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Transition to AC W11-3 What’s Happening?? Quiz Today Labs 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 Circuits First, 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. 0 NI B 2r B=0 outside 4 Magnetism B 2 The magnetic flux in coil #2 due to the current in coil #1. B 2 ~ i1 B 2 ki1 i2 N 2 B 2 M 21 i1 21 i1 emf2 N 2 M 21 t t mutual Inductanc Induction 5/23/2017 5 M N something UNIT: henry i Induction 5/23/2017 6 Minimum Change@t V1 Maximum Change@t Induction 5/23/2017 7 The Same emf1 V1 N1 t FLUX is the same through both coils (windings). emf 2 V2 N 2 t V1 V2 N1 N 2 Induction 5/23/2017 8 Induction 5/23/2017 9 Induction 5/23/2017 10 V2 N 2 V1 N1 V2 V1 (N 2 / N1 ) Powerin Powerout (Lossless) V2 V22 V12 (N 2 / N1 ) 2 I1V1 I 2V2 V2 R R R V1 (N 2 / N1 ) 2 I1 R So V1 R I1 ( N 2 / N1 ) 2 looks like an input resistance! Induction 5/23/2017 11 U=Area=(1/2)LI2 Power Vi i V L t i P Li t U Lii U Li LI i Li DU Lii interval i Induction i I Induction 5/23/2017 12 Energy Stored in a capacitor The energy stored in a capacitor with capacitance C and a voltage V is U=(1/2)cV2 U=(1/2)LI2 Induction 5/23/2017 13 Consider 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. N B 0 ni 0 i l Induction 5/23/2017 14 N B l 1 2 1 N B 2 U Li i 2 2 l B BA L 1 NiBA 2 N B 0 i l Bl Ni U 0 1 Bl 1 B 2lA 1 B 2V U BA 2 0 2 0 2 0 U 1 B2 Energy Density V 2 0 1 Capacitor ED 0 E 2 2 SUPER IMPORTANT ! Induction 5/23/2017 15 Induction 5/23/2017 16 Up and Down and Up and Down and ….. 5/23/2017 Induction 17 Frequency 1 LC 5/23/2017 Induction 18 Alternating Currents & Voltages Some Math First! r ac generator “Output” from the previous diagram 2 But not always! (capacitor) Let’s talk about phase y=f(x)=x2 30 25 20 15 10 5 0 0 1 2 3 4 5 6 y=f(x-2)=(x-2)2 y 30 x2 25 20 15 10 (x-2)2 2 5 0 0 1 2 3 4 5 6 x the “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 switch! The Sine 2 Let’s talk about PHASE f(t)=A sin(t) A=Amplitude (=1 here) sin( t ) cos(t ) 2 f(t)=A sin(t-[/2]) A=Amplitude (=1 here For the future sin( t ) cos(t ) 2 cos(t ) sin( t ) 2 2f AC Applied voltages This graph corresponds to an applied voltage of V cos(t). Because the current and the voltage are together (in-phase) this must apply to a Resisto for which Ohmmmm sa that I~V. phasor oops – the ac phaser i I cos(t ) the resistor v iR IR cos(t ) Phasor diagram Pretty Simple, Huh?? VR IR here comes trouble …. We need the relationship between I (the current through) and vL (the voltage across) the inductor. From the last chapter: i vL L t * unless you have taken calculus. check it out-- means change or difference . (thing) thing final thing initial (t) t final t initial sovL L i t L ( I cos t ) L I (cos(t t ) cos(t )) t t cos(t ) cos(t ) sin(t ) sin(t ) cos(t ) LI cancel vL t When t gets very small, Let's look at what's left : cos (t) goes to 1. sin(t ) sin(t ) vL LI ? ? t r lim 0 sin( ) 1 this leaves v L sin(t ) The resistor voltage looked like a cosine so we would like the inductor voltage to look as similar to this as possible. So let’s look 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) sin( t ) cos(t ) 2 result - inductor vL LI sin(t ) sin( t ) cos(t ) 2 I is the MAXIMUM current in the circuit. sin(t ) cos(t vL LI cos(t 2 2 ) ) Resistor comparing v iR IR cos(t ) vRmax I R inductor vL LI cos(t 2 ) vLMax I L (L) looks like a resistance XL=L Reactance - OHMS For the inductor vLMax I L IX L VL FOR THE RESISTOR vRMax IR VR slightly confusing point We will always use the CURRENT as the basis for calculations and express voltages with respect to the current. What that means? We describe thecurrent as varyingas : i I cos(t) and the voltageas v Vcos(t ) where is thephaseshift between the currentand the voltage. the phasorvL LI cos(t 2 ) vL LI cos(t 2 ) cos(t 2 ) sin(t ) direction t t 2 t remember for ac series circuits In the circuit below, R=30 W and L= 30 mH. If the angular fr 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? R=30 W =3 KHZ E=60V L= 30 mH R=30 W =3 KHZ E=60V L= 30 mH R=30W XL=L=90W The instantaneous voltage across each element is the PROJECTION This is the SAME as the of theofMAXIIMUM voltage sum the onto maximum vectors the horizontal projected ontoaxis! I the horizontal axis. VL IX L VR IR t Source voltage leads the current by the angle . I V Vmax VL IX L Let V IZ t VR IR Z Impedance I Z I R I X L or 2 2 2 2 Z R 2 X L2 VL tan ( ) VR 2 2 R=30 W =3 KHZ E=60V 1 L= 30 mH The drawing is obviously NOT to scale. VL IX L Let V IZ I t VR IR L= 30 mH Z Impedance I Z I R I X L or R 30W 2 2 2 2 Z R X 2 2 2 L =3 KHZ R=30 W E=60V 2 V 60 I 0.632A Z 94.9 X L L 90W Z (30) 2 (90) 2 94.9W VL tan ( ) VR 1 90 0 tan 71 30 1 R 30W X L L 90W Z (30) 2 (90) 2 94.9W 90 0 tan 71 1.25 rad 30 1 V 60 I 0.632A Z 94.9 vL IL sin(t ) 0.63x30000x0.03sin(t ) 567sin(t ) vR 0.63x30x cos(t ) 18.9 cos(t ) Let's look at thesum of these two voltagesthat should look like theAC source. vL IL sin(t ) 0.63x3000x0.03sin(t ) 56.7 sin(t ) vR 0.63x30x cos(t ) 18.9 cos(t ) Let's look at thesum of these two voltagesthat should look like theAC source. wt 56.7Sin(wt) 18.9 cos(wt) SUM 0 0 18.9 18.9 0.2 11.26455106 18.52325832 7.258707 0.4 22.08002001 17.40805279 -4.67197 0.6 32.01522824 15.59884312 -16.4164 0.8 40.67409035 13.16775681 -27.5063 1 47.71140484 10.21171358 -37.4997 1.2 52.84661617 6.84856156 -45.9981 1.4 55.87499969 3.212379001 -52.6626 It Does! 80 2 60 40 20 56.7Sin(wt) 18.9 cos(wt) 0 0 -20 -40 -60 -80 1 2 3 4 5 6 7 8 SUM Enough!