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

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Unformatted text preview: Physics 212 Ph Lecture 9 Today's Concept: Electric Current Ohm’s Law & resistors, Resistors in circuits, Power in circuits Physics 212 Lecture 9, Slide 1 Music Who is the Artist? A) B) C) D) E) Dave Brubeck Thelonius Monk Bill Evans Keith Jarrett Oscar Peterson • Now there’s an example of abstract thinking – composing in real time! Now • Who will be next!? Physics 212 Lecture 9 Some Some Exam Stuff • Exam tomorrw (Wed. Sep. 21) at 7:00 – Covers material through lecture 8 (no resistors) – Bring your ID – Conflict exam at 5:15 – you will need to stay in the room until exam ends • Where do I go? EXAM ROOMS (by discussion section) • Preparation – Practice exams – Exam Prep Exercises – Worked examples Physics 212 Lecture 8, Slide 3 Your Your Comments • “Thank God were are moving on to easier material. Makes my life easier :)” • “I CURRENTly have only a few concerns” • “Can we review for tommorrow's exam?” • “will we have a review section?” • “I was confused with current density” • “Can we do more practice with resistor circuits that have both parallel and series?” • Consider the comments • • • • They help us understand what you are having trouble with Writing, even briefly, helps you to think logically even helps “I bet you put typos in the powerpoints to make it look like you are actually using what students wrote, when you are actually writing them. Checkmate!” “Can we look at schematics for electronics we use everyday? Do they have Can circuits that look like these?” Physics 212 Lecture 8, Slide 4 ELECTRONICS WE USE EVERYDAY (almost) – your iClicker A BIG IDEA REVIEW BIG Coulomb’s Law Force law between point charges kq1q2 ˆ F1,2 2 r1,2 r1,2 F Electric Field E Force per unit charge q Gauss’ Law Law Flux through closed surface is always proportional to charge enclosed Electric Potential Potential energy per unit unit charge 05 Capacitance Relates charge and potential for two conductor system Qenc E dA 0 Va b q1 q2 Electric Field Property of Space Created by Charges Superposition Gauss’ Law Can be used to determine determine E field b U a b E dl q a Q C V r1,2 F1,2 Spheres Spheres Cylinders Infinite Infinite Planes Electric Potential Scalar Function that can can be used to determine E E V Physics 212 Lecture 9, Slide 6 APPLICATIONS APPLICATIONS OF BIG IDEAS Conductors Charges free to move What Determines How They Move? Spheres Cylinders Infinite Planes They move until E = 0 !!! Gauss’ Law E = 0 in conductor determines charge densities on surfaces Field Lines & Equipotentials Capacitor Networks Work Done By E Field b Wa b F dl qE dl b a a Change in Potential Energy U a b Wa b qE dl b a 05 Series: (1/C23)=(1/C2)+(1/C3) Parallel C123 = C1 + C23 Physics 212 Lecture 9, Slide 7 Key Key Concepts for Today: 1) How resistance depends on A, L, , 2) How to combine resistors in series and parallel 3) Understanding resistors in circuits Today’s Plan: 1) Review of resistance & preflights 2) Work out a circuit problem in detail Physics 212 Lecture 9, Slide 8 I A L V Conductivity – high for good conductors. Ohm’s Ohm’s Law: J = E Observables: V = EL I = JA JA I/A I/A = V/L R = Resistance Resistance = 1/ I = V/R V/R I = V/(L/ V/(L/ R= L A Physics 212 Lecture 9, Slide 9 This This is like plumbing! I is like flow rate of water V is like pressure pr R is how hard it is for water to flow in a pipe L R= A To make R big, make L long or A small To make R small, make L short or A big sm sh Physics 212 Lecture 9, Slide 10 Checkpoint Checkpoint 1a Checkpoint 1b Same current through both resistors Compare voltages across resistors