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PDHeng inee r . com Course E-4014
AC Waveforms: Basic Quantities for Non-
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ACWaveforms:BasicQuantitiesPeterBasis
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ACWaveforms:BasicQuantities
AC (Alternating Current) waveforms refer to continually changing timevarying voltagesandcurrents.Themagnitudeanddirectionof thesevoltagesandcurrentsvary inasetmanner
fromaminimumtoamaximumvalue,inaspecifictimeinterval.ThisisincontrasttoDC(Direct
Current), where currents or voltages do not change in magnitude or direction. They are uni
directional. ACwaveformscanbesinusoidal(sinewave),squareortriangularasshowninfigure
1.1.
Sinusoidalwaveformsaretheonesgeneratedbytheutilitycompaniesandareavailablein
ourhomesandplaceofwork.Thesquareandthetriangularwaveformsaremadefromsinusoidal
waveforms with the help of electronic circuitry. The only waveform available in nature is the
sinusoidal.Allotherwaveformsarecomposedofmanysuperimposedsinusoids. Thesinusoidal
waveformistheonlyACwaveformnotaffectedbytheresponseofResistance(R),Inductance(L),
andCapacitance(C).ThismeansthatwhenasinusoidalwaveformexcitesR,L,andC,theoutputis
alsoasinusoidalwaveform.
ThebasicDClawsKirchhoffsVoltageLaw,KirchhoffsCurrentLaw,andOhm'sLawgovern
ACcircuits.AsinDCquantities,theselawswillsolveanyACnetwork.ACquantities,however,are
morecomplicatedthanDCquantities.TosolveforACquantitiesoneneedstousetheconceptof
thephasorandcomplexalgebra,aswillbeshownlaterinthistext.
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1.1 BasicDefinitions
Beforeproceedingwith theanalysisofACnetworks, it is important tobecome familiar
somebasicdefinitions. Using thesinusoidalvoltageshown in figure1.2,quantitiesaredefined
below,whicharetrueforanyACvoltageorcurrent,andsinusoidal,squareortriangularwaves:
PeriodicWaveform awaveformthatcontinuouslyrepeatsitself.
Period(T) thetimeintervalrequiredbetweensuccessiverepetitionsof
the periodic waveform. T designates the period of any
periodic waveform. The best approach in measuring the
periodTistousesuccessivecrossingsofthezeroaxiswitha
positiveslope,asshowninfigure1.2.
Cycle theperiodoftheperiodicwaveform.
Frequency(f) thenumberofcyclesofthewaveforminonesecond.Aunit
of the frequency is 1 cycle per second. This unit is also
known as Hertz [Hz], and it is named after its inventor
HeinrichR.Hertz.
PeakValue themaximumvalueofthewaveformusingthezerolevelas
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reference.Infigure1.2,thepeakvalueisindicatedasVp.
PeaktoPeakValue thedistancebetweenthepositivepeak(maximum)andthe
negativepeak(minimum).ThisvalueisdesignatedasVppin
figure1.2.
InstantaneousValue thevalueofthewaveformatanyinstantoftime.Lowercase
lettersareusedfortheinstantaneousvalue.
It ismost importanttodefineasetofpolaritiesforasinusoidalACvoltagesourceanda
directionforasinusoidalACcurrentsource.Thisisdonebychoosinganinstantoftimeduringthe
positivehalfcycleofthesinusoidalwaveforms.Allsinusoidalsourcesareconsideredinthepositive
half cycle, sodistinctpolarities canbeassigned for thevoltagesanddistinctdirections for the
currentsasinDCquantities.
Thefrequencyandtheperiodarerelatedwiththefollowingequationwherefis inHertz
[Hz]andTisinseconds[s].
f=T
(11)
Thefrequencyofaperiodicwaveformthathasaperiodof20msis:
f=T
=
=50Hz
Whenthefrequencyofaperiodicwaveformis250Hz,theperiodis:
T= = =4x10s=4ms
Theequationthatdescribesthesinusoidalwaveformoffigure1.2is:
V=Vpsin t (12)
where:
Vp isthepeakvalue
sin indicatesasinewave,and
istheangularvelocityofaradiusvectorthatrotatesaboutacenter.This
vectorhasmagnitudeVp,seefigure1.3,andthesinusoidalwaveformcan
beconstructedtakingthelengthoftheverticalprojection,ateveryinstant
ofrotationofthisvector,overonecompleterotation.Theangularvelocity
is calculated as the ratio of distance over time, where the distance is
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measuredinradiansordegreesandthetimeinseconds. Observethatthe
unitoft is [(rad/s)(s)]= [rad],thusconcluding thattheargumentofthe
sinewaveisanangle( = t)inradians.
ThetimetheradiusvectorrequirestomakeacompleterotationisequaltotheperiodT(in
seconds)andthedistancethathastraveledisequalto2 radians(thereare2 radiansinacircle).
Theangularvelocitycanbecalculatedwiththefollowingformula:
T
[
] (13)
Since f=1/T,then:
2f [] (14)
Theangularvelocityofasinewavethathasafrequencyof60Hzis:
=2 f=2(3.14rad)(60Hz)=377rad/s
1.2 PhaseRelations
Infigure1.4thesinusoidalwaveformV=Vpsin tisrepresentedusingtheradianasthe
unitofmeasurement for theabscissa.Themaximum isat /2, theminimumat3/2,andzero
valueat0, and2.
