfiguresfromFig. 1.6:
[a] Nowwehavetomatchthevoltageandcurrentshowninthefirstfigurewiththe
polaritiesshowninFig. 1.6. Rememberthat4AofcurrententeringTerminal2
isthesameas4AofcurrentleavingTerminal1. Weget
(a)v = −20V, i = −4A; (b)v = −20V, i = 4A
(c)v = 20V, i = −4A; (d)v = 20V, i = 4A
[b] UsingthereferencesysteminFig. 1.6(a)andthepassivesignconvention,
p = vi = (−20)(−4) = 80W.Sincethepowerisgreaterthan0,theboxis
absorbingpower.
[c] Fromthecalculationinpart(b),theboxisabsorbing80W.
AP1.6 Applyingthepassivesignconventiontothepowerequationusingthevoltageand
currentpolaritiesshowninFig. 1.5,p = vi. FromEq. (1.3),weknowthatpoweris
thetimerateofchangeofenergy,orp = dw. Ifweknowthepower,wecanfindthe
dt
energybyintegratingEq. (1.3). Tobegin,findtheexpressionforpower:
p = vi = (10,000e−5000t)(20e−5000t) = 200,000e−10,000t = 2×105e−10,000t W
NowfindtheexpressionforenergybyintegratingEq. (1.3):
(cid:1)
t
w(t) = p(x)dx
0
1–4 CHAPTER1. CircuitVariables
Substitutetheexpressionforpower,p,above. Notethattofindthetotalenergy,we
lett → ∞intheintegral. Thuswehave
(cid:2)
(cid:1) ∞ 2×105 (cid:2)(cid:2)∞
w = 2×105e−10,000xdx = e−10,000x(cid:2)
−10,000 (cid:2)
0
0
2×105 2×105 2×105
= (e−∞ −e0) = (0−1) = = 20J
−10,000 −10,000 10,000
AP1.7 AttheOregonendofthelinethecurrentisleavingtheupperterminal,andthus
enteringthelowerterminalwherethepolaritymarkingofthevoltageisnegative.
Thus,usingthepassivesignconvention,p = −vi. Substitutingthevaluesofvoltage
andcurrentgiveninthefigure,
p = −(800×103)(1.8×103) = −1440×106 = −1440MW
Thus,becausethepowerassociatedwiththeOregonendofthelineisnegative,
powerisbeinggeneratedattheOregonendofthelineandtransmittedbythelineto
bedeliveredtotheCaliforniaendoftheline.
Chapter Problems
P1.1 Tobegin,wecalculatethenumberofpixelsthatmakeupthedisplay:
n = (1280)(1024) = 1,310,720pixels
pixels
Eachpixelrequires24bitsofinformation. Since8bitscompriseabyte,eachpixel
requires3bytesofinformation. Wecancalculatethenumberofbytesof
informationrequiredforthedisplaybymultiplyingthenumberofpixelsinthe
displayby3bytesperpixel:
1,310,720pixels 3bytes
n = · = 3,932,160bytes/display
bytes 1display 1pixel
Finally,weusethefactthatthereare106 bytesperMB:
3,932,160bytes 1MB
· = 3.93MB/display
1display 106bytes
Problems 1–5
1
P1.2 c = 3×108 m/s so c = 1.5×108 m/s
2
1.5×108 m 5×106 m 5×106
= so x = = 33.3ms
1s xs 1.5×108
P1.3 Wecansetuparatiotodeterminehowlongittakesthebambootogrow10µmFirst,
recallthat1mm= 103µm. Let’salsoexpresstherateofgrowthofbamboousingthe
unitsmm/sinsteadofmm/day. Useaproductofratiostoperformthisconversion:
250mm 1day 1hour 1min 250 10
· · · = = mm/s
1day 24hours 60min 60sec (24)(60)(60) 3456
Usearatiotodeterminethetimeittakesforthebambootogrow10µm:
10/3456×10−3 m 10×10−6 m 10×10−6
= so x = = 3.456s
1s xs 10/3456×10−3
P1.4 Volume = area × thickness
106 = (10×106)(thickness)
106
⇒ thickness = = 0.10mm
10×106
300×109 dollars 100pennies 1year 1day 1hr 1.5mm 1m
P1.5 · · · · · ·
1year 1dollar 365.25days 24hr 3600s 1penny 1000mm
= 1426m/s
P1.6 Ourapproachisasfollows: Todeterminetheareaofabitonatrack,weneedto
knowtheheightandwidthofthespaceneededtostorethebit. Theheightofthe
spaceusedtostorethebitcanbedeterminedfromthewidthofeachtrackonthe
disk. Thewidthofthespaceusedtostorethebitcanbedeterminedbycalculating
thenumberofbitspertrack,calculatingthecircumferenceoftheinnertrack,and
dividingthenumberofbitspertrackbythecircumferenceofthetrack. The
calculationsareshownbelow.
