# Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Current Conservation of current Batteries Resistance and resistivity Simple.

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Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Current Conservation of current Batteries Resistance and resistivity Simple circuits Chapter 22 Current and Resistance Topics: Sample question: How can the measurement of an electric current passed through a person’s body allow a determination of the percentage body fat? Slide 22-1

Rank the bulbs in the following circuit according to their brightness, from brightest to dimmest. A.A > B > C > D B.A = B > C = D C.A = D > B = C D.B = C > A = D Checking Understanding Slide 22-16

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Answer Rank the bulbs in the following circuit according to their brightness, from brightest to dimmest. A.A > B > C > D B.A = B > C = D C.A = D > B = C D.B = C > A = D Slide 22-17

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. 4.In Trial 1, a battery is connected to a single lightbulb and the brightness noted. Now, in Trial 2, a second, identical, lightbulb is added. How does the brightness of these two bulbs compare to the brightness of the single bulb in Trial 1? A.The brightness is greater. B.The brightness is the same. C.The brightness is less. Additional Questions Slide 22-45

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Answer 4.In Trial 1, a battery is connected to a single lightbulb and the brightness noted. Now, in Trial 2, a second, identical, lightbulb is added. How does the brightness of these two bulbs compare to the brightness of the single bulb in Trial 1? A.The brightness is greater. B.The brightness is the same. C.The brightness is less. Slide 22-46

Definition of a Current Slide 22-9

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Batteries The potential difference between the terminals of a battery, often called the terminal voltage, is the battery’s emf. Slide 22-20

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Simple Circuits The current is determined by the potential difference and the resistance of the wire: Slide 22-13 I = ∆ V chem R _____

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Series and Parallel Slide 22-25 Series Circuit elements in a chain between 2 points Same current flows through circuit elements I 1 = I 2 Electric Potentials add => Delta V total = Delta V 1 + Delta V 2 Parallel Circuit elements on multiple paths connecting the same points Since paths connect the same points, Delta V’s are the same Currents Add => I total = I 1 + I 2

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Charge Carriers  The outer electrons of metal atoms are only weakly bound to the nuclei.  In a metal, the outer electrons become detached from their parent nuclei to form a fluid-like sea of electrons that can move through the solid.  Electrons are the charge carriers in metals. Slide 30-22

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. The Electron Current  We define the electron current i e to be the number of electrons per second that pass through a cross section of the conductor.  The number N e of electrons that pass through the cross section during the time interval Δ  t is Slide 30-23

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. The Electron Density  In most metals, each atom contributes one valence electron to the sea of electrons.  Thus the number of conduction electrons n e is the same as the number of atoms per cubic meter. Slide 30-27

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.  The wire is already full of electrons!  We don’t have to wait for electrons to move all the way through the wire from one plate to another.  We just need to slightly rearrange the charges on the plates and in the wire. Discharging a Capacitor Slide 30-30

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Establishing the Electric Field in a Wire  The figure shows two metal wires attached to the plates of a charged capacitor.  This is an electrostatic situation.  What will happen if we connect the bottom ends of the wires together? Slide 30-32

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.  Within a very brief interval of time (  10  9 s ) of connecting the wires, the sea of electrons shifts slightly.  The surface charge is rearranged into a nonuniform distribution, as shown in the figure. Establishing the Electric Field in a Wire Slide 30-33

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.  The nonuniform distribution of surface charges along a wire creates a net electric field inside the wire that points from the more positive end toward the more negative end of the wire.  This is the internal electric field that pushes the electron current through the wire. Establishing the Electric Field in a Wire Slide 30-34

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Surface charge is distributed on a wire as shown. Electrons in the wire QuickCheck 30.2 A. Drift to the right. B. Drift to the left. C. Move upward. D. Move downward. E. On average, remain at rest. Slide 30-35

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Surface charge is distributed on a wire as shown. Electrons in the wire QuickCheck 30.2 A. Drift to the right. B. Drift to the left. C. Move upward. D. Move downward. E. On average, remain at rest. Slide 30-36 Electric field from nonuniform surface charges is to the right. Force on negative electrons is to the left.

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. A Model of Conduction  Within a conductor in electrostatic equilibrium, there is no electric field.  In this case, an electron bounces back and forth between collisions, but its average velocity is zero. Slide 30-37

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.  In the presence of an electric field, the electric force causes electrons to move along parabolic trajectories between collisions.  Because of the curvature of the trajectories, there is a slow net motion in the “downhill” direction. A Model of Conduction Slide 30-38

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.  The graph shows the speed of an electron during multiple collisions.  The average velocity is the electron drift speed A Model of Conduction Slide 30-39

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. The Current Density in a Wire The current density J in a wire is the current per square meter of cross section: The current density has units of A/m 2. Slide 30-48 The Current Density in a Wire

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. The current density in this wire is QuickCheck 30.4 A. 4  10 6 A/m 2. B. 2  10 6 A/m 2. C. 4  10 3 A/m 2. D. 2  10 3 A/m 2. E. Some other value. Slide 30-49

