ChapterChapter 32A32A –– ACAC CircuitsCircuits

AAA PowerPointPowerPointPowerPoint PresentationPresentationPresentation bybyby PaulPaulPaul E.E.E. Tippens,Tippens,Tippens, ProfessorProfessorProfessor ofofof PhysicsPhysicsPhysics SouthernSouthernSouthern PolytechnicPolytechnicPolytechnic StateStateState UniversityUniversityUniversity

© 2007 Objectives:Objectives: AfterAfter completingcompleting thisthis module,module, youyou shouldshould bebe ableable to:to: •• DescribeDescribe thethe sinusoidalsinusoidal variationvariation inin acac currentcurrent andand voltagevoltage,, andand calculatecalculate theirtheir effectiveeffective values.values. •• WriteWrite andand applyapply equationsequations forfor calculatingcalculating thethe inductiveinductive andand capacitivecapacitive reactancesreactances forfor inductorsinductors andand capacitorscapacitors inin anan acac circuit.circuit. •• Describe,Describe, withwith diagramsdiagrams andand equations,equations, thethe phasephase relationshipsrelationships forfor circuitscircuits containingcontaining resistance,resistance, capacitance,capacitance, andand inductanceinductance.. ObjectivesObjectives (Cont.)(Cont.)

•• WriteWrite andand applyapply equationsequations forfor calculatingcalculating thethe impedanceimpedance,, thethe phasephase angleangle,, thethe effectiveeffective currentcurrent,, thethe averageaverage powerpower,, andand thethe resonantresonant frequencyfrequency forfor aa seriesseries acac circuit.circuit. •• DescribeDescribe thethe basicbasic operationoperation ofof aa stepstep-- upup andand aa stepstep--downdown transformertransformer..

•• WriteWrite andand applyapply thethe transformertransformer equationequation andand determinedetermine thethe efficiencyefficiency ofof aa transformer.transformer. AlternatingAlternating CurrentsCurrents

AnAnAn alternatingalternatingalternating currentcurrentcurrentsuchsuch such asasas thatthatthat producedproducedproduced bybyby aaa generatorgeneratorgenerator hashashas nonono directiondirectiondirection ininin thethethe sensesensesense thatthatthat directdirectdirect currentcurrentcurrent has.has.has. TheTheThe magnitudesmagnitudesmagnitudes varyvaryvary sinusoidallysinusoidallysinusoidallywithwith with timetimetime asasas givengivengiven by:by:by: AC-voltage Emax and current iimax

E = Emax sin  time, t

i = imax sin  RotatingRotating VectorVector DescriptionDescription

TheThe coordinatecoordinate ofof thethe emfemf atat anyany instantinstant isis thethe value of E sin Observe for incremental value of Emaxmax sin Observe for incremental anglesangles inin stepssteps ofof 454500.. SameSame isis truetrue forfor ii..

E E = Emax sin 

 1800 2700 3600 450 900 1350

RadiusR = = E Emaxmax EffectiveEffective ACAC CurrentCurrent TheThe averageaverage currentcurrent iimax I = i inin aa cyclecycle isis zerozero—— max halfhalf ++ andand halfhalf --.. ButBut energyenergy isis expended,expended, regardless of direction. regardless of direction. 2 So the “root-mean- II So the “root-mean- Irms  squaresquare”” valuevalue isis useful.useful. 20.707

TheThe rmsrms valuevalue IIrmsrms isis The effective ac current: sometimessometimes calledcalled thethe ieff = 0.707 imax effectiveeffective currentcurrent IIeffeff:: ACAC DefinitionsDefinitions OneOne effectiveeffective ampereampere isis thatthat acac currentcurrent forfor whichwhich thethe powerpower isis thethe samesame asas forfor oneone ampereampere ofof dcdc current.current. Effective current: i = 0.707 i Effective current: ieffeff = 0.707 imaxmax OneOne effectiveeffective voltvolt isis thatthat acac voltagevoltage thatthat givesgives anan effectiveeffective ampereampere throughthrough aa resistanceresistance ofof oneone ohm.ohm.