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In figure, 1.5 the same sinusoidal waveform is sifted degrees to the left of 0o
. Theequationthatdescribesthesinusoidalwaveforminfigure1.5is:
V=Vpsin(t+) (15)
where: tisinradiansand
isindegrees.
Thereasonforequation(15) isthat ifdegreesareaddedtothesinusoidoffigure1.5,thenitwillbecomethesinusoidoffigure1.4,whichisaregularsinusoidthatstartsat0
o.Equation
(15)isusedforanysinusoidalvoltageorcurrentthatisshifteddegreestotheleftof0o.Considerthesinusoidoffigure1.6,which isshifteddegreestotherightofthe0o.The
equationthatdescribesthesinusoidalwaveforminfigure1.6is:
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V=Vpsin(t ) (16)
where:
tisinradiansand
isindegrees.
Thereasonforequation(16)isthatifdegreesaresubtractedfromthesinusoidoffigure1.6,thenitwillbecomethesinusoidoffigure1.4,whichisaregularsinusoid.Equation(16)isused
foranysinusoidalvoltageorcurrentthatisshifteddegreestotherightof0o.Considerthecosinewaveinfigure1.7(a). Thesinewaveisshiftedtotheleftby90oasseen
fromfigure1.7(b),makingthefollowingistrueregardingsinesandcosines:
cos t=sin(t+90o) (15)
sin t=cos(t 90o) (16)
Sincethesinusoids,infigure1.7,havethesamefrequencyonecansaythatthecosineleads(is
aheadof) thesineby90oorthesine lags (isbehind)thecosineby90
o.Asaresult,thesetwo
sinusoidsare90ooutofphase.
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The phase angle between two waveforms is measured between the points where the
waveforms cross the horizontal axis with a positive going slope. These waveforms are out of
phasebythenumberofdegreesthatseparatedthem.Whentwowaveformscrossthehorizontal
axisatthesamepoint,thesewaveformsareinphase.
Belowisalistofsomeadditionalrelations,whichhelpusfindthephaserelationsbetweenwaveforms:
sin(t)= sin t=sin(t180o) (17)
cos(t)=cos t (18)
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cos t=cos(t180o) (19)
Observe that relation (17) showsanegative sign ina sinusoidalexpressiondealswith the sine
portionoftheexpressionandnotthepeakvalueofthesinusoid.
Whatisthephaserelationshipbetweenthefollowingsinusoidalwaveforms?Seefigure1.8.
v=100sin(314t 30o)
i= 50sin(314t+15o)
ileadsvby45oor vlagsiby45
o.
Althoughbothstatementsabovearecorrect, inpractice,one indicateswhatthecurrent
doeswithrespecttothevoltageanditispreferabletosaythat:
ileadsvby45o
1.3 AverageorDCValue
TheaverageorDCvalueofaperiodicwaveformis:
AVERAGE=RE UNDER THE URE FOR ONE FULL EROD
THE EROD
From the above definition, it is evident that the average value of a pure sinusoidal
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waveformoverone fullcycle iszero,asshown in figure1.9.This isbecausetheareaabovethe
horizontalaxisisequalandoppositeoftheareabelowthehorizontalaxis.Further,sincethenet
areaunderthecurve,overonefullperiodiszero,thentheaveragevalueofthesinusoidiszero.
ThisistrueforanysinusoidalorotherACwaveformwhosecenterisonthehorizontalaxis.
Calculatetheaveragevalueofthesquarewaveshowninfigure1.10:
Heretheareaontopofthehorizontalaxis isequalandoppositeoftheareabelowthe
horizontalaxismakingtheaveragevalueofthesquarewaveequaltozero.
Calculatetheaveragevalueofthesquarewaveshowninfigure1.11:
Theareaontopofthehorizontalaxis isgreaterthantheareabelowthehorizontalaxis,
resultinginanaveragevalueforthissquarewave.
Areaoftopportion: A1=(10V)(1ms)=10Vms
Areaofbottomportion: A2=(2V)(1ms)=2Vms
Average=
=
=4V
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TheaverageorDCvalueofthissquarewaveis4Volts.Thesquarewave,showninfigure
1.11,iscomposedbya4VDClevelanda6Vpeakvaluesquarewaveridingonthat4VDC
level,asseeninfigure1.12.
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1.4 RMSorEffectiveValue
Consider the resistor R with the DC voltage source circuit in figure 1.13. The power
dissipatedbytheresistoris:
PDC= I2R
Figure 1.14 shows the same resistor connected to an AC voltage source. The current
suppliedtotheresistoris:
i=Ipsin t
Thepowerdissipatedbytheresistoris:
PAC
=i2R=(I
psin t)
2R =I
p
2sin
2tR
usingthetrigonometricidentity:
sin2t=
(1cos2t)
inaddition,manipulatingthepowerequationtheresultis:
PAC=IP2[
(1cos2t)]R =
cos2t
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Since the above expression for the power has a cosine term, then the average power
suppliedisonlythefirstterm.Sincetheaveragevalueofasinusoidiszero,thentheabovepower
expressionisasinusoidwithtwicethefrequencythatridesonaDClevel.