L R A L V IR A A2 4 A1 V2 1 V1 4 L2 2 L1 V2 2V1 Physics 212 Lecture 9, Slide 11 Checkpoint Checkpoint 3 The SAME amount of current I passes through three different resistors. R2 has twice the crosssectional area and the same length as R1, and R3 is three times as long as R1 but has the same cross-sectional area as R1. In which case is the CURRENT DENSITY through the resistor the smallest? which case is the CURRENT DENSITY through the resistor the smallest? A. Case 1 B. Case 2 C. Case 3 “The smaller the resistor the smaller the current density” “Has most area for current to spread out in so density is lowest” “Case has the smallest volume of the resistor with same amount of electric current. So, Current density is the smallest in case3.” Physics 212 Lecture 9, Slide 12 Checkpoint Checkpoint 3 The SAME amount of current I passes through three different resistors. R2 has twice the crosssectional area and the same length as R1, and R3 is three times as long as R1 but has the same cross-sectional area as R1. In which case is the CURRENT DENSITY through the resistor the smallest? which case is the CURRENT DENSITY through the resistor the smallest? A. Case 1 B. Case 2 C. Case 3 I J A Same Current J1 J 3 2 J 2 1 J A Physics 212 Lecture 9, Slide 13 Resistor Summary Series Parallel R1 R1 R2 R2 Wiring Each resistor on the th same wire. Each resistor on a different wire. Voltage Different for each resistor. Vtotal = V1 + V2 Same Same for each resistor. Vtotal = V1 = V2 Current Same for each resistor Itotal = I1 = I2 Different for each resistor Itotal = I1 + I2 Decreases 1/Req = 1/R1 + 1/R2 Resistance Increases Req = R1 + R2 Physics 212 Lecture 9, Slide 14 Three resistors are connected to a battery with emf V as shown. The resistances of the resistors are all the same, i.e. R1= R2 = R3 = R. Checkpoint Checkpoint 2a Compare the current through R2 with the current through R3: A. I2 > I3 B. I2 = I3 C. I2 < I3 R2 in series with R3 Current through R2 and R3 is the same I 23 V R2 R3 Physics 212 Lecture 9, Slide 15 R1 = R2 = R3 = R Checkpoint 2b Check Compare the current through R1 with the current through R2 I1 I2 Checkpoint 2c Compare the voltage across R2 with the voltage across R3 V2 V3 Checkpoint 2d Compare the voltage across R1 with the voltage across R2 V1 V2 Physics 212 Lecture 9, Slide 16 Checkpoint Checkpoint 2b R1 = R2 = R3 = R Compare the current through R1 with the current through R2 A. I1/I2 = 1/2 1/2 B. I1/I2 = 1/3 C. I1/I2 = 1 D. I1/I2 = 2 E. I1/I2 = 3 “R2 and R3 make up one resistor, which means the current R2 R3 th would would be multiplied by two. This means that the current through R1 divided by R23 would equal 1/2 ” “I2 must be equal to I3. Therefore, we can reason that I2 and I2 I3 Th th I2 I3 I3 are both one third of I total. Therefore, each resistor has equal current.” “the resistance of the series is twice as much so the current is th th th halved.” Physics 212 Lecture 9, Slide 17 Checkpoint Checkpoint 2b I1 I23 R1 = R2 = R3 = R Compare the current through R1 with the current through R2 A. I1/I2 = 1/2 1/2 B. I1/I2 = 1/3 We know: I 23 V R2 R3 I1 I 23 Similarly: I1 V R1 R2 R3 R1 C. I1/I2 = 1 D. I1/I2 = 2 E. I1/I2 = 3 I1 R2 R3 2 I 23 R1 Physics 212 Lecture 9, Slide 18 Checkpoint Checkpoint 2c V2 V23 V3 R1 = R2 = R3 = R Compare the voltage across R2 with the voltage across R3 A V2 > V3 B V2 = V3 = V C V2 = V3 < V D V2 < V3 Consider loop V23 V V23 V2 V3 R2 R3 V2 V3 V V2 V3 2 Physics 212 Lecture 9, Slide 19 Checkpoint Checkpoint 2d R1 = R2 = R3 = R Compare the voltage across 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 “Since V2 and V3 are in parallel with V1 and V2 and V3 are equal, they all have the same V” “The voltage