Effective voltage: V = 0.707 V Effective voltage: Veffeff = 0.707 Vmaxmax ExampleExample 1:1: ForFor aa particularparticular device,device, thethe househouse acac voltagevoltage isis 120120--VV andand thethe acac currentcurrent isis 1010 AA.. WhatWhat areare theirtheir maximummaximum values?values? i = 0.707 i V = 0.707 V ieffeff = 0.707 imaxmax Veffeff = 0.707 Vmaxmax

i 10 A V 120V i eff V eff max 0.707 0.707 max 0.707 0.707

i = 14.14 A V = 170 V imaxmax= 14.14 A Vmaxmax= 170 V

TheTheThe acacac voltagevoltagevoltage actuallyactuallyactually variesvariesvaries fromfromfrom +170+170+170 VVV tototo -170--170170 VVV andandand thethethe currentcurrentcurrent fromfromfrom 14.114.114.1 AAA tototo –14.1––14.114.1 A.AA.. PurePure ResistanceResistance inin ACAC CircuitsCircuits R Vmax Voltage i A V imax Current

a.c. Source

VoltageVoltageVoltage andandand currentcurrentcurrent areareare ininin phase,phase,phase, andandand OhmOhmOhm’s’’ss lawlawlaw appliesappliesapplies forforfor effectiveeffectiveeffective currentscurrentscurrents andandand voltages.voltages.voltages.

Ohm’s law: Veff = ieffR i I ACAC andand InductorsInductorsI i

Inductor Inductor 0.63I CurrentCurrent CurrentCurrent RiseRise 0.37I DecayDecay

 Time, t  Time, t

TheThe voltagevoltage VV peakspeaks first,first, causingcausing rapidrapid riserise inin ii currentcurrent whichwhich thenthen peakspeaks asas thethe emfemf goesgoes toto zero.zero. VoltageVoltage leadsleads ((peakspeaks beforebefore)) thethe currentcurrent byby 90900.. VoltageVoltage andand currentcurrent areare outout ofof phasephase.. AA PurePure InductorInductor inin ACAC CircuitCircuit

L Vmax Voltage i A V imax Current

a.c.

TheTheThe voltagevoltagevoltage peakspeakspeaks 90909000 before beforebefore thethethe currentcurrentcurrent peaks.peaks.peaks. OneOneOne buildsbuildsbuilds asasas thethethe otherotherother fallsfallsfalls andandand vicevicevice versa.versa.versa.

TheThe reactancereactance maymay bebe defineddefined asas thethe nonresistivenonresistive oppositionopposition toto thethe flowflow ofof acac current.current. InductiveInductive ReactanceReactance

TheThe backback emfemf inducedinduced L byby aa changingchanging currentcurrent A providesprovides oppositionopposition toto V current,current, calledcalled inductiveinductive

reactancereactance XXL.. a.c. SuchSuch losseslosses areare temporarytemporary,, however,however, sincesince thethe currentcurrent changeschanges directiondirection,, periodicallyperiodically rere--supplyingsupplying energyenergy soso thatthat nono netnet powerpower isis lostlost inin oneone cycle.cycle.

InductiveInductive reactancereactance XXL isis aa functionfunction ofof bothboth thethe inductanceinductance andand thethe frequencyfrequency ofof thethe acac current.current. CalculatingCalculating InductiveInductive ReactanceReactance L Inductive Reactance: A V XfLL 2 Unit is the  Ohm's law: ViX a.c. L L

TheThe voltagevoltage readingreading VV inin thethe aboveabove circuitcircuit atat thethe instantinstant thethe acac currentcurrent isis ii cancan bebe foundfound fromfrom thethe inductanceinductance inin HH andand thethe frequencyfrequency inin HzHz..

ViL  (2 fL ) Ohm’s law: VL = ieffXL ExampleExample 2:2: AA coilcoil havinghaving anan inductanceinductance ofof 0.60.6 HH isis connectedconnected toto aa 120120--VV,, 6060 HzHz acac source.source. NeglectingNeglecting resistance,resistance, whatwhat isis thethe effectiveeffective currentcurrent throughthrough thethe coil?coil? XX Reactance:Reactance: XXLL == 22fLfL L = 0.6 H A LL == 22(60(60 Hz)(0.6Hz)(0.6 H)H) V

