Lecture 21PowerPoint PresentationSlide 3Slide 4Slide 5Slide 6All elements in parallel:Example:Example 2 (solution)Driven RLC CircuitSlide 11Slide 12SummaryExampleSlide 15Lecture 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(ωdt)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(ωdt) € iR=vRR=EmRsin(ωdt)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 d , their magnitude is equalto the amplitude of the quantity in question and theirprojection on the vertical axis represents the instantaneous value of the quantity under study.AC Driven Circuits:2) Capacitors: € vC= emf = Emsin(ωdt) € qC= C emf = CEmsin(ωdt) € iC=dqCdt= ωdCEmcos(ωdt) € iC= ωdCEmsin(ωdt + 900) € iC=EmXsin(ωdt + 900)reactance"" 1 whereCXdω= € im=EmX€ looks like i =VRCapacitors “oppose a resistance” to AC (reactance) of frequency-dependent magnitude 1/d C (this idea is true only for maximum amplitudes, the instantaneous story is more complex).AC Driven Circuits:3) Inductors: € vL= emf = Emsin(ωdt)LdtvidtidLvLLLL∫=⇒= € iL= −EmLωdcos(ωdt) € =EmLωdsin(ωdt − 900) € iL=EmXsin(ωdt − 900) € im=EmXdLX ω= whereInductors “oppose a resistance” to AC (reactance) of frequency-dependent magnitude d L(this idea is true only for maximum amplitudes, the instantaneous story is more complex).All elements in parallel:All elements in parallel:~Emsin(dt)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 = :ResistorsLXdω= :Inductors)/(1 :Capacitors CXdω=Example:Example:•Circuit shown is driven by a low frequency (5 Hz) voltage source with an rms amplitude of 7.07 V. Compute the current amplitudes in the resistor, capacitor and inductor (IR, IC and IL, respectively). ~10F10010mHErms = 7.07 Vf = 5 HzRms, peak and peak-to-peak:If E=Emsin(dt)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)~10F10010mHErms = 7.07 Vf = 5 Hz•The three components are in parallel across the driving emf.•Amplitude of voltage across the three components is the same. •So, current amplitude for any given component is inversely proportional to X: Im = Em/X•R = 100 IRV)/(100) = 0.1 A•XC = 1/(C) = 1/(10 IC = (10V)/( )=3.14 mA•XL = L = 10 IL= (10V)/(0.) = 31.85 AEm = 10 V rad/sDriven RLC CircuitDriven RLC CircuitPhase differences between voltage and current! Resistors: = 0Capacitors: = –/2 (I leadsV)Inductors: = +/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 −=−=−=φtanDriven RLC CircuitDriven RLC CircuitSummarySummary € Im=EmZ22)1(CLRZddωω −+=RCLVVV −=φtanRCLddωω−=E = Em sin(dt); I = Imsin(dt - )(We have used XR=R, XC=1/C, XL=L )Example Example •In a given series RLC circuit:• Em = 125 V• Im = 3.2 A• Current LEADS source emf by 0.982 rad.•Determine Z and R.•Z = Em/ Im = (125/3.2) 39.1 •How to find R? Look at phasors!• Emcos = VR = ImR•R = Emcos /Im= (125V)(0.555)/(3.2A) = 21.7 VLVCVR=ImREm=ImZExample Example •In a given series RLC circuit:•VL = 2VR = 2VC.•Determine .RCLVVV −=φtan12=−=RRRVVVHence, = 450VLVCVREmVL-
View Full Document
Unlocking...