EquatetheaveragepowerdeliveredbytheACsourcetothatdeliveredbytheDCsourceto
arrivetothefollowingresult:
PACaverage=PDC
(Ip2R)/2=I
2R
Ip2=2I
2
Thus:
IP=2I (110)
Or:
I=
=0.707IP (111)
Equation(111)istheRMSoreffectivevalueofthesinusoidalcurrent.Itrevealsthatthe
sinusoidal current has an equivalent DC value equal to 0.707 of its peak value and effectively
suppliesthesamepowertoaresistorasaDCcurrentofvalue0.707Ip.
ThetermRMScomesfromRootMeanSquareandshowstheprocessofcalculatingthe
effectivevalueofasinusoidusingCalculus;firstsquarethewaveform,thenfindthearea(mean)
underthesquaredwaveformusingintegrationandthentakethesquarerootofthatarea,hence,thetermRootMeanSquare(RMS).
From this point, the subscript eff will be used for the effective value (RMS) of any
sinusoidalwaveformvoltageorcurrent.Therefore:
Ieff=
=0.707IP (112)
inaddition,inthecaseofavoltage:
Veff=
=0.707VP (113)
All the above equations deal with the effective values of sinusoidal waveforms. The
effective values are different for square and triangular waveforms. For these two types of
waveforms,thefollowingcanbederived:
FortheSquareWaveform:
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EffectiveValue=PeakValue (114)
FortheTriangularWaveform:
EffectiveValue=
(115)
InthecasewhereasinusoidisridingonaDClevel,asshowninfigure1.15,theRMSvalue
iscalledthetotalRMSvalueandequation(116)isused:
TotalRMS=VRMS VD
(116)
VRMSistheRMSvalueofthesinusoiditselfandVDCistheDClevelofthesinusoid.
CalculatethetotalRMSvalueofthesinusoidshowninfigure1.16:
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This1.414Voltpeaksinusoidhasa2VDC level.Equation(116) isusedtocalculatethe
effectivevalueofthiswaveform.TheRMSofthe1.414Vsinusoidis(0.707)(1.414)=1V
TotalRMS= 1 2 = 2.236V
1.5
ComplexNumbers
Imaginarynumbersarose from theneed to find the 1, since the square rootofanynegative number does not exist. By definingj = 1 , a new set of numbers is created, theimaginary numbers. Electrical engineers use the term j instead of the term i, as used in
mathematics.Inelectricalengineering,iisusedtodefinecurrent.
Acomplexnumberhasarealpartandanimaginarypart.Thisnumberrepresentsapointin
a twodimensionalplane called the complexplane.Thehorizontalaxisof thecomplexplane is
calledtherealaxisandisthecollectionofallpointsfrom to+.Theverticalaxisiscalledthe
imaginaryaxisandisthecollectionsofallpointsfrom jto+j. Figure1.17showsthecomplex
planewiththereal[Re]andtheimaginary[Im]axes.Complex numbers are represented using either rectangular form or polar form. The
formatoftherectangularformshowninfigure1.18is:
C=A+jB (117)
Theformatofthepolarformshowninfigure1.19is:
t
v Volts
3.414
FIGURE 1.16
0.586
2
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C=Co (118)
Usingfigure1.18onecanconvertfromrectangulartopolarformwiththefollowingrelations:
C=A B (119)
=tan
(120)
Usingfigure1.20onecanconvertfrompolartorectangularformwiththefollowingrelations:
A=Ccos 121
B C sin 122
1.6 ComplexAlgebra
Complexalgebra is simpleanddoesnot requirememorizingany complicated formulas,
whenthetwosimpleruleslistedbelowarefollowed:
RULE#1: Toaddorsubtractuserectangularcoordinatesonly!
RULE#2: Tomultiplyordivideusepolarcoordinatesonly!
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ADDITION
Using rectangular coordinatesadd the realpartsand the imaginaryparts separately to
formtheresultingcomplexnumber. Whileperformingtheoperation,carrythesignofthereal
andimaginarypartsofthenumbers.
AddthecomplexnumbersC1=5+j3 andC2=2+j8
C3=C1+C2=(5+j3)+(2+j8)
realpartofC3: 5+2 =7
imaginarypartofC3: 3+8 =11
Therefore: C3 = 7+j11
SUBSTRACTION
This operation is similar to addition. Subtract the real parts and the imaginary parts
separatelytoformtheresultingcomplexnumber.
MULTIPLICATION
Using polarcoordinates multiply the individual magnitudes together and add the
individual angles together. The resulting magnitude and angle is the product of the complex
numbers.
FindtheproductofC1=210o,C2=525
oandC3=412
o:
C4=C1C2C3=(210o)(525
o)(412
o)
=(2)(5)(4)(10o+25
o+(12
o))
= 4023o
DIVISION
Division is similar to multiplication. Divide the magnitude of the numerator by the
magnitudeof thedenominatorand from theangleof thenumerator subtract theangleof the
denominator.Theresultantmagnitudeandangleisthecomplexnumber.