XXL == 226226  L 120 V, 60 Hz

Veff 120V ieff  i = 0.531 A ieffeff = 0.531 A X L 226 

ShowShow thatthat thethe peakpeak currentcurrent isis IImaxmax == 0.7500.750 AA Q ACAC andand CapacitanceCapacitanceI i

q Capacitor Capacitor max 0.63 I RiseRise inin CurrentCurrent ChargeCharge 0.37 I DecayDecay  Time, t  Time, t

TheThe voltagevoltage VV peakspeaks ¼¼ ofof aa cyclecycle afterafter thethe currentcurrent ii reachesreaches itsits maximum.maximum. TheThe voltagevoltage lagslags thethe current.current. CurrentCurrent ii andand VV outout ofof phasephase.. AA PurePure CapacitorCapacitor inin ACAC CircuitCircuit C Vmax Voltage i A V imax Current

a.c.

TheTheThe voltagevoltagevoltage peakspeakspeaks 90909000 after afterafter thethethe currentcurrentcurrent peaks.peaks.peaks. OneOneOne buildsbuildsbuilds asasas thethethe otherotherother fallsfallsfalls andandand vicevicevice versa.versa.versa.

TheTheThe diminishingdiminishingdiminishing currentcurrentcurrent iii buildsbuildsbuilds chargechargecharge ononon CCC whichwhich increasesincreases thethe backback emfemf ofof VV which increases the back emf of VCC.. CapacitiveCapacitive ReactanceReactance EnergyEnergy gainsgains andand losseslosses C areare alsoalso temporarytemporary forfor A capacitorscapacitors duedue toto thethe V constantlyconstantly changingchanging acac current.current. a.c. NoNo netnet powerpower isis lostlost inin aa completecomplete cycle,cycle, eveneven thoughthough thethe capacitorcapacitor doesdoes provideprovide nonresistivenonresistive oppositionopposition ((reactancereactance)) toto thethe flowflow ofof acac current.current.

CapacitiveCapacitive reactancereactance XXC isis affectedaffected byby bothboth thethe capacitancecapacitance andand thethe frequencyfrequency ofof thethe acac current.current. CalculatingCalculating InductiveInductive ReactanceReactance C Capacitive Reactance: 1 A V X  Unit is the  C 2 fC

a.c. Ohm's law: VCC iX

TheThe voltagevoltage readingreading VV inin thethe aboveabove circuitcircuit atat thethe instantinstant thethe acac currentcurrent isis ii cancan bebe foundfound fromfrom thethe inductanceinductance inin FF andand thethe frequencyfrequency inin HzHz.. i V  Ohm’s law: V = i X L 2 fL C eff C ExampleExample 3:3: AA 22--FF capacitorcapacitor isis connectedconnected toto aa 120120--V,V, 6060 HzHz acac source.source. NeglectingNeglecting resistance,resistance, whatwhat isis thethe effectiveeffective currentcurrent throughthrough thethe coil?coil? 1 C = 2 F Reactance:Reactance: X C  2 fC A 1 V X  C 2 (60Hz)(2 x 10-6 F) 120 V, 60 Hz XXC == 13301330 

Veff 120V ieff  i = 90.5 mA ieffeff = 90.5 mA X C 1330 

ShowShow thatthat thethe peakpeak currentcurrent isis iimaxmax == 128128 mAmA MemoryMemory AidAid forfor ACAC ElementsElements

An old, but very “E LL i” An old, but very theE effective,effective, wayway toto rememberremember thethe phasephase “I CE” manE differencesdifferences forfor inductorsinductors andand capacitorscapacitors isis ::

““EE LL II”” thethe ““ii CC EE”” ManMan

EmfEmfEmf EEEisisis beforebeforebeforecurrentcurrent current iii ininin inductorsinductorsinductors LLL;;; EmfEmfEmf EEEisisis afterafterafter currentcurrentcurrent iii ininin capacitorscapacitorscapacitors C.C.C. FrequencyFrequency andand ACAC CircuitsCircuits ResistanceResistance RR isis constantconstant andand notnot affectedaffected byby f.f. 1 InductiveInductive reactancereactance XX R,R, XX L XfLL 2 XC  variesvaries directlydirectly withwith 2 fC frequencyfrequency asas expectedexpected sincesince EE   .. i/i/tt XXCC XXLL X CapacitiveCapacitive reactancereactance XCC variesvaries RR inverselyinversely withwith ff sincesince rapidrapid acac allowsallows littlelittle timetime forfor chargecharge toto buildbuild upup onon capacitors.capacitors. ff SeriesSeries LRCLRC CircuitsCircuits

