Lect09 - Physics 212 Lecture 9 Today's Concept: Electric...

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Unformatted text preview: Physics 212 Lecture 9 Today's Concept: Electric Current Ohm’s Law & resistors, Resistors in circuits, Power in circuits 35 30 25 20 15 10 5 0 Confused Confident Physics 212 Lecture 9, Slide 1 Music Who is the Artist? BB A) B) C) D) E) Stephane Grappelli Pearl Django Pearl Django Mark O’Connor’s Hot Swing Trio Bob Wills Hot Club of Cowtown Hot Cowtown Why?? THEY ARE COMING TO TOWN !! THEY SATURDAY, SEPT 25, 8:45PM SATURDAY, http://www.folkandroots.org/site/ (THU – SAT) (THU Catch Prof Kwiat (your 213/214 lecturer next Catch spring) do a swing dance demo before the show.. spring) Great musicians… wide variety of styles Western Swing meets Gypsy Jazz Great Live Show !!!! Physics 212 Lecture 9, Slide 2 Some Exam Stuff • Exam Tomorrow at 7:00 – – – – Covers material through lecture 8 Bring your ID: Rooms determined by discussion section (see link) Conflict exam at 5:15 for those who signed up in gradebook If you have conflicts with both of these, you should have heard from Prof Beck about scheduling https://my.physics.illinois.edu/undergrad/examprep/index.asp Physics 212 Lecture 9, Slide 3 Thanks for your feedback about the course There were, of course, a mixed bag of opinions on what would make the the course better. course However, here’s our initial take on the comments 1) Make sure we have time to devote to the problem for the day 2) Reduce the votes on preflight questions that most students get 3) The homeworks are challenging; it would be good to give more help The homeworks We will probably construct a survey, based on these suggestions from We you, to be given after the exam to get a better quantitative measure of your opinions. measure Physics 212 Lecture 9, Slide 4 A BIG IDEA REVIEW BIG Coulomb’s Law Force law between Force point charges point Electric Field Force per unit charge Gauss’ Law Flux through closed Flux surface is always proportional to charge enclosed charge Electric Potential Potential energy per Potential unit charge 05 Capacitance Capacitance Relates charge and Relates potential for two conductor system kq1q2 ˆ F1,2 = 2 r1,2 r1,2 q1 Electric Field Electric Property of Space Created by Charges Superposition F E≡ q ∫ E idA = Gauss’ Law Can be used to Can determine E field Qenc ε0 ∆U a →b ≡ = − ∫ E ⋅ dl q a b ∆Va →b Q C≡ V q2 r1,2 F1,2 Spheres Spheres Cylinders Infinite Planes Electric Potential Electric Scalar Function that Scalar can be used to determine E determine E = −∇V Physics 212 Lecture 9, Slide 5 APPLICATIONS OF BIG IDEAS APPLICATIONS Conductors Charges free to Charges move move What Determines What How They Move? How Spheres Cylinders Infinite Planes They move until They E = 0 !!! !!! Gauss’ Law E = 0 in conductor in determines charge densities on surfaces densities s lls t tiia en en ot po uiip Eq Eq Fiel d L ine s Field Lines & Field Equipotentials Equipotentials Capacitor Networks Work Done By E Field b b a a Wa →b = ∫ F ⋅ dl = ∫ qE ⋅ dl Change in Potential Energy Change b ∆U a →b = −Wa →b = − ∫ qE ⋅ dl a 05 Series: (1/C23)=(1/C2)+(1/C3) Parallel C123 = C1 + C23 Physics 212 Lecture 9, Slide 6 Key Concepts: 1) How resistance depends on A, L, σ, ρ How 2) How to combine resistors in series and parallel 3) Understanding resistors in circuits Today’s Plan: 1) Review of resistance & preflights Review preflights 2) Work out a circuit problem in detail Physics 212 Lecture 9, Slide 7 σ I A L V Conductivity – high for good conductors. Conductivity Ohm’s Law: J = σ E Observables: V = EL I = JA I/A = σV/L I/A R = Resistance ρ = 1/σ I = V/R I = V/(L/σΑ) R= L