Your comments Having taken ece 110 made it easy for me to grasp most of the concepts. God I can go to bed now. Oh wait I can't cuz EXAM??!?!!111 I know that this course has to move at a fast pace and that there is no time to dilly dally and putts around, but I really really really don't like the idea of having a prelecture and lecture over a topic not covered on an upcoming exam. I find that I can't give the prelecture or the lecture the time they deserve to solidify my understanding, so this prelecture, though important, was gone through very quickly. This stuff is tough, I think this is mostly because we were thrown a ton of unfamiliar variables. Could we go over what all they exactly mean? Current density should be covered a little more thoroughly. PLEASE go over how to calculate voltage drops across a capacitor! Please help. I'm stressing out because the test is next Wednesday and we have to know circuits and all these different values (Q, I, R, ) with resistors and capacitors. For someone who has never done this before, this is hard!!! Electricity & Magnetism Lecture 9, Slide 1
Electric Current Physics 212 Lecture 9 Today s Concept: Ohm s Law, Resistors in circuits Electricity & Magnetism Lecture 9, Slide 2
A Big Idea Review Coulomb s Law Force law between point charges Electric Field Force per unit charge Gauss Law Flux through closed surface is always proportional to charge enclosed kqq 1 2 F1,2 ˆr 2 r1,2 F E q E da 1,2 Q enc 0 q 1 Electric Field Property of Space Created by Charges Superposition Gauss Law Can be used to determine E field F 1,2 r1,2 q 2 Spheres Cylinders Infinite Planes Electric Potential Potential energy per unit charge Capacitance Relates charge and potential for two conductor system ab C U q Q ab b E dl a Electric Potential Scalar Function that can be used to determine E E Electricity & Magnetism Lecture 9, Slide 3
Applications of Big Ideas Conductors Charges free to move What Determines How They Move? They move until E 0! Field Lines & Equipotentials Spheres Cylinders Infinite Planes Gauss Law E 0 in conductor determines charge densities on surfaces Work Done By E Field b b W F dl qe dl ab a a Capacitor Networks Change in Potential Energy b U W qe dl ab ab a Series: (1/C 23 ) (1/C 2 ) + (1/C 3 ) Parallel C 123 C 1 + C 23 Electricity & Magnetism Lecture 9, Slide 4
Current and Resistance Key Concepts: 1) How resistance depends on A, L, s, r 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 Electricity & Magnetism Lecture 9, Slide 5
I s A L Conductivity high for good conductors. Ohm s Law: J s E Observables: EL I JA I/A s/l I /(L/sA) R Resistance r 1/s I /R R L sa Electricity & Magnetism Lecture 9, Slide 6
CheckPoint 3 The SAME amount of current I passes through three different resistors. R 2 has twice the crosssectional area and the same length as, and R 3 is three times as long as but has the same cross-sectional area as. In which case is the CURRENT DENSITY through the resistor the smallest? A. Case 1 B. Case 2 C. Case 3 J I A Same Current J 1 J3 2J2 J 1 A Electricity & Magnetism Lecture 9, Slide 7
This is just like Plumbing! I is like flow rate of water is like pressure R is how hard it is for water to flow in a pipe R L sa To make R big, make L long or A small To make R small, make L short or A big Electricity & Magnetism Lecture 9, Slide 8
CheckPoint 1a CheckPoint 1b Same current through both resistors Compare voltages across resistors R IR L A L A Electricity & Magnetism Lecture 9, Slide 9
Resistor Summary Series Every loop with also has R 2 R 2 Parallel There is a loop that contains ONLY and R 2 Wiring oltage Each resistor on the same wire. Different for each resistor. total 1 + 2 R 2 Each resistor on a different wire. Same for each resistor. total 1 2 Current Resistance Same for each resistor I total I 1 I 2 Increases R eq + R 2 Different for each resistor I total I 1 + I 2 Decreases 1/R eq 1/ + 1/R 2 Electricity & Magnetism Lecture 9, Slide 10
CheckPoint 2a Three resistors are connected to a battery with emf as shown. The resistances of the resistors are all the same, i.e. = R 2 = R 3 = R. Compare the current through R 2 with the current through R 3 : A. I 2 > I 3 B. I 2 = I 3 C. I 2 < I 3 R 2 in series with R 3 Current through R 2 and R 3 is the same I 23 R 2 + R 3 Electricity & Magnetism Lecture 9, Slide 11
Checkpoint 2b I 1 I 23 CheckPoint 2b Compare the current through with the current through R 2 A B C D E 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 R 2 R 3 R I We know: Similarly: 23 R + R I 1 2 R 1 3 I I 1 2 3 23 I 1 R I 23 + R R 1 R2 + R R 1 3 2 Electricity & Magnetism Lecture 9, Slide 13