VT Series ac circuit R A a.c. L C

VL VR VC

ConsiderConsiderConsider ananan inductorinductorinductorLLL,,,aa a capacitorcapacitorcapacitorCCC,,,andandand aaa resistorresistorresistorRRRallallall connectedconnectedconnected ininin seriesseriesserieswithwith with ananan acacac sourcesourcesource... TheTheThe instantaneousinstantaneousinstantaneous currentcurrentcurrent andandand voltagesvoltagesvoltages cancancan bebebe measuredmeasuredmeasured withwithwith meters.meters.meters. PhasePhase inin aa SeriesSeries ACAC CircuitCircuit TheThe voltagevoltage leadsleads currentcurrent inin anan inductorinductor andand lagslags currentcurrent inin aa capacitor.capacitor. InIn phasephase forfor resistanceresistance RR.. V = Vmax sin 

VL  180V 0 2700 3600 450 900 1350 VR VC

RotatingRotating phasorphasor diagramdiagram generatesgenerates voltagevoltage waveswaves forfor eacheach elementelement RR,, LL,, andand CC showingshowing phasephase relations.relations. CurrentCurrent ii isis alwaysalways inin phasephase withwith VVR.R. PhasorsPhasors andand VoltageVoltage

AtAt timetime tt == 0,0, supposesuppose wewe readread VVLL,, VVRR andand VVCC forfor anan acac seriesseries circuit.circuit. WhatWhat isis thethe sourcesource voltagevoltage VVT??

Source voltage Phasor Source voltage Diagram VVT VL VL - VC 

VR VR VC

WeWe handlehandle phasephase differencesdifferences byby findingfinding thethe vectorvector sumsum ofof thesethese readings.readings. VVTT == VVi.. TheThe angleangle  isis thethe phasephase angleangle forfor thethe acac circuit.circuit. CalculatingCalculating TotalTotal SourceSource VoltageVoltage

Source voltage TreatingTreating asas vectors,vectors, wewe find:find:

VVT 22 VL - VC VVVV()  TRLC V VV R tan  L C VR

NowNow recallrecall that:that: VV == == ;; andand VV == RR iRiR;; LL iXiXLL CC iViVCC SubstitutionSubstitution intointo thethe aboveabove voltageVvoltageV equationequation gives:gives:

22 ViRXXTLC() ImpedanceImpedance inin anan ACAC CircuitCircuit

X Impedance 22 ViRXXTLC() ZZ L - XC  ImpedanceImpedance ZZ isis defined:defined: R 22 ZRXX()LC

V OhmOhm’’ss lawlaw forfor acac currentcurrent ViZi or T andand impedance:impedance: T Z

TheThe impedanceimpedanceisis thethe combinedcombined oppositionopposition toto acac currentcurrent consistingconsisting ofof bothboth resistanceresistance andand reactance.reactance. ExampleExample 3:3: AA 6060-- resistor,resistor, aa 0.50.5 HH inductor,inductor, andand anan 88--FF capacitorcapacitor areare connectedconnected inin seriesseries withwith aa 120120--V,V, 6060 HzHz acac source.source. CalculateCalculate thethe impedanceimpedance forfor thisthis circuit.circuit. 1 XfLX2 and 0.5 H LC2 fC A 8 F X L 2 (60Hz)(0.6 H) = 226 120 V  1 X 332 C 2 (60Hz)(8 x 10-6 F) 60 Hz 60 

222 2 ZRXX(LC )  (60  ) (226 332 )

Thus,Thus, thethe impedanceimpedance is:is: Z = 122  ExampleExample 4:4: FindFind thethe effectiveeffective currentcurrent andand thethe phasephase angleangle forfor thethe previousprevious example.example. XXLL == 226226 XXCC == 332332  RR == 6060 ZZ == 122122  V 120 V i T 0.5 H eff Z 122  A 8 F i = 0.985 A ieffeff = 0.985 A 120 V NextNext wewe findfind thethe phasephase angleangle:: 60 Hz 60  X Impedance XXL –– XXC == 226226 –– 332332 == --106106  ZZ L - XC XX  RR == 6060  tan  L C R R ContinuedContinued ...... ExampleExample 44 (Cont.):(Cont.): FindFind thethe phasephase angleangle forfor thethe previousprevious example.example.