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Findthequotient C3=C1/C2whenC1=10045o and
C2=2010o:
C3= 10045o /2010
o =535
o
Always followtheaboveruleswhenperformingcomplexoperationsandalwaysconvertthe complex numbers in the appropriate coordinate system before performing complex
operations.
Find C3=C1C2whereC1=3+j4 andC2=432o:
Beforemultiplication,get thenumberC1 intopolar coordinatesas requiredby the rule
governingmultiplication.
C1=3 4 5 and tan
53o
Therefore:C1=553o
and
C3=C1C2=(553o)(432
o) = (5)(2)(53
o+32
o =1085
o
1.7 Phasors
Theradiusvectorshowninfigure1.3,repeatedhereinfigure1.21,iscalledaphasor.This
phasorisusedtorepresentasinusoidinthecomplexplane.Phasorssimplifyalgebraicoperations
ofsinusoids,withthesame
frequency,
usingcomplexalgebra.
Thisradiusvectorisusedwithtwomodifications.Thefirstmodificationisthatitcannotbe
permittedtorotatewithangularvelocity ,butitwillremainstationaryatt=0.Asaresult,the
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relativepositionthatthephasorshavewitheachotherisrevealed. Thisisaccomplishedthrough
thephasordiagram.Thesecondmodificationisthattheeffectivevalueofthesinusoidisusedfor
the magnitude of the phasor, since the effective value iswidely used. The phase angle of the
sinusoidwillbetheangleofthephasor. Ifthesinusoidhasaphaseangleofdegreesthenthephasoratt=0islocateddegreesfromthehorizontal(real)axis.
Usingtheabovemodificationsthephasorofthegeneralsinusoidalvoltagev=Vpsin(t)is:
V=Veff (123)
inaddition,thatofthegeneralsinusoidalcurrenti=Ipsin(t)is:
I=Ieff 124
Example1
Writethefollowingsinusoidalwaveformsinthephasordomain:
a)v=28.28sin(754t+35o)
b)i=10cos754t
Solution:
a)Theeffectivevalueofthesinusoidis: (0.707)(28.28)=20
Thephasorthereforeis: V=2035o
b)i=10cos754t=10sin(754t+90o)
Theeffectivevalueofthesinusoidis: (0.707)(10)=7.07
Thephasorthereforeis: I=7.0790o
Example2
Drawthephasordiagramofthephasorsasinexample1andcalculatethephaseangledifferencebetweenthephasors.
Solution:
Thephasordiagramisshowninfigure1.23.Thecurrentisaheadofthevoltageby55oand
thereforeileadsvby55o.
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1.8 EffectonResistance(R),Inductance(L)andCapacitance(C)
R
ConsidertheresistorRshowninfigure1.23,wherethesinusoidalvoltagevR=Vpsin tis
presentacrossitsterminals. Usingphasorquantitiesthisvoltageis VR=V0o. Theresistance
value isarealnumberanditsrepresentation inthecomplexplanewillbeR0o.UsingOhm's
lawandphasorsthecurrentthroughtheresistoris:
I = R =
R = I0
HereVandIrepresenttheeffectivevaluesofthesinusoidalvoltageandcurrent,respectively.In
thetimedomain,thecurrentthroughtheresistorisequaltoi=Ipsin t.Therefore:
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Forapureresistorthecurrentthroughandthevoltageacrossareinphase(pointingin
thesamedirection).
Thephasordiagramfortheresistorvoltageandcurrent isshown in figure1.24andthetimedomainquantitiesare shown in figure1.25.A finalpoint regarding resistance is that the
valueoftheresistanceisnotaffectedbythefrequency.
L
Theinductorvoltageandcurrentarerelatedbythefollowingrelationship:
VL = L
(125)
WhereL,representstheinductanceandhastheunitofHenry,[H].
Throughtheinductorinfigure1.26thesinusoidalcurrentiL=Ipsin tflows. Thevoltageacross
theinductoris:
VL = L
=L
(IPsint) = LIPcost =VPcost = VP sin(t+900) (126)
where:
Vp= LIp
Observethatthevoltageoftheinductoris90oaheadofthecurrent.Therefore:
Forapure inductor,thecurrentthroughandthevoltageacrossare90ooutofphase.
Thecurrentlagsthevoltageby90o.
Thephasordiagram for the inductorvoltageandcurrent is shown in figure1.27and the time
domainquantitiesareshowninfigure1.28.
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TherelationVp= LIpshowsthatthepeakvalueofthevoltagedependsonthevaluesof andL.
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Since this relation is Ohm's law and the units of voltage and current are volts and amperes
respectively,thentheunitofthequantity LmustbeOhms.Thequantity Liscalledinductive
reactancefromthewaytheinductorreactstosinusoidalquantities,andistheoppositionofthe
inductor to the flow of current. The inductive reactance has the unit of Ohms [] and is
representedwithXLas:
XL= L (127)
Equation (127) is a straightline equation, which passes from the origin. Figure 1.29
showsthegraphofXLvs. .TheslopeofthelineisLandasthevalueoftheinductorincreases,
theslopeofthelinebecomessteeper.When =0(DC),thevalueofthereactanceiszeroOhms
and the inductor isa shortcircuit.As approaches infinity, the reactance isalsoapproaching
infinity.Atveryhighfrequencies,theinductorisanopencircuit,asseeninfigure1.30.
