Lecture 21Lecture 21Physics 2102Jonathan DowlingAlternating Current CircuitsAlternating Current CircuitsAlternating Current:To keep oscillations going we need to drive the circuit with an external emf that producesa current that goes back and forth.Notice that there are two frequencies involved: one at which the circuit wouldoscillate “naturally”. The other is the frequency at which we drive theoscillation. However, the “natural” oscillation usually dies off quickly(exponentially) with time. Therefore in the long run, circuits actuallyoscillate with the frequency at which they are driven. (All this is truefor the gentleman trying to make the lady swing back and forth in thepicture too).We have studied that a loop of wire,spinning in a constant magnetic fieldwill have an induced emf that oscillates with time, ! E = Emsin("dt)That is, it is an AC generator.Alternating Current:AC’s are very easy to generate, they are also easy to amplify anddecrease in voltage. This in turn makes them easy to send in distributiongrids like the ones that power our homes. Because the interplay of AC and oscillating circuits can be quitecomplex, we will start by steps, studying how currents and voltagesrespond in various simple circuits to AC’s.AC Driven Circuits:1) A Resistor:0=!Rvemf ! vR= emf = Emsin("dt) ! iR=vRR=EmRsin("dt)Resistors behave in AC very much as in DC, current and voltage are proportional (as functions of time in the case of AC),that is, they are “in phase”.For time dependent periodic situations it is useful torepresent magnitudes using “phasors”. These are vectorsthat rotate at a frequency ωd , their magnitude is equalto the amplitude of the quantity in question and theirprojection on the vertical axis represents theinstantaneous value of the quantity under study.AC Driven Circuits:2) Capacitors: ! vC= emf = Emsin("dt) ! qC= C emf = CEmsin("dt) ! iC=dqCdt="dCEmcos("dt) ! iC="dCEmsin("dt + 900) ! iC=EmXsin("dt + 900)reactance"" 1 whereCXd!= ! im=EmX! looks like i =VRCapacitors “oppose a resistance” to AC (reactance) of frequency-dependent magnitude 1/ωd C (this idea is true only for maximum amplitudes, the instantaneous story is more complex).AC Driven Circuits:3) Inductors: ! vL= emf = Emsin("dt)LdtvidtidLvLLLL!="= ! iL= "EmL#dcos(#dt) ! =EmL"dsin("dt # 900) ! iL=EmXsin("dt # 900) ! im=EmXdLX!= whereInductors “oppose a resistance” to AC (reactance) of frequency-dependent magnitude ωd L(this idea is true only for maximum amplitudes, the instantaneous story is more complex).All elements in parallel:All elements in parallel:~Emsin(ωdt)CRLILIRICVR, VC, VLOnce again: • VR is always in phase with IR; • VL leads IL by 900 • IC leads VC by 900• “ELI the ICE man...”Impedance ( or reactance):RX = :ResistorsLXd!= :Inductors)/(1 :Capacitors CXd!=Example:Example:• Circuit shown is driven by alow frequency (5 Hz) voltagesource with an rmsamplitude of 7.07 V.Compute the currentamplitudes in the resistor,capacitor and inductor (IR, ICand IL, respectively).~10µF100Ω10mHErms = 7.07 Vf = 5 HzRms, peak and peak-to-peak:If E=Emsin(ωdt)Then the peak-to-peak emf is 2Em;The amplitude of the emf is Em (peak)The rms emf is Em/√2Example 2 (solution)Example 2 (solution)~10µF100Ω10mHErms = 7.07 Vf = 5 Hz• The three components are in parallelacross the driving emf.• Amplitude of voltage across thethree components is the same.• So, current amplitude for any givencomponent is inverselyproportional to X: Im = Em/X• R = 100ΩIR = (10V)/(100Ω) = 0.1 A• XC = 1/(ωC) = 1/(10π.10−5)= 3184Ω IC = (10V)/(3184Ω)=3.14 mA• XL = ωL = 10π.10−2 = 0.314 Ω IL= (10V)/(0.314 Ω) = 31.85 AEm = 10 Vω = 10π rad/sDriven RLC CircuitDriven RLC CircuitPhase differences between voltage and current!Resistors: φ = 0Capacitors:φ = –π/2 (I leadsV)Inductors: φ = +π/2 (V leads I)Series circuit: current is the samein all devices.“Taking a walk” we see that the emfsin the various devices should add upto that of the AC generator. But theemfs are out of phase with each other.How to add them up? Use phasors.Current in circuit:Emf in devices:Resulting emf:Driven RLC CircuitDriven RLC CircuitApplying Pythagoras’ theorem to the picture: ! E2= VR2+ (VL" VC)2 ! E2= (i XR)2+ (i XL" i XC)2 ! E2= i2(XR)2+ (XL" XC)2[ ] ! i =EXR2+ (XL" XC)2Which resembles “i=E/R” ! i =EZ22)(Z ,impedance"" called is CLRXXXZ !+=Also for the phase:RCLRCLRCLXXXXiXiXiVVV !=!=!="tanDriven RLC CircuitDriven RLC CircuitSummarySummary ! Im=EmZ22)1(CLRZdd!!"+=RCLVVV !="tanRCLdd!!1"=E = Em sin(ωdt); I = Imsin(ωdt - φ)(We have used XR=R, XC=1/ωC, XL=ωL )ExampleExample• In a given series RLCcircuit:• Em = 125 V• Im = 3.2 A• Current LEADS sourceemf by 0.982 rad.• Determine Z and R.• Z = Em/ Im = (125/3.2) Ω = 39.1 Ω• How to find R? Look at phasors!• Emcos φ = VR = ImR• R = Emcos φ /Im=(125V)(0.555)/(3.2A) = 21.7 ΩVLVCVR=ImREm=ImZφExampleExample• In a given series RLCcircuit:• VL = 2VR = 2VC.• Determine φ.RCLVVV !="tan12=!=RRRVVVHence, φ = 450VLVCVREmφVL-
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