60   XXL –– XXC == 226226 –– 332332 == --106106  -106 XX  ZZ RR == 6060  tan  L C R

106 tan  = -60.500 60 = -60.5

TheTheThe negativenegativenegativephasephase phase angleangleangle meansmeansmeans thatthatthat thethethe acacac voltagevoltagevoltage lagslagslags thethethe currentcurrentcurrent bybyby 60.560.560.500... ThisThisThis isisis knownknownknown asasas aaa capacitivecapacitivecapacitivecircuit.circuit. circuit. ResonantResonant FrequencyFrequency

BecauseBecauseBecauseinductanceinductance inductancecausescauses causes thethethe voltagevoltagevoltage tototo leadleadlead thethethe currentcurrentcurrent andandand capacitancecapacitancecapacitance causescausescauses ititit tototo laglaglagthethe the current,current,current, theytheythey tendtendtend tototo cancelcancelcanceleacheach each otherotherother out.out.out.

X ResonanceResonance (Maximum(Maximum Power)Power) L XXLL == XXCC occursoccurs whenwhen XXL == XXC X R ZRXX 22()  R C LC

1 1 ResonantResonant rr 2 fL  fr  XXLL == XXCC ff 2 fC 2 LC  ExampleExample 5:5: FindFind thethe resonantresonant frequencyfrequency forfor thethe previousprevious circuitcircuit example:example: LL == .5.5 H,H, CC == 88 FF

1 Resonance X = X fr  L C 2 LC 0.5 H

1 A f  8 F 2 (0.5H)(8 x 10-6 F 120 V

Resonant f = 79.6 Hz ? Hz 60  Resonant frr = 79.6 Hz

AtAtAt resonantresonantresonant frequency,frequency,frequency, theretherethere isisis zerozerozero reactancereactancereactance (((onlyonlyonly resistanceresistanceresistance))) andandand thethethe circuitcircuitcircuit hashashas aaa phasephasephase angleangleangle ofofof zero.zero.zero. PowerPower inin anan ACAC CircuitCircuit

NoNoNo powerpowerpower isisis consumedconsumedconsumed bybyby inductanceinductanceinductance ororor capacitance.capacitance.capacitance. ThusThusThus powerpowerpower isisis aaa functionfunctionfunction ofofof thethethe componentcomponentcomponent ofofof thethethe impedanceimpedanceimpedance alongalongalong resistance:resistance:resistance:

X Impedance InIn termsterms ofof acac voltage:voltage: ZZ L - XC PP == iViV cos cos  R InIn termsterms ofof thethe resistanceresistance R:R: PP lostlost inin RR onlyonly PP == ii22RR

TheThe fractionfraction CosCos  isis knownknown asas thethe powerpower factor.factor. ExampleExample 6:6: WhatWhat isis thethe averageaverage powerpower lossloss forfor thethe previousprevious example:example: VV == 120120 V,V,  == --60.560.50,, ii == 90.590.5 A,A, andand RR == 6060 ..

2 PP == ii2RR == (0.0905(0.0905 A)A)2(60(60  Resonance XL = XC 0.5 H AverageAverage PP== 0.4910.491 WW A 8 F TheThe powerpower factorfactor is:is: CosCos 60.560.50 120 V

CosCos == 0.4920.492 oror 49.2%49.2% ? Hz 60 

TheTheThe higherhigherhigher thethethe powerpowerpower factor,factor,factor, thethethe moremoremore efficientefficientefficientisis is thethethe circuitcircuitcircuit ininin itsitsits useuseuse ofofof acacac power.power.power. TheThe TransformerTransformer AA transformertransformer isis aa devicedevice thatthat usesuses inductioninduction andand acac currentcurrent toto stepstep voltagesvoltages upup oror down.down.

