Physics 1202: Lecture 11 Today s Agenda

Transcription

1 Physcs 122: Lecture 11 Today s Agenda Announcements: Team problems start ths Thursday Team 1: Hend Ouda, Mke Glnsk, Stephane Auger Team 2: Analese Bruder, Krsten Dean, Alson Smth Offce hours: Monday 2:3-3:3 Thursday 3:-4: Homework #5: due ths comng Frday Mdterm 1: Thursday March 1 st (n class) Revew sesson Tuesday Feb. 27 (+ Team problems) Mdterm sample + To-Know sheet on web by end of week Chapter 23: nducton Inductance of solenod + n seres & parallel Chapter 24: AC crcuts AC voltage, current + phaser and RMS values C & L n AC crcuts + RC & RL crcuts Faraday's Law B B n q B N S v S N B v 1

2 23-7: Inductance Consder the loop at the rght. swtch closed current starts to flow n the loop. \ magnetc feld produced n the area enclosed by the loop. \ flux through loop changes X X XX X X X X XX X X XX \ emf nduced n loop opposng ntal emf Self-Inducton: the act of a changng current through a loop nducng an opposng current n that same loop. Inductance The magnetc feld produced by the current n the loop shown s proportonal to that current. I The flux, therefore, s also proportonal to the current. We defne ths constant of proportonalty between flux and current to be the nductance L. We can also defne the nductance L, usng Faraday's Law, n terms of the emf nduced by a changng current. 2

3 23-7: Inductance The magnetc feld produced by the current n the loop shown s proportonal to that current. I The flux, therefore, s also proportonal to the current. So, the nductance s Long Solenod: Calculaton N turns total, radus r, Length l (n: number of turns per unt length) r l N turns For a sngle turn, The total flux through solenod s gven by: Inductance of solenod can then be calculated as: Ths (as for R and C) depends only on geometry (materal) 3

4 23-9: Energy of an Inductor How much energy s stored n an nductor when a current s flowng through t? a I Start wth loop rule: b e R = -RI I Multply ths equaton by I: L e L = -L DI / Dt P L s the rate at whch energy s beng stored n the nductor: LI L LDI DI The total U stored n the nductor when the current = I s the shaded trangle: I Where s the Energy Stored? Clam: (wthout proof) energy s stored n the Magnetc feld tself (just as n the Capactor / Electrc feld case). To calculate ths energy densty, consder the unform feld generated by a long solenod: l The nductance L s: r N turns Energy U: We can turn ths nto an energy densty by dvdng by the volume contanng the feld: 4

5 23-1: Transformers Devce to change e (or the voltage) 2 cols wrapped around ron core Prmary (P) and secondary (S) B-feld nsde core Tme varyng current n P (wth N P ) Faraday's Law: DFB e P = - N P Dt Tme varyng flux nduces emf n secondary col S (wth N S ) DFB e S = - N S Dt Same varyng flux \ Inductors n seres & parallel Lke resstors: bascally wres Inductors n seres the current s the same n both L 1 and L 2 the voltage drops add V L 1 Inductors n parallel the voltage drop s the same n both L 1 and L 2 the currents add V L 2 L 1 L 2 5

6 C R e ~ L w Nkola Tesla o o Inventor o Key fgure n development of o AC electrcty o Hgh-voltage transformers o Transport of electrcal power va AC transmsson lnes o Beat Edson s dea of DC transmsson lnes 6

7 24-1: Alternatng voltage and current We begn by consderng smple crcuts wth one element (R,C, or L) n addton to the drvng emf. Begn wth R: Loop eqn gves: e m - e m e Voltage across R n phase wth current through R ~ xt e m / R - e m / R t R R Note: ths s always, always, true always Phaser To vsualze the phase relatonshps between V & I Phasors vectors whose length s the maxmum V or I, rotate around an orgn wth the angular speed of the oscllatng current. The nstantaneous value of V or I Projecton of phasor on the y axs. Lke crcular moton n Pearson Educaton, Inc. 7

8 RMS Values Average values for I,V are not that helpful they are zero But power s not! Recall analogy wth flud e m e m / R - e m - e m / R xt RMS Values Thus we ntroduce the dea of the Root of the Mean Squared. In general, So Average Power s. 12 volts s the rms value of household ac 8

9 Lecture 11, ACT 1a Consder a smple AC crcut wth a R purely resstve load. For ths crcut the voltage source s e = 1V sn (2p5(Hz)t) and R = 5 W. What s the average e current n the crcut? ~ A) 1 A B) 5 A C) 2 A D) 2 A E) A Chapter 11, ACT 1b Consder a smple AC crcut wth a R purely resstve load. For ths crcut the voltage source s e = 1V sn (2p5(Hz)t) and R = 5 W. What s the average power e n the crcut? ~ A) W B) 2 W C) 1 W D) 1 2 W 9

10 24-2: Capactors n AC Crcuts Now consder C: Loop eqn gves: C e ~ e m Voltage across C lags current through C by one-quarter cycle (9 ). w Ce m Is ths always true? YES - e m xt -w Ce m t Snce 24-2: Capactors n AC Crcuts the rms current n the capactor s related to ts capactance and to the frequency 1

11 24-2 Capactve reactance In analogy wth resstance, we wrte: wth 24-2 Phaser for C V and I n a capactor are not n phase: V lags by 9º e m w Ce m - e m xt -w Ce m t 11

12 Lecture 11, ACT 2 A crcut consstng of capactor C and voltage source e s constructed as shown. The graph shows the voltage presented to the capactor as a functon of tme. Whch of the followng graphs best represents the tme dependence of the current n the crcut? e t (a) (b) (c) t t t 24-3 RC Crcuts Current I across R and C are not n phase maxmum I s not the sum of the maxmum of I R and the maxmum of I C they do not peak at the same tme. Ths phasor dagram V across C and across R are at 9º n the dagram; f added as vectors, we fnd the maxmum 12

13 24-3 RC Crcuts Ths has the exact same form as V = IR : f we defne the mpedance, Z: 24-3 RC Crcuts There s a phase angle between the voltage and the current, as seen n the dagram. The power n the crcut s gven by: Because of ths, the factor cos φ s called the power factor. 13

14 24-4: nductors n AC Crcuts Now consder L: Loop eqn gves: e ~ L e m Voltage across L leads current through L by onequarter cycle (9 ). e m / w L - e m xt - e m / w L tx Yes, yes, but how to remember? Snce 24-4: nductors n AC Crcuts the rms current n the capactor s related to ts capactance and to the frequency 14

15 24-4 Inductve reactance In analogy wth resstance, we wrte: wth Just as wth capactance, we can defne nductve reactance: 24-4 Phaser for L V and I n an nductor are not n phase: V leads by 9º e m e m / w L - e m xt - e m / w L tx 15

16 RL Crcuts Current I across R and L are not n phase maxmum I s not the sum of the maxmum of I R and the maxmum of I L they do not peak at the same tme. Ths phasor dagram V across L and across R are at 9º n the dagram; f added as vectors, we fnd the maxmum 24-4 RL Crcuts Ths has the exact same form as V = IR : f we defne the mpedance, Z: 16

17 24-4 RL Crcuts As before, there s a phase angle between V and I The power n the crcut s Because of ths, the factor cos φ s called the power factor. R: V n phase wth C: V lags by 9 L: V leads by 9 Phasors A phasor s a vector whose magntude s the maxmum value of a quantty (eg V or I) and whch rotates counterclockwse n a 2-d plane wth angular velocty w. Recall unform crcular moton: The projectons of r (on the vertcal y axs) execute snusodal oscllaton. y w x y 17

18 Suppose: ß Phasors for L,C,R w wt w wt wt w w - dependence n AC Crcuts The maxmum current & voltage are related va the mpedence Z Currents AC-crcuts as a functon of frequency: 18

19 Problem Soluton Method: Fve Steps: 1) Focus on the Problem - draw a pcture what are we askng for? 2) Descrbe the physcs - what physcs deas are applcable - what are the relevant varables known and unknown 3) Plan the soluton - what are the relevant physcs equatons 4) Execute the plan - solve n terms of varables - solve n terms of numbers 5) Evaluate the answer - are the dmensons and unts correct? - do the numbers make sense? Recap of Today s Topc : Announcements: Team problems start ths Thursday Team 1: Hend Ouda, Mke Glnsk, Stephane Auger Team 2: Analese Bruder, Krsten Dean, Alson Smth Offce hours: Monday 2:3-3:3 Thursday 3:-4: Homework #5: due ths comng Frday Mdterm 1: Thursday March 1 st (n class) Revew sesson Tuesday Feb. 27 (+ Team problems) Mdterm sample + To-Know sheet on web by end of week Chapter 23: nducton Inductance of solenod + n seres & parallel Chapter 24: AC crcuts AC voltage, current + phaser and RMS values C & L n AC crcuts + RC & RL crcuts 19