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PHY-2054 J. B. Bindell CHAPTER 22 – ALTERNATING CURRENT PART 1 UPCOMING ITEMS FOR YOUR CONSIDERATION No problem session on Monday (Not much to do!) Watch for a Mastering Physics Assignment Quiz next Friday on AC … will it ever end? Today we start AC Circuits We will review the exam when they are returned. EXAM STRUCTURE Question Number 1 Section 003 Section 004 Multiple Choice (5) Multiple Choice (5) 2-Wire Problem – Different currents in each class and different from the quiz 2-Wire Problem – Different currents in each class and different from the quiz 3 (similar or exactly from past.) Inverted coil problem. Coil problem from class. 4 (New) Charged particle orbit. Electron & proton in same orbit. 2 (seen before) We will retain this schedule. PHY2054 Problem Solving/Office Hours Schedule Room MAP-318 Bindell Monday 8:30-9:15AM Tuesday Bindell Dubey 11:00-12:00PM 12:00-1:00PM Wednesday 8:30-9:15AM 10:30 - 11:15 AM* Thursday Friday 8:30-9:15AM 10:30-11:30AM 1:30-2:45PM These sessions will be used both for office hours and problem solving. Students from any section of 2054 are invited to stop by for assistance in course materials (problems, etc.) Note: There will be times when the room may not be available. In that case we will use our individual offices. * In Office Dr. Dubey's hours are for problem solving only. If something else is going on in the room, come to my office! If I am not there …. come to my office. OK … HOW WAS THE TEST? A. B. C. D. Easy OK Difficult Impossible HOW DID YOU DO? A. B. C. D. E. great less than stellar ok poor bombed THIS TIME I DID A. B. C. better than last time about the same as last time worse than last time AC GENERATOR “OUTPUT” FROM THE PREVIOUS DIAGRAM 2 NUCLEAR DC /AC HOME GENERATORS FUEL WHAT WORKS ON AC? 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(wt) A=Amplitude (=1 here) f(t)=A sin(wt-[/2]) A=Amplitude (=1 here) sin( wt ) cos(wt ) 2 FOR THE FUTURE sin( wt ) cos(wt ) 2 cos(wt ) sin( wt ) 2 w 2f AC APPLIED VOLTAGES This graph corresponds to an applied voltage of V cos(wt). Because the current and the voltage are together (in-phase) this must apply to a Resistor for which Ohmmmm said that I~V. PHASOR OOPS – THE AC PHASER i I cos(wt ) THE RESISTOR v iR IR cos(wt ) 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 wt ) L I (cos(wt t ) cos(wt )) t t cos(wt ) cos(wt ) sin(wt ) sin(wt ) cos(wt ) LI cancel vL t When t gets very small, cos (wt) goes to 1. Let's look at what's left : sin(wt ) sin(wt ) vL wLI w t ? ? r lim 0 sin( ) 1 THIS LEAVES vL w sin(wt ) 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(wt) A=Amplitude (=1 here) f(t)=A sin(wt-[/2]) A=Amplitude (=1 here) sin( wt ) cos(wt ) 2 RESULT vL wLI sin(wt ) sin( wt ) cos(wt ) 2 sin(wt ) cos(wt vL wLI cos(wt 2 2 ) ) RESISTOR v iR IR cos(wt ) vRmax I R COMPARING INDUCTOR vL wLI cos(wt 2 vLMax I wL (wL) looks like a resistance XL=wL Reactance - OHMS ) SLIGHTLY CONFUSING POINT We will 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(wt) and the voltageas v Vcos(wt ) where is thephaseshift between the currentand the voltage. BACK TO THE PHASOR THING WHAT ABOUT THE CAPACITOR?? C: q vc c vc 1 q 1 1 i I cos(wt ) t c t c c Without repeating what we did, the question is what function will have a f/t = cosine? Obviously, the sine! So, using the same process that we used for the inductor, 1 vc I sin(wt ) wC 1 Xc (ohms) wC CAPACITOR PHASOR DIAGRAM 1 vc I sin(wt ) wC 1 Xc (ohms) wC I vC cos(wt ) wC 2 NOTICE THAT The voltage lags the current by 90 deg I and V are represented on the same graph but are different quantities. SUMMARY