PHY2049: Chapter 3118AC Circuits with RLC ComponentsÎEnormous impact of AC circuits Power delivery Radio transmitters and receivers Tuners Filters TransformersÎBasic components R L C Driving emfÎNow we will study the basic principlesPHY2049: Chapter 3119AC Circuits and Forced OscillationsÎRLC + “driving” EMF with angular frequency ωdÎGeneral solution for current is sum of two termssinmdtεεω=sinmddi qLRi tdt Cεω++=“Transient”: Fallsexponentially & disappears“Steady state”:Constant amplitudeIgnore/2costR Lie tω−′∼PHY2049: Chapter 3120Steady State SolutionÎAssume steady state solution of form Imis current amplitude φ is phase by which current “lags” the driving EMF Must determine Imand φÎPlug in solution: differentiate & integrate sin(ωt-φ)()sinmdiI tωφ=−() () ()cos sin cos sinmmd d m d d m ddIILIRt t tCωωτ φ ω φ ω φ ε ωω−+ −− −=sinmdi qLRi tdt Cεω++=()sinmdiI tωφ=−()cosdm ddiItdtωωφ=−()cosmddIqtωφω=− −SubstitutePHY2049: Chapter 3121Steady State Solution for AC Current (2)ÎExpand sin & cos expressionsÎCollect sinωdt&cosωdtterms separatelyÎThese equations can be solved for Imand φ (next slide)()()1/ cos sin 01/ sin cosddmd d m mLC RIL C IRωωφφωωφ φε−−=−+=()()sin sin cos cos sincos cos cos sin sindd ddd dtt ttttωφωφωφωφωφωφ−= −−= +High school trig!cosωdttermssinωdtterms() () ()cos sin cos sinmmd d m d d m ddIILIRt t tCωωτ φ ω φ ω φ ε ωω−+ −− −=PHY2049: Chapter 3122ÎSolve for φ and ImÎR, XL, XCand Z have dimensions of resistanceÎLet’s try to understand this solution using “phasors”Steady State Solution for AC Current (3)1/tanddLCLCXXRRωωφ−−=≡mmIZε=()22LCZRXX=+−LdXLω=1/CdXCω=Inductive “reactance”Capacitive “reactance”Total “impedance”PHY2049: Chapter 3123Graphical Representation: R, XL, XC, ZtanLCXXRφ−=LCXX−φ()22LCZRXX=+−PHY2049: Chapter 3124Understanding AC Circuits Using PhasorsÎPhasor Voltage or current represented by “phasor” Phasor rotates counterclockwise with angular velocity = ωd Length of phasor is amplitude of voltage (V) or current (I) y component is instantaneous value of voltage (v) or current (i)εmImωdt −φ()sinmdiI tωφ=−sinmdtεεω=iεCurrent “lags” voltage by φPHY2049: Chapter 3125AC Source and Resistor OnlyÎVoltage isÎRelation of current and voltage Current is in phase with voltage (φ = 0)IRiεR~sin /RdRRiI tI VRω==VRsinRRdviRV tω==ωdtPHY2049: Chapter 3126AC Source and Capacitor OnlyÎVoltage is ÎDifferentiate to find currentÎRewrite using phaseÎRelation of current and voltageΓCapacitive reactance”: Current “leads” voltage by 90°sinCdqCV tω=VCICiεC~/cosdC didqdt CV tωω==/sinCCdvqCV tω==()sin 90dC diCV tωω=+°()sin 90CdiI tω=+°ωdt + 90ωdt1/CdXCω=PHY2049: Chapter 3127AC Source and Inductor OnlyÎVoltage is ÎIntegrate di/dt to find current: ÎRewrite using phaseÎRelation of current and voltageΓInductive reactance”: Current “lags” voltage by 90°()//sinLddi dt V L tω=VLILiεL~()/cosLd diV L tωω=−/sinLLdv Ldi dt V tω==()()/sin 90Ld diV L tωω=−°ωdt − 90()sin 90LdiI tω=−°ωdtLdXLω=PHY2049: Chapter 3128What is Reactance?Think