σA Physics 212 Lecture 9, Slide 8 This is just like plumbing! I is like flow rate of water V is like pressure R is how hard it is for water to flow in a pipe is L R= σA To make R big, make L long or A small big, long To make R small, make L short or A big To small, short Physics 212 Lecture 9, Slide 9 Preflight 2 Preflight 4 Same current through Same both resistors both Compare voltages Compare across resistors across L R∝ A 70 60 V = IR ∝ 50 40 30 L A 60 50 40 30 20 20 10 10 0 70 0 A2 = 4 A1 ⇒ V2 = 1 V1 4 L2 = 2 L1 ⇒ V2 = 2V1 Physics 212 Lecture 9, Slide 10 10 Preflight 12 BB 70 I J≡ A Same Current 60 J1 = J 3 = 2 J 2 50 40 30 1 J∝ A 20 10 0 Physics 212 Lecture 9, Slide 11 11 Resistor Summary Series Parallel R1 R1 R2 R2 Wiring Voltage Current Resistance Each resistor on the same wire. Each resistor on a different wire. Different for each resistor. Vtotal = V1 + V2 Same for each resistor. Vtotal = V1 = V2 Same for each resistor Itotal = I1 = I2 Different for each resistor Itotal = I1 + I2 Decreases 1/Req = 1/R1 + 1/R2 Increases Req = R1 + R2 Physics 212 Lecture 9, Slide 12 12 Preflight 6 100 80 60 40 20 0 R2 in series with R3 Current through R2 and R3 is the same I 23 = V R2 + R3 Physics 212 Lecture 9, Slide 13 13 R1 = R2 = R3 = R Preflight 7 Compare the current through Compare R1 with the current through R2 I1 I2 Preflight 9 Compare the voltage Compare across R2 with the across with voltage across R3 voltage V2 V3 Preflight 10 Compare the voltage across Compare R1 with the voltage across R2 V1 V2 Physics 212 Lecture 9, Slide 14 14 I1 50 40 I23 30 20 10 0 R1 = R2 = R3 = R Preflight 7 Compare the current through Compare R1 with the current through R2 I1/I2 = 1/2 I1/I2 = 1/3 We know: I 23 = V R2 + R3 I1 = I 23 Similarly: I1 = V R1 R2 + R3 R1 I1/I2 = 1 I1/I2 = 2 I1/I2 = 3 I1 R2 + R3 = =2 I 23 R1 Physics 212 Lecture 9, Slide 15 15 BB V2 V3 V23 50 40 30 20 R1 = R2 = R3 = R 10 Preflight 9 Compare the voltage Compare across R2 with the across with voltage across R3 voltage A V2 > V3 B V2 = V3 = V C V2 = V3 < V D 0 Consider loop V23 = V V2 < V3 V23 = V2 + V3 R2 = R3 ⇒ V2 = V3 V V2 = V3 = 2 Physics 212 Lecture 9, Slide 16 16 V1 BB 50 40 V23 30 20 R1 = R2 = R3 = R 0 Preflight 10 Compare the voltage across Compare R1 with the voltage across R2 A V1 = V2 = V B V1 = ½ V2 = V C V1 = 2V2 = V D V1 = ½ V2 = 1/5 V 1/5 E V1 = ½ V2 = ½ V 10 R1 in parallel with series in combination of R2 and R3 combination V 1= V23 R2 = R3 ⇒ V2 = V3 V1 = 2V2 = V V23 = V2 + V3 = 2V2 Physics 212 Lecture 9, Slide 17 17 R2 R1 Calculation In the circuit shown: V = 18V, R1 = 1Ω, R2 = 2Ω, R3 = 3Ω, and R4 = 4Ω. V R3 R4 What is V2, the voltage across R2? • Conceptual Analysis: – – Ohm’s Law: when current I flows through resistance R, the potential drop V is given by: V = IR. Resistances are combined in series and parallel combinations • Rseries = Ra + Rb • (1/Rparallel) = (1/Ra) + (1/Rb) • Strategic Analysis – – – Combine resistances to form equivalent resistances Evaluate voltages or currents from Ohm’s Law Expand circuit back using knowledge of voltages and currents Physics 212 Lecture 9, Slide 18 18 R2 R1 V Calculation In the circuit shown: V = 18V, R1 = 1Ω, R2 = 2Ω, R3 = 3Ω, and R4 = 4Ω. BB R3 What is V2, the voltage across R2? R4 • Combine Resistances: R1 and R2 are connected: (A) in series (B) in parallel (B) (C) neither in series nor in parallel (C) Parallel Combination Parallel Ra Series Combination Series Ra Rb Parallel: Can make a loop that contains only those two resistors Rb Series : Every loop with resistor 1 also has resistor 2. Physics 212 Lecture 9, Slide 19 19 R2 R1 V Calculation In the circuit shown: V = 18V, R1 = 1Ω, R2 = 2Ω, R3 = 3Ω, and R4 = 4Ω. BB R3 R4 What is V2, the voltage across R2? • We first will combine resistances R2 + R3 + R4: • Which of the following is true? (A) R2, R3 and R4 are connected in series (B) R2, R3, and R4 are connected in parallel (C) R3 and R4 are connected in series (R34) which is connected in parallel with R2 (D) R2 and R4 are connected in series (R24) which is connected in parallel with R3 (E) R2 and R4 are connected in parallel (R24) which is connected in parallel with R3 Physics 212 Lecture 9, Slide 20 20 R2 R1 Calculation In the circuit shown: V = 18V, R1 = 1Ω, R2 = 2Ω, R3 = 3Ω, and R4 = 4Ω. V BB R3 What is V2, the voltage across R2? R4 R2 and R4 are connected in series (R24) which is connected in parallel with R3 Redraw the circuit using the equivalent resistor R24 = series combination of R2 and R4. R1 V R1 R3 R3 V V R3 R24 R24 (A) (A) R1 (B) R24 (C) Physics 212 Lecture 9, Slide 21 21 Calculation R1 V In the circuit shown: V = 18V, R1 = 1Ω, R2 = 2Ω, R3 = 3Ω, and R4 = 4Ω. R3 BB R24 What is V2, the voltage across R2? • Combine Resistances: R2 and R4 are connected in series = R24 R3 and R24 are connected in parallel = R234 What is the value of R234? (A) R234 = 1 Ω (A) (B) R234 = 2 Ω (B) R2 and R4 in series (1/Rparallel) = (1/Ra) + (1/Rb) (C) R234 = 4 Ω (C) (D) R234 = 6 Ω (D) R24 = R2 + R4 = 2Ω + 4Ω = 6Ω 1/R234 = (1/3) + (1/6) = (3/6) Ω−1 R234 = 2 Ω Physics 212 Lecture 9, Slide 22 22 R1 Calculation In the circuit shown: V = 18V, R1 = 1Ω, R2 = 2Ω, R3 = 3Ω, and R4 = 4Ω. V I1 = I234 R234 R24 = 6Ω R234 = 2Ω What is V2, the voltage across R2? R1 and R234 are in series. R1234 = 1 + 2 = 3 Ω Our next task is to calculate the total current in the circuit Ohm’s Law tells us V = I1234 R1234 I1234 = V/R1234 = 18 / 3 = 6 Amps Physics 212 Lecture 9, Slide 23 23 Calculation V I1234 R1234 In the circuit shown: V = 18V, R1 = 1Ω, R2 = 2Ω, R3 = 3Ω, and R4 = 4Ω. BB R24 = 6Ω R234 = 2Ω I1234 = 6 A What is V2, the voltage across R2? R1 a I234 = I1234 Since R1 in series w/ R234 V R234 = I1 = I234 V234 = I234 R234 =6x2 b = 12 Volts • What is Vab, the voltage across R234 ? (A) Vab = 1 V (B) Vab = 2 V (B) (C) Vab = 9 V (D) Vab = 12 V (C) (D) (E) Vab = 16 V (E) Physics 212 Lecture 9, Slide 24 24 Calculation R1 R1 V V R234 R3 R24 V = 18V R1 = 1Ω R2 = 2Ω R3 = 3Ω R4 = 4Ω. BB R24 = 6Ω R234 = 2Ω I1234 = 6 Amps I234 = 6 Amps V234 = 12V What is V2? Which of the following are true? A) V234 = V24 B) I234 = I24 C) Both A+B R3 and R24 were combined in parallel to get R234 D) None Voltages are same! Ohm’s Law I24 = V24 / R24 = 12 / 6 = 2 Amps Physics 212 Lecture 9, Slide 25 25 Calculation R1 V I1234 R3 R1 V I24 R2 R3 R24 R4 Which of the following are true? A) V24 = V2 B) I24 = I2 C) Both A+B R2 and R4 where combined in series to get R24 D) None V = 18V R 1 = 1Ω R2 = 2Ω R3 = 3Ω R4 = 4Ω. BB R24 = 6Ω R234 = 2Ω I1234 = 6 Amps I234 = 6 Amps V234 = 12V V24 = 12V I24 = 2 Amps What is V2? Currents are same! Ohm’s Law The Problem Can Now The Be Solved! Be V2 = I2 R2 = 2 x2 = 4 Volts! Physics 212 Lecture 9, Slide 26 26 I1 Quick Follow-Ups R1 R2 I2 R1 I3 V = R3 a V R234 R4 b • What is I3 ? (A) I3 = 2 A (B) I3 = 3 A (B) V3 = V234 = 12V (C) I3 = 4 A (C) V = 18V R1 = 1Ω R2 = 2Ω R3 = 3Ω R4 = 4Ω. BB R24 = 6Ω R234 = 2Ω V234= 12V V2 = 4V I1234 = 6 Amps I3 = V3/R3 = 12V/3Ω = 4A • What is I1 ? We know I1 = I1234 = 6 A We NOTE: I2 = V2/R2 = 4/2 = 2 A I1 = I2 + I3 Make Sense??? Physics 212 Lecture 9, Slide 27 27 ...
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This note was uploaded on 02/09/2012 for the course PHYSICS 212 taught by Professor Mestre during the Spring '11 term at University of Illinois at Urbana–Champaign.

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