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Chapter 22
Current and Resistance
Topics:
• Current
• Conservation of current
• Batteries
• Resistance and resistivity
• Simple circuits
Sample question:
How can the measurement of an electric current passed through a
person’s body allow a determination of the percentage body fat?
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Slide 22-1
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Slide 22-3
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Slide 22-4
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Slide 22-5
Circuit Current Model => on whiteboards
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Slide 21-16
Checking Understanding
Rank the bulbs in the following circuit according to their brightness,
from brightest to dimmest.
A.
B.
C.
D.
A>B>C>D
A=B>C=D
A=D>B=C
B=C>A=D
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Slide 22-16
Answer
Rank the bulbs in the following circuit according to their brightness,
from brightest to dimmest.
A.
B.
C.
D.
A>B>C>D
A=B>C=D
A=D>B=C
B=C>A=D
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Slide 22-17
Additional Questions
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.
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Slide 22-45
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.
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Slide 22-46
Questions on Circuit Tutorial Worksheet 1
=> on whiteboards
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Slide 21-16
Definition of a Current
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Slide 22-9
Batteries
The potential difference
between the terminals of a
battery, often called the
terminal voltage, is the
battery’s emf.
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Slide 22-20
Simple Circuits
The current is determined by
the potential difference and
the resistance of the wire:
∆V
_____
chem
I = R
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Slide 22-13
Series and Parallel
Series
• Circuit elements in a chain between 2 points
• Same current flows through circuit elements
I1 = I2
• Electric Potentials add => Delta Vtotal = Delta V1 + Delta V2
Parallel
• Circuit elements on multiple paths connecting the same
points
• Since paths connect the same points,
Delta V’s are the same
• Currents Add => Itotal = I1 + I2
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Slide 22-25
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
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The Electron Current
 We define the
electron current ie
to be the number of
electrons per second
that pass through a
cross section of the
conductor.
 The number Ne of
electrons that pass
through the cross
section during the
time interval t is
Slide 30-23
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The Electron Density
 In most metals,
each atom
contributes one
valence electron to
the sea of
electrons.
 Thus the number
of conduction
electrons ne is the
same as the
number of atoms
per cubic meter.
Slide 30-27
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Discharging a Capacitor
 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.
Slide 30-30
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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
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Establishing the Electric Field in a Wire
 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.
Slide 30-33
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Establishing the Electric Field in a Wire
 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.
Slide 30-34
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QuickCheck 30.2
Surface charge is distributed on a wire as shown.
Electrons in the wire
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
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QuickCheck 30.2
Surface charge is distributed on a wire as shown.
Electrons in the wire
A. Drift to the right.
B. