Lect09 - Physics 212 Le cture9 Today's C pt: once Ele ctric...

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Unformatted text preview: Physics 212 Le cture9 Today's C pt: once Ele ctric C nt urre Ohm Law & re ’s sistors, Re sistors in circuits, Powe in circuits r 50 40 30 20 10 0 Confused Avg = 3.4 Confident Physics 212 Le cture9, S 1 lide Music Who is the Artist? A) B) C) D) E) Grateful Dead Jefferson Airplane CCR The Band ZZ Top Why? Last tim I playe Van Morrison tune fromthis album e d s Haveyou se n them ? e ovie Be m m e r?? st usic ovie ve 1978: Martin S corce se C lapton, Dylan. Dr. John, Joni Mitche ll, Van theMan, Ne Young, … il Physics 212 Le cture9, S 2 lide Physics 212 Le cture9 Today's C pt: once Ele ctric C nt urre Ohm Law & re ’s sistors, Re sistors in circuits, Powe in circuits r 50 40 30 20 10 0 Confused Avg = 3.4 Confident Physics 212 Le cture9, S 3 lide S eExamS om tuff xt dne • Examne We sday at 7:00 – – – – C rs m rial through le ove ate cture8 Bring your I D: Room de rm d by discussion se s te ine ction (se link) e C onflict e at 5:15 – sign up in your grade xam book if you ne d to e If you haveconflicts with both of the , you should havehe fromProf Erre about se ard de sche duling online.physics.uiuc.edu/courses/phys212/fall09/ExamPrepHe1.html Physics 212 Le cture9, S 4 lide Thanks for your fe dback about thecourse e The we , of course a m d bag of opinions on what would m thecoursebe r. re re , ixe ake tte Howe r, he ’s what I will takeaway fromall thecom e ve re m nts 1) Makesurewehavetim to de e voteto theproble for theday m 2) Re ducethevote on pre s flight que stions that m stude got ost nts 3) S orry.. Thele cturenote we m be gue at what I would do.. Base on pre s re y st ss d flights, it is be r for m to changethele tte e ctureto addre what difficultie I se .. There le ss se al ctureslide s areposte on thewe afte thele d br cture …… are Physics 212 Le cture9, S 5 lide Ke Conce y pts: 1) How re pe How sistancede nds on A, L, σ , ρ 1) How to com re bine sistors in se s and paralle rie l 1) 1) Unde rstanding re sistors in circuits Today’s Plan: 1) Re w of re vie sistance& pre flights 1) Work out a circuit proble in de m tail Physics 212 Le cture9, S 6 lide I σ A L V Observables: Conductivity – high for good conductors. Ohm’s Law: J = σ E I/A = σ V/L I/A I = V/R V/R I = V/(L/σ Α) R= L σA Physics 212 Le cture9, S 7 lide V = EL I = JA R = Resistance ρ = 1/σ 1/ This is just likeplum bing! This I is likeflow rateof wate r V is likepre ssure R is how hard it is for wate to flow in a pipe r L R= σA To m R big, m L long or A sm ake ake all To m R sm m L short or A big all, ake To ake Physics 212 Le cture9, S 8 lide Pre flight 2 S ecurre through both am nt re sistors re Pre flight 4 C parevoltage across om s re sistors re 70 60 50 40 30 20 10 L R∝ A 70 60 50 40 30 20 10 0 L V = IR ∝ A 0 A2 = 4 A1 ⇒ V2 = 1 V1 4 L2 = 2 L1 ⇒ V2 = 2V1 Physics 212 Le cture9, S 9 lide Pre flight 12 BB 60 I J≡ A S eCurre am nt 50 J1 = J 3 = 2 J 2 40 30 20 10 0 1 J∝ A Physics 212 Le cture9, S 10 lide 10 Resistor Summary Series R1 R2 R2 Each resistor on a different wire. S ef or e re am ach sistor. Vtotal = V1 = V2 Different for each resistor I =I +I total De ase 2 cre 1 s 1/Re = 1/R1 + 1/R2 q Parallel R1 Wiring Voltage Each resistor on the same wire. Different for each resistor. Vtotal = V1 + V2 Same for each resistor I =I =I Itotal 1 2 ncreases Req = R1 + R2 Current Resistance Physics 212 Le cture9, S 11 lide 11 Pre flight 6 100 80 60 40 20 0 R2 in se s with R3 rie C nt through R2 and R3 urre is thesam e is I 23 = V R2 + R3 Physics 212 Le cture9, S 12 lide 12 R1 = R2 = R3 = R Pre flight 7 C parethecurre through R1 with the om nt curre through R2 nt curre I 1/I 2 = 1/2 I 1/I 2 = 1/3 I 1/I 2 = 1 I 1/I 2 = 2 I 1/I 2 = 3 Pre flight 9 C parethevoltageacross R2 om with thevoltageacross R3 with V2 > V3 V2 = V3 = V Pre flight 10 C parethevoltageacross R1 with om thevoltageacross R2 t he V1 = V2 = V V1 = ½ V2 = V V1 = 2V2 = V V2 = V3 < V V2 < V3 V1 = ½ V2 = 1/5 V 1/5 V1 = ½ V2 = ½ V Physics 212 Le cture9, S 13 lide 13 60 50 40 30 20 10 0 R1 = R2 = R3 = R Pre flight 7 C parethecurre through R1 with the om nt curre through R2 nt curre I 1/I 2 = 1/2 I 1/I 2 = 1/3 I 1/I 2 = 1 I 1/I 2 = 2 I 1/I 2 = 3 Weknow: I 23 = V R2 + R3 I1 = I 23 V R1 S ilarly: im I1 = R2 + R3 R1 I1 R2 + R3 = =2 I 23 R1 Physics 212 Le cture9, S 14 lide 14 BB 60 50 40 30 20 R1 = R2 = R3 = R 10 0 Pre flight 9 C parethevoltageacross R2 om with thevoltageacross R3 with A B C D V2 > V3 V2 = V3 = V V2 = V3 < V V2 < V3 C onside loop r V23 = V V23 = V2 + V3 R2 = R3 ⇒ V2 = V3 V V2 = V3 = 2 Physics 212 Le cture9, S 15 lide 15 BB 50 40 30 20 R1 = R2 = R3 = R 10 0 Pre flight 10 C parethevoltageacross R1 with om thevoltageacross R2 t he A B C D E V1 = V2 = V V1 = ½ V2 = V V1 = 2V2 = V V1 = ½ V2 = 1/5 V 1/5 V1 = ½ V2 = ½ V R1 in paralle with se s com l rie bination of R2 and R3 of V 1= V23 R2 = R3 ⇒ V2 = V3 V1 = 2V2 = V V23 = V2 + V3 = 2V2 Physics 212 Le cture9, S 16 lide 16 C alculation R1 V R3 R4 R2 I n thecircuit shown: V = 18V, R1 = 1Ω, R2 = 2Ω, R3 = 3Ω, and R4 = 4Ω. once • C ptual Analysis: – – Ohm Law: whe curre I f lows through re ’s n nt sistanceR, thepote ntial drop V is give by: V = I R. n Re sistance arecom d in se s and paralle com s bine rie l binations rie • Rse s = Ra + Rb l • (1/ Rparalle) = (1/ Ra) + (1/ Rb) trate • S gic Analysis om sistance to forme s quivale re nt sistance s – C binere s nts ’s – Evaluatevoltage or curre fromOhm Law dge s nts – Expand circuit back using knowle of voltage and curre Physics 212 Le cture9, S 17 lide 17 C alculation R1 V R3 R4 R2 I n thecircuit shown: V = 18V, R1 = 1Ω, R2 = 2Ω, R3 = 3Ω, and R4 = 4Ω. BB om sistance s: • C bineRe R1 and R2 areconne d: cte (A) in se s (B) in paralle (C ne r in se s nor in paralle rie l ) ithe rie l Paralle Com l bination Paralle Ra S rie C bination e s om Ra Rb Rb Parallel: Can make a loop that contains only those two resistors Series : Every loop with resistor 1 also has resistor 2. Physics 212 