1in 25,400µm
Widthoftrack = = 329.87µm/track
77tracks in
1.4MB8bits 1side
Bitsonatrack = = 72,727.273bits/track
2sides byte 77tracks
Circumferenceofinnertrack = 2π(1/2(cid:1)(cid:1))(25,400µm/in) = 79,796.453µm
79,796.453µm
Widthofbitoninnertrack = = 1.0972µm/bit
72,727.273bits
Areaofbitoninnertrack = (1.0972)(329.87) = 361.934µm2
1–6 CHAPTER1. CircuitVariables
P1.7 C/m3 = 1.6022×10−19 ×1029 = 1.6022×1010C/m3
C/m = (1.6022×1010)(5.4×10−4) = 8.652×106C/m
(cid:4) (cid:5)
C m
Therefore,(8.652×106) ×avevel = i
m s
1400
Thus,averagevelocity = ×10−6 = 161.81µm/s
8.652
20×10−6 C/s
P1.8 [a] n = = 1.25×1014 elec/s
1.6022×10−19 C/elec
5280ft 12in 2.54cm 104µm
[b] m = 3303mi· · · · = 5.32×1012µm
1mi 1ft 1in 1cm
n
Therefore, = 23.5
m
Thenumberofelectrons/secondisapproximately23.5timesthenumberof
micrometersbetweenSydneyandSanFrancisco.
P1.9 FirstweuseEq. (1.2)torelatecurrentandcharge:
dq
i = = 20cos5000t
dt
Therefore,dq = 20cos5000tdt
Tofindthecharge,wecanintegratebothsidesofthelastequation. Notethatwe
substitutexforq ontheleftsideoftheintegral,andy fortontherightsideofthe
integral:
(cid:1) (cid:1)
q(t) t
dx = 20 cos5000ydy
q(0) 0
Wesolvetheintegralandmakethesubstitutionsforthelimitsoftheintegral,
rememberingthatsin0 = 0:
(cid:2)
sin5000y (cid:2)t 20 20 20
q(t)−q(0) = 20 (cid:2) = sin5000t− sin5000(0) = sin5000t
(cid:2)
5000 5000 5000 5000
0
Butq(0) = 0byhypothesis,i.e.,thecurrentpassesthroughitsmaximumvalueat
t = 0,soq(t) = 4×10−3sin5000tC= 4sin5000tmC
P1.10 w = qV = (1.6022×10−19)(9) = 14.42×10−19 = 1.442aJ
Problems 1–7
3600s
P1.11 p = (6)(100×10−3) = 0.6W; 3hr· = 10,800s
1hr
(cid:1) (cid:1)
t 10,800
w(t) = pdt w(10,800) = 0.6dt = 0.6(10,800) = 6480J
0 0
P1.12 AssumewearestandingatboxAlookingtowardboxB.Then,usingthepassive
signconventionp = vi,sincethecurrentiisflowingintothe+terminalofthe
voltagev. Nowwejustsubstitutethevaluesforv andiintotheequationforpower.
Rememberthatifthepowerispositive,Bisabsorbingpower,sothepowermustbe
flowingfromAtoB.Ifthepowerisnegative,Bisgeneratingpowersothepower
mustbeflowingfromBtoA.
[a] p = (20)(15) = 300W 300WfromAtoB
[b] p = (100)(−5) = −500W 500WfromBtoA
[c] p = (−50)(4) = −200W 200WfromBtoA
[d] p = (−25)(−16) = 400W 400WfromAtoB
P1.13 [a]
p = vi = (−20)(5) = −100W
Powerisbeingdeliveredbythebox.
[b] Leaving
[c] Gaining
P1.14 [a] p = vi = (−20)(−5) = 100W,sopowerisbeingabsorbedbythebox.
[b] Entering
[c] Losing
P1.15 [a] InCarA,thecurrentiisinthedirectionofthevoltagedropacrossthe12V
battery(thecurrentiflowsintothe+terminalofthebatteryofCarA).
Thereforeusingthepassivesignconvention, p = vi = (−40)(12) = −480
W.
Sincethepowerisnegative,thebatteryinCarAisgeneratingpower,soCarB
musthavethe”dead”battery.
1–8 CHAPTER1. CircuitVariables
(cid:1)
t 60s
[b] w(t) = pdx; 1.5min = 1.5· = 90s
1min
0
(cid:1)
90
w(90) = 480dx
0
w = 480(90−0) = 480(90) = 43,200J = 43.2kJ
(cid:1)
t
P1.16 p = vi; w = pdx
0
Sincetheenergyistheareaunderthepowervs. timeplot,letusplotpvs. t.