is the same the beginning and end of the the parallel split. Since the resistors have the same resistance, then R1=R2+R3.” “V1=Vbattery V2+V3=Vbattery” V1=Vbattery V2+V3=Vbattery Physics 212 Lecture 9, Slide 20 Checkpoint Checkpoint 2d V1 V23 R1 = R2 = R3 = R Compare the voltage across 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 R1 in parallel with series combination of R2 and R3 V 1 V23 R2 = R3 , I 2 = I 3 V2 = V3 V23 V2 V3 2V2 V1 2V2 V Physics 212 Lecture 9, Slide 21 R2 R1 V Calculation Calculation In the circuit shown: V = 18V, R1 = 1R2 = 2R3 = 3and R4 = 4 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: “packing” Evaluate voltages or currents from Ohm’s Law Expand circuit back using knowledge of voltages and currents: “unpacking” Physics 212 Lecture 9, Slide 22 R2 R1 V Calculation Calculation In the circuit shown: V = 18V, R1 = 1R2 = 2R3 = 3and R4 = 4 R3 What is V2, the voltage across R2? R4 • Combine Resistances: R1 and R2 are connected: (A) (A) in series (B) in parallel ll (C) neither in series nor in parallel ith Parallel Combination Ra Series Combination 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 Physics 212 Lecture 9, Slide 23 R2 R1 V Calculation Calculation In the circuit shown: V = 18V, R1 = 1R2 = 2R3 = 3and R4 = 4 R3 R4 What is V2, the voltage across R2? • We first will combine resistances R2 + R3 + R4: fi • 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 24 R2 R1 V Calculation Calculation In the circuit shown: V = 18V, R1 = 1R2 = 2R3 = 3and R4 = 4 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) R1 (B) R24 (C) Physics 212 Lecture 9, Slide 25 Calculation Calculation R1 V In the circuit shown: V = 18V, R3 R1 = 1R2 = 2R3 = 3and R4 = 4 R24 • Combine Resistances: What is V2, the voltage across R2? 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 (B) R234 = 2 R2 and R4 in series (1/Rparallel) = (1/Ra) + (1/Rb) (C) R234 = 4 (D) R234 = 6 R24 = R2 + R4 = 2 1/R234 = (1/3) + (1/6) = (3/6) R234 = 2 Physics 212 Lecture 9, Slide 26 R1 V I1 = I234 Calculation Calculation In the circuit shown: V = 18V, R234 R1 = 1R2 = 2R3 = 3and R4 = 4 R24 = 6R234 = 2 What is V2, the voltage across R2? R1 and R234 are in series. R1234 = 1 + 2 = 3 Our next task is to calculate the total current in the circuit V Ohm’s Law tells us = I1234 R1234 I1234 = V/R1234 = 18 / 3 = 6 Amps Physics 212 Lecture 9, Slide 27 Calculation: Calculation: begin unpacking V In the circuit shown: V = 18V, I1234 R1234 R1 a V R234 = I1 = I234 R1 = 1R2 = 2R3 = 3and R4 = 4 R24 = 6R234 = 2I1234 = 6 A What is V2, the voltage across R2? I234 = I1234 Since R1 in series w/ R234 V234 = I234 R234 =6x2 b = 12 Volts • What is Vab, the voltage across R234 ? (A) Vab = 1 V (B) Vab = 2 V (C) Vab = 9 V (D) Vab = 12 V (E) Vab = 16 V Physics 212 Lecture 9, Slide 28 Calculation Calculation R1 R1 V V R234 R3 R24 V = 18V R1 = 1 R2 = 2 R3 = 3 R4 = 4 R24 = 6 R234 = 2 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 29 Calculation Calculation R1 V I1234 R3 R1 V R24 I24 R2 R3 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 R1 = 1 R2 = 2 R3 = 3 R4 = 4 R24 = 6 R234 = 2 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 Be Solved! V2 = I 2 R 2 = 2 x2 = 4 Volts! Physics 212 Lecture 9, Slide 30 I1 V R1 Quick FollowQuick Follow-Ups R2 I2 R1 I3 R3 = V R4 (B) I3 = 3 A V3 = V234 = 12V R234 b • What is I3 ? (A) (A) I3 = 2 A a (C) I3 = 4 A I3 = V3/R3 = 12V/3 = 4A V = 18V R1 = 1 R2 = 2 R3 = 3 R4 = 4 R24 = 6 R234 = 2 V234= 12V V2 = 4V I24 = 2 Amps I1234 = 6 Amps • What is I1 ? We We know I1 = I1234 = 6 A NOTE: I2 = V2/R2 = 4/2 = 2 A 4/2 I1 = I2 + I 3 Make Sense??? Physics 212 Lecture 9, Slide 31 ...
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