Le cture9, S 18 lide 18 C alculation R1 V R3 R4 R2 I n thecircuit shown: V = 18V, R1 = 1Ω, R2 = 2Ω, R3 = 3Ω, and R4 = 4Ω. BB om sistance s: • C bineRe R2 and R4 areconne d: cte (A) in se s (B) in paralle (C ne r in se s nor in paralle rie l ) ithe rie l S rie C bination e s om Ra Rb Series : Every loop with resistor 1 also has resistor 2. Physics 212 Le cture9, S 19 lide 19 C alculation R1 V R3 R4 R2 I n thecircuit shown: V = 18V, R1 = 1Ω, R2 = 2Ω, R3 = 3Ω, and R4 = 4Ω. BB om sistance s: • C bineRe R2 and R4 areconne d in se s = R24 = 2 + 4 = 6 Ω cte rie Re draw the circuit using thee quivale re nt sistor R24 = se s com rie bination of R2 and R4. R1 V R3 R24 V R1 R3 R24 V R1 R3 R24 (A) (B) (C ) Physics 212 Le cture9, S 20 lide 20 C alculation R1 I n thecircuit shown: V = 18V, V R3 R24 R1 = 1Ω, R2 = 2Ω, R3 = 3Ω, and R4 = 4Ω. BB om sistance s: • C bineRe R2 and R4 areconne d in se s = R24 cte rie R3 and R24 areconne d in paralle = R234 cte l What is thevalueof R234? What (A) R234 = 1 Ω (B) R234 = 2 Ω (C R234 = 4 Ω (D) R234 = 6 Ω ) (1/Rparalle) = (1/ Ra) + (1/ Rb) l 1/R234 = (1/3) + (1/6) = (3/6) Ω− 1 R234 = 2 Ω Physics 212 Le cture9, S 21 lide 21 C alculation R1 I n thecircuit shown: V = 18V, V I =I 1 234 R1 = 1Ω, R2 = 2Ω, R3 = 3Ω, and R4 = 4Ω. R234 R1 and R234 arein se s. R1234 = 1 + 2 = 3 Ω rie V =I 1234 R1234 Ohm Law te us I 1234 = V/R1234 ’s lls = 18 / 3 = 6 Am ps Physics 212 Le cture9, S 22 lide 22 C alculation V I 1234 I n thecircuit shown: V = 18V, R1234 R1 = 1Ω, R2 = 2Ω, R3 = 3Ω, and R4 = 4Ω. BB R1 V =I =I 1 234 a I 234 = I 1234 S R1 in se s w/ R234 ince rie R234 b V234 = I 234 R234 = 6x2 = 12 Volts • What is Vab, thevoltageacross R234 ? (A) Vab = 1 V (B) Vab = 2 V (C Vab = 9 V (D) Vab = 12 V (E) Vab = 16 V ) Physics 212 Le cture9, S 23 lide 23 C alculation R1 V R234 V R3 R24 R1 V = 18V R1 = 1Ω R2 = 2Ω R3 = 3Ω R4 = 4Ω. BB Which of thefollowing aretrue ? A) V234 = V24 B) I 234 = I 24 C Both A+B ) D) None S R3 and R24 whe com d in paralle to ge R234 Voltage aresam ! ince re bine l t s e Ohm Law ’s I 24 = V24 / R24 = 12 / 6 = 2 Am ps Physics 212 Le cture9, S 24 lide 24 C alculation R1 V R3 R24 V R3 R4 R1 R2 V = 18V R1 = 1Ω R2 = 2Ω R3 = 3Ω R4 = 4Ω. BB Which of thefollowing aretrue ? A) V24 = V2 B) I 24 = I 2 C Both A+B ) D) None S R2 and R4 whe com d in se s to ge R24 C nts aresam ! ince re bine rie t urre e Ohm Law ’s V2 = I 2 R2 = 2 x2 = 4 Volts! Physics 212 Le cture9, S 25 lide 25 Quick Follow-Ups R1 V R3 R4 b R2 V R1 a = V = 18V R1 = 1Ω R2 = 2Ω R3 = 3Ω R4 = 4Ω. BB R234 • What is I 3 ? (A) I 3 = 2 A (B) I 3 = 3 A (C) I 3 = 4 A I 3 = V3/R3 = 12V/3Ω = 4A V3 = Vab = 12V • What is I 1 ? Weknow I = I We = 6A NOTE: I 2 = V2/R2 = 4/2 = 2 A I1= I2+ I3 MakeS nse e ??? Physics 212 Le cture9, S 26 lide 26 ...
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This note was uploaded on 03/05/2011 for the course PHYS 212 taught by Professor Kim during the Spring '08 term at University of Illinois, Urbana Champaign.

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