2 3 23 R 2 R 3 R CheckPoint 2c Compare the voltage across R 2 with the voltage across R 3 A B C D 2 > 3 2 3 2 3 < 2 < 3 I 2 2R2 I 3 3R3 I 2 = I 3 (Series) R 2 = R 3 (Problem statement) 2 3 23 + 23 2 3 2 3 2 Electricity & Magnetism Lecture 9, Slide 14
1 23 CheckPoint 2d Compare the voltage across with the voltage across R 2 A B C D E 1 2 1 ½ 2 1 2 2 1 ½ 2 1/5 1 ½ 2 ½ R R 2 R 3 R in parallel with series combination of R 2 and R 3 1 23 2 R3 2 3 23 2 + 3 2 2 2 1 2 Electricity & Magnetism Lecture 9, Slide 15
Calculation R 2 R 3 R 4 In the circuit shown: 18, 1W, R 2 2W, R 3 3W, and R 4 4W. What is 2, the voltage across R 2? Conceptual Analysis: Ohm s Law: when current I flows through resistance R, the potential drop is given by: IR. Resistances are combined in series and parallel combinations R series R a + R b (1/R parallel ) (1/R a ) + (1/R b ) 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 Electricity & Magnetism Lecture 9, Slide 16
Calculation R 2 R 3 R 4 In the circuit shown: 18, 1W, R 2 2W, R 3 3W, and R 4 4W. What is 2, the voltage across R 2? Combine Resistances: and R 2 are connected: A) in series B) in parallel C) neither in series nor in parallel Parallel Combination R a Series Combination R a R b R b Parallel: Can make a loop that contains only those two resistors Series : Every loop with resistor 1 also has resistor 2. Electricity & Magnetism Lecture 9, Slide 17
Calculation R 2 R 3 R 4 In the circuit shown: 18, 1W, R 2 2W, R 3 3W, and R 4 4W. What is 2, the voltage across R 2? We first will combine resistances R 2, R 3, R 4 : Which of the following is true? A) R 2, R 3 and R 4 are connected in series B) R 2, R 3, and R 4 are connected in parallel C) R 3 and R 4 are connected in series (R 34 ) which is connected in parallel with R 2 D) R 2 and R 4 are connected in series (R 24 ) which is connected in parallel with R 3 E) R 2 and R 4 are connected in parallel (R 24 ) which is connected in parallel with R 3 Electricity & Magnetism Lecture 9, Slide 18
Calculation R 2 R 3 R 4 In the circuit shown: 18, 1W, R 2 2W, R 3 3W, and R 4 4W. What is 2, the voltage across R 2? R 2 and R 4 are connected in series (R 24 ) which is connected in parallel with R 3 Redraw the circuit using the equivalent resistor R 24 series combination of R 2 and R 4. R 3 R 3 R 3 R 24 R 24 R 24 (A) (B) (C) Electricity & Magnetism Lecture 9, Slide 19
Calculation In the circuit shown: 18, 1W, R 2 2W, R 3 3W, and R 4 4W. R 3 R 24 What is 2, the voltage across R 2? Combine Resistances: What is the value of R 234? R 2 and R 4 are connected in series R 24 R 3 and R 24 are connected in parallel R 234 A) R 234 1 W B) R 234 2 W C) R 234 4 W D) R 234 6 W R 2 and R 4 in series R 24 R 2 + R 4 2W + 4W 6W (1/R parallel ) (1/R a ) + (1/R b ) 1/R 234 (1/3) + (1/6) (3/6) W -1 R 234 2 W Electricity & Magnetism Lecture 9, Slide 20
Calculation In the circuit shown: 18, I 1 I 234 R 234 234 1W, R 2 2W, R 3 3W, and R 4 4W. R 24 6W R 234 2W What is 2, the voltage across R 2? and R 234 are in series. 234 1 + 2 3 W Our next task is to calculate the total current in the circuit Ohm s Law tells us: I 1234 /234 I 1234 234 18 / 3 6 Amps Electricity & Magnetism Lecture 9, Slide 21
Calculation I 1234 234 In the circuit shown: 18, 1W, R 2 2W, R 3 3W, and R 4 4W. R 24 6W R 234 2W I 1234 6 A What is 2, the voltage across R 2? a I 234 I 1234 Since in series with R 234 I 1 I 234 R 234 234 b 234 I 234 R 234 6 x 2 12 olts What is ab, the voltage across R 234? A) ab 1 B) ab 2 C) ab 9 D) ab 12 E) ab 16 Electricity & Magnetism Lecture 9, Slide 22
Calculation R 234 R 3 R 24 Which of the following are true? A) 234 24 B) I 234 I 24 C) Both A+B D) None 18 1W R 2 2W R 3 3W R 4 4W R 24 6W R 234 2W I 1234 6 Amps I 234 6 Amps 234 12 What is 2? R 3 and R 24 were combined in parallel to get R 234 oltages are same! Ohm s Law I 24 24 / R 24 12 / 6 2 Amps Electricity & Magnetism Lecture 9, Slide 23
Calculation R 3 R 24 Which of the following are true? A) 24 2 B) I 24 I 2 C) Both A+B D) None I 1234 R 2 R 3 R 4 R 2 and R 4 where combined in series to get R 24 Currents are same! I 24 18 1W R 2 2W R 3 3W R 4 4W. R 24 6W R 234 2W I 1234 6 Amps I 234 6 Amps 234 12 24 12 I 24 2 Amps What is 2? The Problem Can Now Be Solved! Ohm s Law 2 I 2 R 2 2 x 2 4 olts! Electricity & Magnetism Lecture 9, Slide 24
Quick Follow-Ups I 1 I 2 R 2 I 3 R 3 R 4 What is I 3? A) I 3 2 A B) I 3 3 A C) I 3 4 A 3 = 234 = 12 I 3 3 /R 3 12/3W 4A What is I 1? We know I 1 I 1234 6 A a b R 234 18 1W R 2 2W R 3 3W R 4 4W R 24 6W R 234 2W 234 = 12 2 = 4 I 1234 = 6 Amps NOTE: I 2 2 /R 2 4/2 2 A I 1 I 2 + I 3 Make Sense? Electricity & Magnetism Lecture 9, Slide 25