Thecurrentandvoltageoftheinductorinthephasordomainare:
IL=I
0
o
and
VL=V
90
o
Vand Iaretheeffectivevaluesofthesinusoidalvoltageandcurrentrespectively.UsingOhm's
law:
VL=XLIL
andsolvingforXL:
XL =
=
= XL90 = jXL (128)
Itisevidentfromequation(128)thattheinductivereactanceisavectorinthecomplex
planewithafixedmagnitudeofXLandatanangleof90o.Asseeninfigure1.31itisavector(not
aphasor,since itdoesnotrepresentasinusoidalfunction) inthepositive imaginaryaxisofthecomplexplane.
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C
Thecapacitorcurrentandvoltagearerelatedbythefollowingrelationship:
IC = C
(129)
WhereC,representsthecapacitanceandhastheunitofFarad,[F]
Acrossthecapacitor in figure1.32thesinusoidalvoltagevC=Vpsin t ispresent. The
currentthroughthecapacitoris:
IC = C
= C
(VPsint) =CVPcos t = IPsin(t +90o)
where: Ip= CVp
Observethatthecurrentofthecapacitoris90o
aheadofthevoltage.Therefore:
Forapurecapacitor,thecurrentthroughandthevoltageacrossare90ooutofphase.
Thecurrentleadsthevoltageby90o.
Thephasordiagramforthecapacitorcurrentandvoltage isshown infigure1.33,andthetime
domainquantitiesareshowninfigure1.34.
RearrangetheexpressionIp= CVptoget:
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==
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Theunitofthequantity1/C istheOhmsincetheaboveequation isOhm's law. Thequantity
1/C is called the capacitive reactance and is the opposition of the capacitor to the flow of
charge.ThecapacitivereactancehastheunitofOhms[]andisrepresentedwithXC:
XC =
(130)
thus:
Vp=XCIpFigure1.35showsthegraphofXCvs. ( =2f).When approaches0(DC),thevalueof
thereactanceapproachesinfinityandthecapacitorisanopencircuit. As approachesinfinity,
thereactanceisapproachingzero. Inaddition,atveryhighfrequencies,thecapacitor isashort
circuit,asseeninfigure1.36.
Thecurrentandvoltageofthecapacitorinthephasordomainare:
IC=I90o
and
VC=V0o
Vand Iaretheeffectivevaluesofthesinusoidalvoltageandcurrentrespectively.UsingOhm'slaw:
VC=XCIC
solvingforXC: XC =
=
= XC 90o (131)
Itisevidentfromequation(131)thatthecapacitivereactanceisavectorinthecomplex
planewithafixedmagnitudeofXCandanangleof90o. Asseeninfigure1.37,itisavector(notaphasor, since itdoesnot representasinusoidal function) in thenegative imaginaryaxisof the
complexplane.
1.9 AveragePowerandPowerFactor
Considerthegeneralloadshowninfigure1.37wherev=Vpsintand i=Ipsin(t+).
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Thepowertotheloadis: P=vi=VPsint IPsin(t VP IP sin t sin(t +)
Usingthetrigonometricidentity: sin sin
then:
P =
cos
cos (2t + 132
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Theplotsofv,i,andpareshownusingthesamesetofaxesasinfigure1.39.Observethat
thepowercurve isasinusoidwithtwicethe frequencyofthecurrentorvoltage.Observealso
thatpartofthepowersinusoid is locatedbelowthehorizontalaxis.Asshown inthesectionof
reactive power, the part of the sinusoid located below the horizontal axis represents power
returnedtothesourceanditdependsonthesineofangle.Theportionofthepowercurve,on
topoftheaxis,ispowerthatwillbedissipatedbyresistance.Inequation(132),thefirsttermisaconstantfactorandrepresentstheaveragevalueof
thepower.
cos
cos Veff Ieffcos
Therefore,theaveragepoweris:
P=Veff Ieffcos 133
Theangleisthephaseangledifferencebetweenthevoltageandthecurrentandthetermcosisthepowerfactoroftheload.Therefore:
p.f.= cos (134)
R
Inapurelyresistivecircuit,thevoltageandcurrentareinphase.Therefore,=0oandtheaveragepoweris:
P=VeffIeffcos0o=VeffIeff [W]
Thepowerfactorofapurelyresistivecircuitis1sincecos0o=1.
L
In a purely inductive circuit, the phase angle difference between the voltage and the
currentis90o.Therefore,=90oandtheaveragepoweris:
P=VeffIeffcos90o=0W
The power factor of a purely inductive circuit is 0, since cos90o = 0 and there is no power
dissipationbyan ideal inductor.Since inan inductorthecurrent lags,thevoltagean inductive
networkwillhavealaggingpowerfactor.
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C
In a purely capacitive circuit, the phase angle difference between the voltage and the
currentis90o.=90oandtheaveragepowerisasfollows:
P=VeffIeffcos90o
=0W
The power factor of a purely capacitive circuit is 0 since cos90o
= 0 and there is no power
dissipationbyanidealcapacitor.Sinceinacapacitorthecurrentleads,thevoltageinacapacitive
networkwillhavealeadingpowerfactor.
