Transition to ACWhat’s Happening??So far we have looked atThe Toroid–Mutual InductanceNote the form:Consider “AC” voltageThe transformerSlide 9Slide 10Input/Output Impedance (Resistance)Remember that a Capacitor stored ENERGY? U=(1/2)CV2SO …The Energy stored is in the Magnetic FieldEnergy Storage in InductorUp and Down and Up and Down and …..Slide 17FrequencySlide 19Alternating Currents & VoltagesSome Math First!Slide 22Slide 23Slide 24Slide 25Slide 26ac generator“Output” from the previous diagramSlide 29But not always! (capacitor)Let’s talk about phase y=f(x)=x2y=f(x-2)=(x-2)2the “rule”The SineLet’s talk about PHASESlide 36For the futureAC Applied voltagesphasoroops – the ac phaserthe resistorPhasor diagramhere comes trouble ….From the last chapter:check it out---so-Slide 47this leavesresult - inductorcomparingFor the inductorslightly confusing pointthe phasorSlide 54remember for ac series circuitsSlide 56Slide 57Slide 58Slide 59Slide 60Slide 61It Does!Enough!Transition to AC•W11-3What’s Happening??Quiz TodayLabs look pretty crappy–There is an education issue here!Exam #3 Next Wednesday Study Session Monday Morning if there is interest!Next topic is AC CircuitsFirst, let’s do a bit of a clicker review …Inductors and how they workHow they work in circuits … purely DC circuits but not necessarily constant.There are a few topics that we have missed and I want to go over them today.Then, if time, you can start the AC unit.Magnetism4rNIB20B=0 outside01/14/19Induction 5i2122~#1. coilincurrent the todue #2 coilinflux magnetic TheiBBtiMtNemfiMNkiBB12121221212212mutual Inductance01/14/19Induction 6iNMsomethingUNIT: henry01/14/19Induction 7 V1Maximum Change@tMinimum Change@t01/14/19Induction 8FLUX is the same throughboth coils (windings).2211222111NVNVtNVemftNVemfThe Same01/14/19Induction 901/14/19Induction 1001/14/19Induction 112 21 1212 12 2 22 2 1 2 11 1 2 2 221 2 111212 1(N / ) (Lossless)(N / )(N / )looks like an input resistance!( / )in outV NV NVVNPower PowerV V V NI V I V VR R RV NIRSoVRIN N==== = = ==� �=� �� �01/14/19Induction 12intervaliLiUiLiUtiLiPtiLVViPoweriiLiLIiLiDUIInductionU=Area=(1/2)LI201/14/19Induction 13Energy Stored in a capacitorThe energy stored in a capacitor with capacitance C and a voltage V isU=(1/2)LI2U=(1/2)cV201/14/19Induction 14Consider a solenoid with N turns that is very long. We assume that the field is uniform throughout its length, ignoring any “end effects”. For a long enough solenoid, we can get away with it for the following argument. Maybe.ilNniB0001/14/19Induction 150022212121BlNiilNBNiBAUBAilNLiUlNLBBB20020202021ED 21VUDensity 212121ECapacitorBEnergyVBlABBABlUSUPER IMPORTANT !01/14/19Induction 16Up and Down and Up and Down and …..01/14/19 Induction 17Frequency01/14/19 Induction 18LC1Alternating Currents & VoltagesSome Math First!rac generator“Output” from the previous diagram2But not always! (capacitor)Let’s talk about phasey=f(x)=x2y=f(x-2)=(x-2)2x2(x-2)2yx2the “rule”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The SineLet’s talk about PHASEPHASEf(t)=A sin(t)A=Amplitude (=1 here)f(t)=A sin(t-[/2])A=Amplitude (=1 here))cos()2sin( ttFor the futureftttt2)sin()2cos()cos()2sin(AC Applied voltagesThis 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.phasoroops – the ac phaser)cos( tIithe resistor)cos( tIRiRvPhasor diagram IRVRPretty Simple, Huh??here comes trouble ….We need the relationship between I (the current through)and vL (the voltage across) the inductor.From the last chapter:tiLvL* unless you have taken calculus.check it out---initialfinalinitialfinal tt(t) thingthing(thing).differenceor change means so- 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this leaves)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)cos()2sin( tt)sin( tLIvL)2cos()2cos()sin(tLIvttLresult - inductorI is the MAXIMUMcurrent in thecircuit.comparingResistor inductor)cos( tIRiRv)2cos( tLIvL RIvR max LIvMaxLL) looks like a resistanceXL=LReactance - OHMSFor the inductor LLMaxLVIXLIv FOR THE RESISTORRMaxRVIRv slightly confusing pointWe 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the phasor)2cos( tLIvL)2cos( tLIvLtt2t)sin()2cos( ttdirectionremember for ac series circuitsIn 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