of it as a frequency-dependent resistanceShrinks with increasing ωd1CdXCω=LdXLω=(" " )RXR=Grows with increasing ωdIndependent of ωdPHY2049: Chapter 3129QuizÎThree identical EMF sources are hooked to a singlecircuit element, a resistor, a capacitor, or an inductor. The current amplitude is then measured vs frequency. Which curve corresponds to an inductivecircuit? (1) a (2) b (3) c (4) Can’t tell without more infofdImacbLdXLω=For inductor, higher frequency gives higherreactance, therefore lower currentPHY2049: Chapter 3130AC Source and RLC CircuitÎVoltage is R, L, C all presentÎRelation of current and voltage Current “lags” voltage by φ Impedance: Due to R, XCand XLÎCalculate Imand φ See next slideVRVCsinmdtεεω=VLεmImωdt −φ()sinmdiI tωφ=−IPHY2049: Chapter 3131AC Source and RLC Circuit (2)ÎSolution using trig (or geometry) Magnitude = Im, lags emf by phase φ)() ()2222/1/tan1/mmLC d dLC d dIZXX L CRRZRXX R L Cεωωφωω=−−===+− =+ −()sinmdiI tωφ=−PHY2049: Chapter 3132AC Source and RLC Circuit (3)ÎWhen XL= XC, then φ = 0 Current in phase with emf, “Resonant circuit”: Z = R (minimum impedance, maximum current)ÎWhen XL< XC, then φ < 0 Current leads emf, “Capacitive circuit”: ωd< ω0ÎWhen XL> XC, then φ > 0 Current lags emf, “Inductive circuit”: ωd> ω0tanLCXXRφ−=01/dLCωω==()22LCZRXX=+−PHY2049: Chapter 3133Graphical Representation: VR, VL, VC, εmtanLCLCRXX VVRVφ−−==LCVV−φmε()22mRLCVVVε=+−PHY2049: Chapter 3134RLC Example 1ÎBelow are shown the driving emf and current vs time of an RLC circuit. We can conclude the following Current “leads” the driving emf (φ<0) Circuit is capacitive (XC> XL) ωd< ω0εItPHY2049: Chapter 3135QuizÎWhich one of these phasor diagrams corresponds to an RLC circuit dominated by inductance? (1) Circuit 1 (2) Circuit 2 (3) Circuit 3εmεmεm1 32Inductive: Current lags emf, φ>0PHY2049: Chapter 3136QuizÎWhich one of these phasor diagrams corresponds to an RLC circuit dominated by capacitance? (1) Circuit 1 (2) Circuit 2 (3) Circuit 3εmεmεm1 32Capacitive: Current leads emf, φ<0PHY2049: Chapter 3137RLC Example 2ÎR = 200Ω, C = 15μF, L = 230mH, εm= 36v, fd= 60 Hz ωd= 120π = 377 rad/s Natural frequency ()()601/ 0.230 15 10 538rad/sω−=×=377 0.23 86.7LX =× =Ω()61/ 377 15 10 177CX−=××=Ω()22200 86.7 177 219Z =+−=Ω/ 36/219 0.164AmmIZε== =XL< XCCapacitive circuit186.7 177tan 24.3200φ−−⎛⎞==−°⎜⎟⎝⎠Current leads emf(as expected)PHY2049: Chapter 3138RLC Example 2 (cont)ÎÎÎÎ0.164 86.7 14.2VLmLVIX==×=0.164 177 29.0VCmCVIX==×=0.164 200 32.8VRmVIR== ×=()2232.8 14.2 29.0 36.0mε+− ==LCVV−φmεPHY2049: Chapter 3139RLC Example 3ÎCircuit parameters: C = 2.5μF, L = 4mH, εm= 10v ω0= (1/LC)1/2= 104rad/s Plot Imvs ωd/ ω0R = 5ΩR = 10ΩR = 20ΩImResonance0dωω=0/dωωPHY2049: Chapter 3140Power in AC CircuitsÎInstantaneous power emitted by circuit: P = i2RÎMore useful to calculate power averaged over a cycle Use <…> to indicate average over a cycleÎDefine RMS quantities to avoid ½ factors in AC
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