ac waveforms

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PDHeng inee r . com Course E-4014

AC Waveforms: Basic Quantities for Non-

Electrical Engineers

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Copyright 2009 Peter Basis, PE

ACWaveforms:BasicQuantitiesPeterBasis


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Copyright 2009 Peter Basis,PE

2

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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3

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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5

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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16


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17

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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27

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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28

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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29

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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34

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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39

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.

Documents

ac waveforms