Example3
Calculatethepowerfactorofaloadthathasavoltageacrossitsterminalsv=100sin(314t 35o),
whilethecurrentthroughisi=5x103
sin(314t+15o).
Solution:
Thecurrentleadsthevoltageby50oandthereforethisloadhasaleadingpowerfactor:
p.f.=cos50o
=0.643leading.
1.10 ReactivePower
Considerequation(132)againandtakeitafewstepsfurther.
P =
cos
cos (2t +
Usingthetrigonometricidentity:
cos( + )=cos cos sin sin
Aftermanipulations,wehavethefollowing:
P=VeffIeffcos(1 cos2t) + VeffIeffsin sin 2t 135
Equation(135)willpresentafewinterestingresultsregardingR,LandC.
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L
Thediscussionstartedintheprevioussectionandcontinuesinthissectionbyconsidering
thegeneralcase,showninfigure1.40,ofaloadthathasthefollowingvoltageandcurrent:
v=Vpsint and i=Ipsin(t+)
Iftheloadisanidealinductor,thephaseangleofthecurrentis= 90owhenthecurrentlags the voltage by 90
o. Substituting this phase angle in equation (135) gives the following
results:
P=VeffIeffcos90(1 cos2t) + VeffIeffsin90 sin 2t VeffIeff sin 2t 136
sincecos(90o)=0andsin(90
o)= 1
Thesinusoidofequation(136)isshown infigure1.41.This isasinusoid,withtwicethe
frequencyofthecurrentorvoltage,apeakvalueequal toVeffIeff,andhasnoaveragevalueas
seenfrombothequation(136)andfigure1.41.Observethatoveronefullcycle,theareaontop
ofthehorizontalaxisisequaltotheareabelowtheaxis.Theareaontoprepresentsthepower
suppliedbythesourceandtheareabelowrepresentsthepowerreturnedbytheinductor.Thissuggeststhatthe ideal inductordoesnotdissipateanyenergyandthatthere isanexchangeof
power between the source and the inductor every quarter cycle of the current or voltage
sinusoid.
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Thepowerexchangedbetweenthesourceandtheinductoriscalledreactivepower.The
symbolusedforreactivepowerisQandtheunitofmeasureistheVAR(VoltAmpereReactive).
Reactivepoweriscalculatedasfollows:
Q=VeffIeffsin [VAR] (137)
isthephaseanglebetweenthevoltageandthecurrent. Fortheinductorthephaseangleis90oandthereactivepowerisasfollows:
QL=VeffIeff [VAR]
In conclusion, an ideal inductor does not consume any average power, but receives
reactive power from the source and returns the same to the source without dissipating any
energy.
C
Considernowthattheloadisanidealcapacitorasshowninfigure1.42. Thephaseangle
ofthecurrentis=90o,sincethecurrentleadsthevoltageby90oinacapacitor.Substitutingthisphaseangleinequation(135)givesthefollowing:
P=VeffIeffcos90(1 cos2t) + VeffIeffsin90 sin 2t VeffIeff sin 2t 138
sincecos90o=0andsin90
o=1.
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Thesinusoidofequation(138)isshown infigure1.43.This isasinusoid,withtwicethe
frequencyofthecurrentorvoltage,apeakvalueequaltoVeffIeff,andhasnoaveragevaluejust
liketheoneoftheinductor.Observethatoveronefullcycletheareaontopofthehorizontalaxis
isequaltotheareabelowtheaxis. Theidealcapacitordoesnotdissipateanyenergy. Thereisan
exchangeofpowerbetweenthesourceandthecapacitoreveryquartercycleofthecurrentor
voltage sinusoid. This power is the reactive power of equation (137). For the capacitor, the
phaseangleis90o
andthereactivepoweris:
QC=VeffIeff [VAR]
In conclusion, an ideal capacitor does not consume any average power but receives
reactive power from the source and returns the same to the source without dissipating any
energy.
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R
Considernowthattheloadisaresistorasshowninfigure1.44. Thephaseangleofthe
currentis=0o,sincecurrentandvoltageareinphaseinaresistor.Substitutingthisphaseangleinequation(135)givesthefollowing:
P=VeffIeffcos0(1 cos2t) + VeffIeffsin0 sin 2t VeffIeff VeffIeffcos2t (139)sincecos0
o=1andsin0
o=0
Thesinusoidofequation(139)isshowninfigure1.45.Itisaninvertedcosinewave,with
twicethefrequencyofthecurrentorvoltage,apeakvalueequaltoVeffIeff,and averagevalue
equaltoVeffIeff. Observethatthepowercurveiscompletelyabovethehorizontalaxis.Thismeans
thatallthepowerdeliveredtotheresistorwillbedissipatedasheat. Asaresult,theresistorsees
onlyaveragepowerandusingequation(137)thereactivepowerforaresistoris:
VeffIeffsin0o
=0 VAR.
Example4
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ThesourcevoltageofanetworkisV=200o
andthecurrentsuppliedtothenetworkisI=10
30o.Calculatetheaveragepowerandthereactivepowersuppliedtothenetwork.
