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Physics 2112 Unit 9: Electric Current Today’s Concept: Electric Current Electricity & Magnetism Lecture 9, Slide 1.

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Presentation on theme: "Physics 2112 Unit 9: Electric Current Today’s Concept: Electric Current Electricity & Magnetism Lecture 9, Slide 1."— Presentation transcript:

1 Physics 2112 Unit 9: Electric Current Today’s Concept: Electric Current Electricity & Magnetism Lecture 9, Slide 1

2 A Big Idea Review Electric Potential Potential energy per unit charge Electric Potential Scalar Function that can be used to determine E Gauss’ Law Flux through closed surface is always proportional to charge enclosed Gauss’ Law Can be used to determine E field Spheres Cylinders Infinite Planes Electricity & Magnetism Lecture 9, Slide 2 Capacitance Relates charge and potential for two conductor system Electric Field Force per unit charge Electric Field Property of Space Created by Charges Superposition Coulomb’s Law Force law between point charges q2q2 q1q1

3 Conductors Charges free to move Spheres Cylinders Infinite Planes Gauss’ Law Field Lines & Equipotentials Field Lines Equipotentials Work Done By E Field Change in Potential Energy Capacitor Networks Series:  1  C 23  1  C 2  1  C 3  Parallel C 123  C 1  C 23 Applications of Big Ideas What Determines How They Move? They move until E  0 ! E  0 in conductor determines charge densities on surfaces Electricity & Magnetism Lecture 9, Slide 3

4 Today’s Plan: 1) 1)Review of resistance & preflights 2) 2)Work out a circuit problem in detail Key Concepts: 1) 1)How resistance depends on A, L, s, r 2) 2)How to combine resistors in series and parallel 3) 3)Understanding resistors in circuits Electricity & Magnetism Lecture 9, Slide 4

5 I A V L s V  EL I  JA Observables: R R  L AA Ohm’s Law: J   E Conductivity – high for good conductors. IA  VLIA  VL I  V  L  A  I  VRI  VR R  Resistance R  1  Electricity & Magnetism Lecture 9, Slide 5

6 R  R  L AA I is like flow rate of water V is like pressure R is how hard it is for water to flow in a pipe Civil Engineering Analogy 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 6 Battery is like a pump

7 Same current through both resistors Compare voltages across resistors 1 CheckPoint: Two Resistors 2 Electricity & Magnetism Lecture 9, Slide 7

8 Electricity & Magnetism Lecture 9, Slide 8 CheckPoint: Current Density The SAME amount of current I passes through three different resistors. R 2 has twice the cross-sectional area and the same length as R 1, and R 3 is three times as long as R 1 but has the same cross-sectional area as R 1. In which case is the CURRENT DENSITY through the resistor the smallest?

9 Circuit Unit 9, Slide 9 VBVB R1R1 R2R2 a b c d E is conservative force  go completely around circuit V f = V o V a bc d a +V B -IR 1 -IR 2

10 Voltage Current Resistance Series Parallel Different for each resistor. V total  V 1  V 2 Increases R eq  R 1  R 2 Same for each resistor I total  I 1  I 2 Same for each resistor. V total  V 1  V 2 Decreases 1  R eq  1  R 1  1  R 2 Wiring Each resistor on the same wire. Each resistor on a different wire. Different for each resistor I total  I 1  I 2 R1R1 R2R2 R1R1 R2R2 Resistor Summary Electricity & Magnetism Lecture 9, Slide 10

11 CheckPoint: Resistor Network 1 Electricity & Magnetism Lecture 9, Slide 11 Three resistors are connected to a battery with emf V as shown. The resistances of the resistors are all the same, i.e. R 1 = R 2 = R 3 = R. Compare the current through R 2 with the current through R 3 : A. A. I 2 > I 3 B. B. I 2 = I 3 C. C. I 2 < I 3

12 CheckPoint: Resistor Network 2 Electricity & Magnetism Lecture 9, Slide 12 Three resistors are connected to a battery with emf V as shown. The resistances of the resistors are all the same, i.e. R 1 = R 2 = R 3 = R. Compare the current through R 1 with the current through R 2 : A. A. I 1 /I 2 =1/2 B. B. I 1 /I 2 =1/3 C. C. I 1 = I 2 D. D. I 1 /I 2 =2 E. E. I 1 /I 2 =3

13 CheckPoint: Resistor Network 3 Electricity & Magnetism Lecture 9, Slide 13 Three resistors are connected to a battery with emf V as shown. The resistances of the resistors are all the same, i.e. R 1 = R 2 = R 3 = R. Compare the voltage across R 2 with the voltage across R 3 : A. A. V 2 > V 3 B. B. V 2 = V 3 = V C. C. V 2 = V 3 < V D. D. V 2 < V 3

14 R 1  R 2  R 3  R CheckPoint 2 Compare the current through R 1 with the current through R 2 I 1 I 2 CheckPoint 3 Compare the voltage across R 2 with the voltage across R 3 V 2 V 3 CheckPoint 4 Compare the voltage across R 1 with the voltage across R 2 V 1 V 2 Electricity & Magnetism Lecture 9, Slide 14

15 CheckPoint: Resistor Network 4 Electricity & Magnetism Lecture 9, Slide 15 Three resistors are connected to a battery with emf V as shown. The resistances of the resistors are all the same, i.e. R 1 = R 2 = R 3 = R. Compare the voltage across R 1 with the voltage across R 2. A. A. V 1 = V 2 = V B. B. V 1 = 1/2 V 2 = V C. C. V 1 = 2V 2 = V D. D. V 1 =1/2 V 2 =1/5 V E. E. V 1 =1/2 V 2 = 1/2 V

16 Example 9.1 In the circuit shown: V  18V, R 1  1  R 2  2  R 3  3  and  R 4  4  What is V 2, the voltage across R 2 ? 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 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 V R1R1 R2R2 R4R4 R3R3 Electricity & Magnetism Lecture 9, Slide 16

17 Example 9.1 V R1R1 R2R2 R4R4 R3R3 Electricity & Magnetism Lecture 9, Slide 17 V R1R1 R 234 V R 1234 V R1R1 R2R2 R 24 R3R3

18 Quick Follow-Ups Electricity & Magnetism Lecture 9, Slide 18 What is I 3 ? A) I 3  2 A B) I 3  3 A C) I 3  4 A V R1R1 R 234 a b V R1R1 R2R2 R4R4 R3R3  V 3 = V 234 = 12V What is I 1 ? We know I 1  I 1234  6 A V  18V R 1  1  R 2  2  R 3  3  R 4  4  R 24  6  R 234  2  V 234 = 12V V 2 = 4V I 1234 = 6 Amps I 3  V 3 /R 3  12V/3   4A I1  I2  I3I1  I2  I3 Make Sense? I1I1 I2I2 I3I3

19 Power In Resistors Unit 9, Slide 19 Electro-Motive Force (EMF), Energy provided to make charges move, units of V Not a force!! = VI (for a battery) = I 2 R (for resistor)

20 Example 9.2 In the circuit shown: V  18V, R 1  1  R 2  2  R 3  3  and  R 4  4  How much electrical energy does the battery put into the circuit every second in the previous problem? V R1R1 R2R2 R4R4 R3R3 How much electrical energy does each resistor turn into thermal energy every second?

21 Parallel and Series (with color) Unit 9, Slide 21 V R1R1 R2R2 R4R4 R3R3. If every electron that goes through one element must go through another, those two are in series.. If two sides of two elements can be connected by different colored lines, those two are in parallel.. If two points are connected by a line not containing any circuit elements those point are at the same potential.


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