AnAnAn acacac sourcesourcesource ofofof emfemfemf Transformer EE isis connectedconnected toto Epp is connected to a.c. primaryprimaryprimary coilcoilcoil withwithwith NNNp p R turns.turns.turns. SecondarySecondarySecondary hashashas NN turnsturns andand emfemf ofof Nss turns and emf of E .. Np Ns EEss. Induced   Induced E N E N emf’semf’s are: are: PPt SSt TransformersTransformers (Continued):(Continued):

Transformer  EPPN a.c. t

Np Ns  ESSN R t

RecognizingRecognizing thatthat //tt isis thethe samesame inin eacheach coil,coil, wewe dividedivide firstfirst relationrelation byby secondsecond andand obtain:obtain: E N TheThe transformertransformer P  P equation:equation: ESSN ExampleExample 7:7: AA generatorgenerator producesproduces 1010 AA atat 600600 V.V. TheThe primaryprimary coilcoil inin aa transformertransformer hashas 2020 turns.turns. HowHow manymany secondarysecondary turnsturns areare neededneeded toto stepstep upup thethe voltagevoltage toto 24002400 V?V?

ApplyingApplying thethe I = 10 A; Vp = 600 V transformer equation: transformer equation: a.c. VN N N P  P 20 p s VNSS turns R

NV (20)(2400V) N PS N = 80 turns S NSS = 80 turns VP 600V

ThisThisThis isisis aaa stepstepstep-up--upup transformertransformertransformer;;; reversingreversingreversing coilscoilscoils willwillwill makemakemake ititit aaa stepstepstep-down--downdown transformer.transformer.transformer. TransformerTransformer EfficiencyEfficiency ThereThere isis nono powerpower gaingain inin steppingstepping upup thethe voltagevoltage sincesince voltagevoltage isis increasedincreased byby reducingreducing current.current. InIn anan idealideal transformertransformer withwith nono internalinternal losses:losses:

Ideal Transformer AnAn idealideal a.c. transformer:transformer:

Np Ns iP ES EEPPii SS or R isPE

TheTheThe aboveaboveabove equationequationequation assumesassumesassumes nonono internalinternalinternal energyenergyenergy losseslosseslosses dueduedue tototo heatheatheat ororor fluxfluxflux changes.changes.changes. ActualActualActual efficienciesefficienciesefficienciesareare are usuallyusuallyusually betweenbetweenbetween 909090 andandand 100%.100%.100%. ExampleExample 7:7: TheThe transformertransformer inin Ex.Ex. 66 isis connectedconnected toto aa powerpower lineline whosewhose resistanceresistance isis 1212 .. HowHow muchmuch ofof thethe powerpower isis lostlost inin thethe transmissiontransmission line?line? VV == 24002400 VV I = 10 A; Vp = 600 V SS 12  EPPi a.c. EEPPiii SS S ES Np Ns (600V)(10A) 20 iS 2.50 A turns 2400V R

22 2 PPlostlost == ii RR == (2.50(2.50 A)A) (12(12 )) PPlostlost == 75.075.0 WW

PPin == (600(600 V)(10V)(10 A)A) == 60006000 WW %Power%Power%Power LostLostLost === (75(75(75 W/6000W/6000W/6000 W)(100%)W)(100%)W)(100%) === 1.25%1.25%1.25% SummarySummary

Effective current: i = 0.707 i Effective current: ieffeff = 0.707 imaxmax

Effective voltage: V = 0.707 V Effective voltage: Veffeff = 0.707 Vmaxmax

Inductive Reactance: Capacitive Reactance: 1 XfL2 Unit is the X  Unit is the L C 2 fC Ohm's law: ViX L L Ohm's law: VCC iX SummarySummary (Cont.)(Cont.) VV 22tan  L C VVVVTRLC() VR

XX 22tan  L C ZRXX()LC R

V 1 ViZi or T f  T Z r 2 LC SummarySummary (Cont.)(Cont.) PowerPower inin ACAC Circuits:Circuits: InIn termsterms ofof acac voltage:voltage: InIn termsterms ofof thethe resistanceresistance R:R:

PP == iViV cos cos PP == ii22RR  Transformers:Transformers:

EP N P  EEPPii SS ESSN CONCLUSION:CONCLUSION: ChapterChapter 32A32A ACAC CircuitsCircuits