Physics 2102Jonathan DowlingLecture 30: FRI 27 MAR 09Lecture 30: FRI 27 MAR 09Ch. 31.4Ch. 31.4––7: Electrical Oscillations, LC7: Electrical Oscillations, LCCircuits, Alternating CurrentCircuits, Alternating CurrentEXAM 03: 6PM THU 02 APR LOCKETT 5 EXAM 03REVIEW: 6PM WED 01 APR NICHOLSON 130The exam will cover: Ch.28 (secondhalf) through Ch.32.1-3 (displacementcurrent, and Maxwell's equations). Theexam will be based on: HW08 – HW11.Final Day to Drop Course: TODAY FRI 27 MARThe formula sheet for the exam can be found here:http://www.phys.lsu.edu/classes/spring2009/phys2102/formulasheet.pdfYou can see examples of old exam IIIs here:http://www.phys.lsu.edu/classes/spring2009/phys2102/Test3.oldtests.pdfDamped LCR OscillatorDamped LCR OscillatorIdeal LC circuit without resistance: oscillations goon for ever; ω = (LC)–1/2Real circuit has resistance, dissipates energy:oscillations die out, or are “damped”Math is complicated! Important points:– Frequency of oscillator shifts away fromω = (LC)-1/2– Peak CHARGE decays with time constant =– τQLCR=2L/R– For small damping, peak ENERGY decays withtime constant– τULCR= L/R0 4 8 12 16 200.00.20.40.60.81.0UEtime (s)Umax=Q22Ce!RtLCRL22If we add a resistor in an circuit (see figure) we mustmodify the energy equation, because now energy isbeing dissipated on the resistor: .E BRLdUi RdtqU U U= != + =Damped Oscillations in an CircuitRCL222 2Li dU q dq diLi i RC dt C dt dt+ " = + = !( )222/222 210. This is the same equation as thatof the damped harmonics osc 0, which has theillator: The angular f solution re( ) c que:os .bt mmd x dxm b kxdt dtxdq di d q d q dqi L R qdt dt dt dt dtt x e tC! "#= $ = $ + + =+ + =%= +( )2/2222ncy For the damped circuit the solution is:The angular freque1 ( ) cos . .4ncy . 4Rt LRqk bRCt Qe tmLLC Lm! " !!#% %##+== =%(31-6)/ 2Rt LQe!/ 2Rt LQe!( )q tQQ!( )q t( )/ 2( ) cosRt Lq t Qe t! "#$= +2214RLC L!"= #/222The equations above describe a harmonic oscillator with an exponentially decayingamplitude . The angular frequency of the damped oscillator1 is always smaller than the angular 4Rt LQeRLC L!"#= "221frequency of the 1undamped oscillator. If the term we can use the approximation .4LCRL LC!! !=#$!!RC= RC !!!!!!!!!!!!!!!!RL= L / R!!!!!!!!!!!!!RCL= 2L / RSummarySummary• Capacitor and inductor combination producesan electrical oscillator, natural frequencyof oscillator is ω=1/√LC• Total energy in circuit is conserved:switches between capacitor (electric field)and inductor (magnetic field).• If a resistor is included in the circuit, thetotal energy decays (is dissipated by R).Alternating Current:To keep oscillations going we need todrive the circuit with an external emfthat produces a current that goesback and forth.Notice that there are two frequenciesinvolved: one at which the circuit wouldoscillate “naturally”. The other is the frequency at which wedrive theoscillation.However, the “natural” oscillation usually dies offquickly (exponentially) with time. Therefore in thelong run, circuits actually oscillate with the frequencyat which they are driven. (All this is true for thegentleman trying to make the lady swing back andforth in the picture 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).Power Station Transmission linesErms =735 kV , I rms = 500 AHome110 VT1T2Step-uptransformerStep-downtransformerR = 220Ω1000 km=!2heat rmsThe resistance of the power line . is fixed (220 in our example). Heating of power lines . This parameter is also fixed (55 MW in our exR RAP I R!= "=Energy Transmission Requirements!trans rms rmsheat transheat rmsample).Power transmitted (368 MW in our example).In our example is almost 15 % of and is acceptable.To keep we must keep as low as possible. The onlP IP PP I= Erms rmsy way to accomplish thisis by . In our example 735 kV. To do that we need a device that can change the amplitude of any ac voltage (either increase or decrease).=increasing E E(31-24)Ptrans=iV= “big”Pheat=i2R= “small”Solution:Big V!The transformer is a device that can change the voltage amplitude of any ac signal. It consists of two coils with a different number of turns wound around a common iron core. The TransformerThe coil on which we apply the voltage to be changed is called the " " andit has turns. The transformer output appears on the second coil, which is knownas the "secondary" and has turnsPSNNprimary. The role of the iron core is to ensure that the magnetic field lines from one coil also pass through the second. We assume that if voltage equal to is applied across the primary then a voltagPV e appears on the secondary coil. We
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