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. The current density in this wire is QuickCheck 30.4 A. 4  10 6 A/m 2. B. 2  10 6 A/m 2. C. 4  10 3 A/m 2. D. 2  10 3 A/m 2. E. Some other value. Slide 30-50

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. The current in the fourth wire is QuickCheck 30.6 A. 16 A to the right. B. 4 A to the left. C. 2 A to the right. D. 2 A to the left. E. Not enough information to tell. Slide 30-57

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. The current in the fourth wire is QuickCheck 30.6 A. 16 A to the right. B. 4 A to the left. C. 2 A to the right. D. 2 A to the left. E. Not enough information to tell. Slide 30-58

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Conductivity and Resistivity  Conductivity, like density, characterizes a material as a whole.  The current density J is related to the electric field E by:  The resistivity tells us how reluctantly the electrons move in response to an electric field: Slide 30-59

Conductivity and Resistivity This woman is measuring her percentage body fat by gripping a device that sends a small electric current through her body. Because muscle and fat have different resistivities, the amount of current allows the fat-to-muscle ratio to be determined. Slide 30-66

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Batteries and Current  A battery is a source of potential difference Δ V bat.  The battery creates a potential difference between the ends of the wire.  The potential difference in the wire creates an electric field in the wire.  The electric field pushes a current I through the wire.  The current in the wire is: I  Δ V wire /R Graph V for circuit Slide 30-74

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.  Ohm’s law is limited to those materials whose resistance R remains constant—or very nearly so—during use.  The materials to which Ohm’s law applies are called ohmic.  The current through an ohmic material is directly proportional to the potential difference; doubling the potential difference doubles the current.  Metal and other conductors are ohmic devices. Ohm’s Law Slide 30-75

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. The current through a wire is measured as the potential difference ΔV is varied. What is the wire’s resistance? QuickCheck 30.11 A. 0.01. B. 0.02. C. 50. D. 100. E. Some other value. Slide 30-76

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. The current through a wire is measured as the potential difference Δ V is varied. What is the wire’s resistance? QuickCheck 30.11 A. 0.01. B. 0.02. C. 50. D. 100. E. Some other value. Slide 30-77

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Nonohmic Materials  Some materials and devices are nonohmic, meaning that the current through the device is not directly proportional to the potential difference.  Diodes, batteries, and capacitors are all nonohmic devices. Slide 30-78

Resistivity The resistance of a wire depends on its dimensions and the resistivity of its material: Slide 22-22

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Wire 2 is twice the length and twice the diameter of wire 1. What is the ratio R 2 /R 1 of their resistances? QuickCheck 30.10 A. 1/4. B. 1/2. C. 1. D. 2. E. 4. Slide 30-72

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Wire 2 is twice the length and twice the diameter of wire 1. What is the ratio R 2 /R 1 of their resistances? QuickCheck 30.10 A. 1/4. B. 1/2. C. 1. D. 2. E. 4. Slide 30-73

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Checking Understanding A battery is connected to a wire, and makes a current in the wire. i.Which of the following changes would increase the current? ii.Which would decrease the current? iii.Which would cause no change? A.Increasing the length of the wire B.Keeping the wire the same length, but making it thicker C.Using a battery with a higher rated voltage D.Making the wire into a coil, but keeping its dimensions the same E.Increasing the temperature of the wire Slide 22-15

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. The filament of a 100-W bulb carries a current of 0.83 A at the normal operating voltage of 120 V. A.What is the resistance of the filament? B.If the filament is made of tungsten wire of diameter 0.035 mm, how long is the filament? Example Problem Slide 22-28

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. If you use wire of the same diameter operating at the same temperature, should you increase or decrease the length of the wire from the value calculated in the previous example in order to make a 60 W light bulb? (Hint: The bulb is dimmer. What does this tell us about the current?) Conceptual Example Problem Slide 22-29

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Energy and Power in Resistors Slide 22-31

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. A resistor is connected to a 3.0 V battery; the power dissipated in the resistor is 1.0 W. The battery is traded for a 6.0 V battery. The power dissipated by the resistor is now A.1.0 W B.2.0 W C.3.0 W D.4.0 W Checking Understanding Slide 22-32

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Answer A resistor is connected to a 3.0 V battery; the power dissipated in the resistor is 1.0 W. The battery is traded for a 6.0 V battery. The power dissipated by the resistor is now A.1.0 W B.2.0 W C.3.0 W D.4.0 W Slide 22-33

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. An electric blanket has a wire that runs through the interior. A current causes energy to be dissipated in the wire, warming the blanket. A new, low-voltage electric blanket is rated to be used at 18 V. It dissipates a power of 82 W. What is the resistance of the wire that runs through the blanket? Example Problem Slide 22-34

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. For the electric blanket of the previous example, as the temperature of the wire increases, what happens to the resistance of the wire? How does this affect the current in the wire? The dissipated power? Conceptual Example Problem: Electric Blankets Slide 22-35

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