Solution:
Thephaseangledifferencebetweenthevoltageandthecurrentis30o,andtheeffective
valuesofthesinusoidsarealreadygiveninthephasors.Therefore:
P=VeffIeffcos = (20)(10)cos30o
=173.2 W
Q=VeffIeffsin = (20)(10)sin30o
=100 VAR
1.11
Apparent
Power
Apparent power is the product of the applied voltage and produced current. It is the
power that is apparently delivered to the load like in DC circuits. However, the power factor
decideshowmuchpower isdissipatedbya loadand then there is the continualexchangeof
reactive power between the source and loads that contain reactive elements. Therefore, the
productoftheappliedvoltageandproducedcurrentisnotalwaysthepowerdeliveredtoaload,
unless thepower factor is1 (purely resistive load).Thisproduct isusefulasapower ratingof
electricalcomponentsandsystems.
ApparentpowerisrepresentedbySandhastheunitoftheVoltAmpere(VA),sinceitis
theproductofvoltageandcurrent.
S=VeffIeff [VA] (140)
Now fromthedefinitionofapparentpower theaveragepowerandthereactivepower
canbewrittenas:
P=VeffIeffcos=Scos [W] (141)
Q=VeffIeffsin=Ssin [VAR] (142)
Fromequation(141)solveforcos.
Powerfactor=pf=cos=S
(143)
Thepowerfactorofanetworkistheratiooftheaveragepowertotheapparentpower.
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Example5
Calculatetheapparentpowerdeliveredtothenetworkofexample4.
Solution:
S=VeffIeff=(20)(10)=200VA
1.12 Impedance
The previous sections discussed the effect on R, L and C caused by the sinusoidal
waveform. Fromthesediscussions,thefollowingareconcluded:
TheResistanceisarealnumberlocatedonthepositiverealaxisofthecomplexplane:
R=R0o.
The inductorhasan InductiveReactance,an imaginarynumber locatedon thepositive
imaginaryaxisofthecomplexplane:
XL=XL90o=jXL=jL.
ThecapacitorhasaCapacitiveReactance,animaginarynumber,locatedonthenegative
imaginaryaxisofthecomplexplane:
XC=XC 90o= jXC= j/C
since j=1/j then XC=1/jC
Theplotsofresistance,inductiveandcapacitivereactances,onthecomplexplane,areshownin
figure1.46.Here[Re]istherealaxisand[Im]istheimaginaryaxis.
AnyelementbyitselforanycombinationsoftheseelementsiscalledtheImpedanceofa
circuit.ThequantityofimpedanceisrepresentedwiththecapitalletterZandthediagraminfigure
1.46 is called the Impedance diagram. The unit of the impedance is the Ohm [], since both
resistanceandreactanceareinOhms.
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Impedance isameasureofhowmuchthecircuitwill impede(oppose)currentflow.The
impedanceisnotaphasorbutavectorquantityinthecomplexplaneandcanberepresentedin
eitherrectangularorpolarcoordinates.Iftheimpedancediagramisinthefirstquadrant,thenthe
circuit is inductivewitha laggingpowerfactor.Thecurrentwillthen lagthevoltagebyanangle
equaltotheimpedanceangle.Iftheimpedancediagramisinthefourthquadrant,thenthecircuit
iscapacitivewithaleadingpowerfactor.Thecurrentwillleadthevoltagebyanangleequaltothe
impedanceangle.The total impedanceofelements connected in series isequal to sumof the
individualimpedances.
Example6
Drawtheimpedancediagramfortheseriesconnectionshowninfigure1.47.
Solution:
Thetotalimpedanceis:
ZT=Z
1+Z
2+Z
3
=690o + 4 90
o +100
o
=j6 j4+10
=10+j2 =10.211.3o
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Theimpedancediagramisshowninfigure1.48.
The circuit has both an inductive and a capacitive reactance. However, the inductive
reactance is largerthanthecapacitivemakingthe impedance inductive.Here,thepower
factorislaggingandtheimpedancediagramislocatedinthe1st
quadrant.
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1.13 PowerTriangleandComplexPower
In the previous sections, the three components of AC power were defined to be the
averagepowerP,thereactivepowerQ,andtheapparentpowerS,whicharerelatedasfollows:
S=VeffIeff P=Scos and Q=Ssin
isthephaseanglebetweenthevoltageandthecurrent.
Theabove relationsconclude that theACpowerquantitiesarevectors,which forma
righttriangleandasvectorsarerelatedbythefollowingrelationship:
S=P+Q (144)
Forinductiveloadstheapparentpower S=P+jQL andthepowertriangleisinthefirstquadrant
asseeninfigure1.49.Forcapacitiveloadstheapparentpower S=P jQCandthepowertriangle
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isinthe4th
quadrant,asseeninfigure1.50.
When both inductive and capacitive loads are present, the total reactive power is
determinedfromthedifferenceofQL QC.Thepowertriangleiseitherinthe1st
quadrant,fora
positive reactive power (inductive) or in the 4th
quadrant for a negative reactive power
(capacitive). Thepowertrianglecanbedeterminedfromtheimpedancetrianglebymultiplying
each component of the impedance diagram by the current squared (I2
). This is shown in theimpedancediagraminfigure1.51.MultiplicationbyI
2yieldsthepowertriangleoffigure1.52.