Notethatinconstructingtheplotabove,weusedthefactthat60hr= 216,000s=
216ks
p(0) = (6)(15×10−3) = 90×10−3 W
p(216ks) = (4)(15×10−3) = 60×10−3 W
1
w = (60×10−3)(216×103)+ (90×10−3 −60×10−3)(216×103) = 16.2kJ
2
P1.17 [a] Tofindthepowerat625µs,wesubstitutethisvalueoftimeintoboththe
equationsforv(t)andi(t)andmultiplytheresultingnumberstogetp(625µs):
v(625µs) = 50e−1600(625×10−6) −50e−400(625×10−6) = 18.394−38.94 = −20.546V
i(625µs) = 5×10−3e−1600(625×10−6) −5×10−3e−400(625×10−6)
= 0.0018394−0.003894 = −0.0020546A
p(625µs) = (−20.546)(−0.0020546) = 42.2mW
[b] Tofindtheenergyat625µs,weneedtointegratetheequationforp(t)from0to
625µs. Tostart,weneedanexpressionforp(t):
p(t) = v(t)i(t) = (50)(5×10−3)(e−1600t −e−400t)(e−1600t −e−400t)
Problems 1–9
1
= (e−3200t −2e−2000t +e−800t)
4
Nowweintegratethisexpressionforp(t)togetanexpressionforw(t). Note
wesubstitutexfortontherightsideoftheintegral.
(cid:1)
1 t
w(t) = (e−3200x −2e−2000x +e−800x)dx
4
0
(cid:7) (cid:8)(cid:2)
1 e−3200x e−2000x e−800x (cid:2)(cid:2)t
= + − (cid:2)
4 −3200 1000 800 (cid:2)
0
(cid:7) (cid:4) (cid:5)(cid:8)
1 e−3200t e−2000t e−800t 1 1 1
= + − − + −
4 −3200 1000 800 −3200 1000 800
(cid:7) (cid:8)
1 e−3200t e−2000t e−800t
= + − +5.625×10−4
4 −3200 1000 800
Finally,substitutet = 625µsintotheequationforw(t):
1
w(625µs) = [−4.2292×10−5 +2.865×10−4 −7.5816×10−4 +5.625×10−4]
4
= 12.137µJ
[c] Tofindthetotalenergy,welett → ∞intheaboveequationforw(t). Notethat
thiswillcauseallexpressionsoftheforme−nt togotozero,leavingonlythe
constantterm5.625×10−4. Thus,
1
w = [5.625×10−4] = 140.625µJ
total 4
P1.18 [a] v(20ms) = 100e−1sin3 = 5.19V
i(20ms) = 20e−1sin3 = 1.04A p(20ms) = vi = 5.39W
[b] p = vi = 2000e−100tsin.2150t
(cid:3) (cid:6)
1 1
= 2000e−100t − cos300t
2 2
= 1000e−100t −1000e−100tcos300t
(cid:1) (cid:1)
∞ ∞
w = 1000e−100tdt− 1000e−100tcos300tdt
(cid:2)
0 e−100t(cid:2)(cid:2)∞ 0
= 1000 (cid:2)
−100(cid:2)
(cid:9) 0 e−100t (cid:10)(cid:2)(cid:2)(cid:2)∞
−1000 [−100cos300t+300sin300t] (cid:2)
(100)2 +(300)2 (cid:2)
(cid:3) 100 (cid:6) 0
= 10−1000 = 10−1
1×104 +9×104
w = 9 J
1–10 CHAPTER1. CircuitVariables
P1.19 [a] 0s≤ t < 1s:
v = 5V; i = 20tA; p = 100tW
1s< t ≤ 2s:
v = 0V; i = 20A; p = 0W
2s≤ t < 3s:
v = 0V; i = 20A; p = 0W
3s< t ≤ 4s:
v = −5V; i = 80−20tA; p = −400+100tW
4s≤ t < 5s:
v = −5V; i = 80−20tA; p = −400+100tW
5s< t ≤ 6s:
v = 5V; i = −120+20tA; p = −600+100tW
6s≤ t < 7s:
v = 5V; i = −120+20tA; p = −600+100tW
t > 7s:
v = 0V; i = 20A; p = 0W
[b] Calculatetheareaunderthecurvefromzerouptothedesiredtime:
w(1) = 1(1)(100) = 50J
2
w(6) = 1(1)(100)− 1(1)(100)+ 1(1)(100)− 1(1)(100) = 0J
2 2 2 2
w(10) = 1(1)(100)− 1(1)(100)+ 1(1)(100)− 1(1)(100)+ 1(1)(100) = 50J
2 2 2 2 2
P1.20 [a] p = vi = (100e−500t)(0.02−0.02e−500t) = (2e−500t −2e−1000t)W
dp
= −1000e−500t +2000e−1000t = 0 so 2e−1000t = e−500t
dt
2 = e500t so ln2 = 500t thus pismaximumatt=1.4ms
