Alternating Current Part 1Slide 2This weekSlide 4ac generator – Similar to the motor but really different … whatever that means!“Output” from the previous diagramSlide 7DC /ACSlide 9But not always! (capacitor)so, Let’s talk about phase y=f(x)=x2y=f(x-2)=(x-2)2the “rule”The SineLet’s talk about PHASEA sine wave shifted by P radians isSlide 17For the futureAC Applied voltagesoops – the ac phaserthe resistorPhasor diagramhere comes trouble ….From the last chapter:check it out---so-Slide 27this leavesBREAKSlide 30Where bee we?Keepeth in Mindresult - inductorcomparingFor the inductorslightly confusing pointthe phasorSlide 38remember for ac series circuitsSlide 40Slide 41Slide 42Slide 43Slide 44Slide 45it does!What about the capacitor??capacitor phasor diagramNOTICE THATSUMMARYFor AC CircuitsResonanceac circuits look complicatedstoppeth hereSlide 55PHY-2054We will do the lab session on the inductor’s time constant.We will discuss what happens when the applied voltage is a sinusoid.We will do some experiments on AC circuitsWe will have a quiz on Friday.NOTE: Start reading chapter 23! Start reading chapter 23! – AC CircuitsMONDAY MONDAY WEDNESDAY FRIDAY29 AC AC AC5 WAVES EXAM III OPTICS12 OPTICS OPTICS OPTICS19 OPTICS OPTICS OPTICS26 EXAM IVDate and Time of Final is being investigated.2x2(x-2)2yx2f(x-b): shift a distance b in the POSITIVE directionf(x+b): shift a distance n in the NEGATIVE direction.The signs switchswitch!2f(t)=A sin(t)A=Amplitude (=1 here)f(t)=A sin(t-[/2])A=Amplitude (=1 here))cos()2sin( ttA. cosineB. - sineC. -cosineD.sineE. tangentftttt2)sin()2cos()cos()2sin(This graph correspondsto an applied voltageof V cos(t).Because the currentand the voltage aretogether (in-phase) thismust apply to a Resistorfor which Ohmmmm saidthat I~V.)cos( tIi)cos( tIRiRvIRVRPretty Simple, Huh??We need the relationship between I (the current through)and vL (the voltage across) the inductor.tiLvL* unless you have taken calculus.initialfinalinitialfinal tt(t) thingthing(thing).differenceor change means  ttttttLILvttttILttILtiLLv)cos()sin()sin()cos()cos())cos()(cos()cos(Whent gets very small,cos (t) goes to 1.canceltttLIvL)sin()sin(:left sat what'look sLet'??r1)sin(lim0)sin( tvLThe resistor voltage looked like a cosine so we would like theinductor voltage to look as similar to this as possible. So let’slook at the following graph again (~10 slides back):f(t)=A sin(t)A=Amplitude (=1 here)f(t)=A sin(t-[/2])A=Amplitude (=1 here))cos()2sin( ttBREAKWhere bee we?Equipment didn’t work on Monday but it should be working today. You finished all of the calculations in the previous hand-out.Today we will begin with a look at LR circuits:LR with a square wave input so you can determine the time constant.LR with AC so you can look at phase relationships as well as inductive reactanceAdd a capacitor and look at an RLC circuit to determine resonance conditions as well as phase relationships.The previous will take at least one additional session. Maybe two!Keepeth in MindNote the appearance of a new WebethAssignment.Quiz on FridayExam Next Wednesday – Magnetism  AC circuitsMonday – we will try to begin optics. Some of the AC may spill over into that sessionStarting Monday, Dr. Dubey will take over the class as lead instructor. She is a better teacher than I am.START STUDYING FOR EXAM III!!!START STUDYING FOR EXAM III!!!)cos()2sin( tt)sin( tLIvL)2cos()2cos()sin(tLIvttLI is the MAXIMUMcurrent in thecircuit.RESISTOR INDUCTOR)cos( tIRiRv)2cos( tLIvL RIvR max LIvMaxLL) looks like a resistanceXL=LReactance - OHMS LLMaxLVIXLIv FOR THE RESISTORRMa xRVIRv We will always use the CURRENT as the basis for calculations and express voltages with respect to the current. What that means? voltage. theandcurrent ebetween thshift phase theis where)tVcos(vas voltage theandt)cos( Ii:as varyingascurrent thedescribe We)2cos( tLIvL)2cos( tLIvLtt2t)sin()2cos( ttdirectionIn the circuit below, R=30  and L= 30 mH. If the angular frequency of the 60 volt AC source is is 3 K-Hz WHAT WE WANT TO DO:(a)calculate the maximum current in the circuit(b)calculate the voltage across the inductor(c)Does Kirchoff’s Law Work?E=60VR=30 L= 30 mH=3 KHZE=60VR=30 L= 30 mH=3 KHZtIR=30XL=L=90IRVRLLIXV The instantaneous voltage acrosseach element is the PROJECTIONof the MAXIIMUM voltage ontothe horizontal axis!This is the SAME as the sum of themaximum vectors projected ontothe horizontal axis.tIIRVRLLIXV axVVmSource voltage leadsthe current by the angle .22222222IImpedance ZIZV LLXRZorXIRIZLet)(tan1RLVVE=60VR=30 L= 30 mH=3 KHZtIIRVRLLIXV 22222222IImpedance ZIZV LLXRZorXIRIZLet)(tan1RLVVE=60VR=30 L= 30 mH=3 KHZ0122713090tan9.94)90()30(9030ZLXRLThe drawing is obviouslyNOT to scale.AZVI 632.09.9460radZLXRL 25.1713090tan9.94)90()30(90300122AZVI 632.09.9460source. AC thelikelook should that voltages two theseof sum at thelook sLet')cos(9.18)cos(3063.0)sin(567)sin(03.03000063.0)sin(ttxxvttxxtLIvRLsource. AC thelikelook should that voltages two theseof sum at thelook sLet')cos(9.18)cos(3063.0)sin(7.56)sin(03.0300063.0)sin(ttxxvttxxtLIvRLwt 56.7Sin(wt) 18.9 cos(wt) SUM0 0 18.9 18.90.211.2645510618.523258327.2587070.422.0800200117.40805279 -4.671970.632.0152282415.59884312 -16.41640.840.6740903513.16775681 -27.5063147.7114048410.21171358 -37.49971.252.84661617 6.84856156 -45.99811.455.874999693.212379001 -52.66262)cos(111:tIcictqctvcqvCccWithout repeating what we did, the question is what function will havea f/t = cosine?