Sincethethreepowersformarighttriangle,thePythagoreanTheoremrelatesthem.
S=P Q (145)
Example7
Constructthepowertriangleofthenetworkoffigure1.53.
Solution:
ZT=6+j8=1053.1o
I=E/ZT=(500o)/(1053.1
o)=553.1
o
P=I2R=(5)
26=150W
QL=I2XL=(5)
28=200VAR
S=I
2
ZT=(5)
2
10=250VAandalternatively,fromequation(145):
S=150 200 = 250VA
Thepowertriangle isshown infigure1.54.Theangle isthe impedanceanglebut itcan
alsobecalculatedas:
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cos1
(150/250)=53.1o
Theapparentpowerofanetworkcanbe calculated invector formusingthe following
equation:
S=VI*
(146)HereVisthevoltageappliedacrossthenetworkandI
*isthecomplexconjugateofthecurrent
supplied to the network. Equation (146) is called the complex power of the network. The
complexconjugateofacomplexnumber isfoundbysimplychangingthesignofthe imaginary
part intherectangularcoordinatesystemorbychangingthesignoftheangle inthepolarco
ordinatesystem.
Example8
Calculatethecomplexpowerforthenetworkofexample7.
Solution:
Fromexample7:
E=500o
and I=553.1o
Therefore:
I*=553.1
o
and
S=(500o)(553.1
o)= 25053.1
o VA
1.14 TotalP,QandS
Tocalculatethetotalaveragepowerofanetwork,addtheindividualaveragepowersof
thebranches.This is independentofwhether thebranchesareconnected in seriesorparallel
connections.Inasimilarfashion,tofindthetotalreactivepowerofanetwork,addtheindividual
reactive powers of the branches. Inductive reactive power is positive and capacitive reactive
powerisnegative. Afterboththetotalaveragepowerandthereactivepowerarecalculatedthen
useequation(145) tocalculatethetotalapparentpowerofanetwork.Thisisdemonstratedintheexamplethatfollows.
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Example9
a) CalculatethetotalP,QandSforthenetworkoffigure1.55.
b) Drawthepowertriangleandcalculatethepowerfactor.
Solution:
a) Thetotalaveragepoweris:
PT=400W+200W+500W+100W=1200W
Thetotalreactivepoweris:
QT= 100VAR+400VAR 1000VAR
= 700VAR
=700VARcapacitive
Thetotalapparentpoweris:
S=1200 700 = 1389.24VA
b) Thepowertriangle isshown in figure1.56. It isacapacitivenetworkandtherefore the
powertriangleislocatedinthe4th
quadrant.
Thepowerfactoris: pf=P/S=1200/1384.24=0.864
and =cos1
0.864=30.3o
1.15 PowerFactorCorrection
Mostof the real life loadsare inductive loadswith laggingpower factor.Power factor
correctionistheprocessofintroducingreactiveelements(capacitors)tobringthepowerfactor
closertounity(resistivenetwork).TheimprovementofthepowerfactortounitymeansthatS=
PandQ=0.Asa result,thereare lower levelsofSand thecurrent isminimum.Thus,power
factorcorrectionminimizesthecurrentrequirementsbycalculatingthecapacitorneededtobe
placedinparallelwiththesystemthatrequirespowerfactorcorrection.
Example10
A loadconnectedtoa120V,60Hzsupplyis2kWresistiveand 1.25kVARinductive.Calculate
thecapacitancerequiredforunitypowerfactor.
Solution:
Thepowertriangleisshowninfigure1.57.Theapparentpoweris:
S=2000 1250 = 2358.5VA
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thepowerfactoris:
cos =(2000)/(2358.5)=0.848lagging
inaddition,thecurrentsuppliedbythesourceis:
I=S/V=(2358.5VA)/(120V)=19.65A
Withtheunitypowerfactorthepowertrianglebecomesthestraight line infigure1.58,
andS=P=2000VA.Forunitypowerfactoraddacapacitivereactivepowerequaltothe
inductivereactivepowerseeninfigure1.57.ThereforeQC=1250VARand:
QC =
X XC =
=
= 11.52
C =
X =
X
= =
X . X X .= 230.4 F
Then,connecta230.4 Fcapacitor inparallelandthepowerfactorbecomesunity.Thecurrentsuppliedbythesourceisnow:
I=S/V=(2000VA)/(120V)=16.67A
thisislessthanthecurrentsuppliedwithoutthepowerfactorcorrection.
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1.16 Summary
Inthiscourse,wediscusssomebasicquantitiesassociatedwithACwaveforms.
Wepresentsinusoids,squareandtriangularwaves.
Wedefinethefollowingterms:periodicwaveform,period,cycle,frequency,peakvalue,peak
topeakvalue,instantaneousvalue,angularvelocity,averagevalue,RMSoreffectivevalue
andtotalRMS.
Westudyphaserelations,phasors,complexnumbersandcomplexalgebraandhowithelps
withphasors.
Thetermslags,leadsandinphaseandoutofphasearepresented.
TheeffectoftheR,LandC elementsisdiscussed.
Averagepower,powerfactor,reactivepower,apparentpower,impedanceandimpedance
triangle,powertriangleandcomplexpower,totalP,QandSandfinallypowerfactor
correctionarepresented.
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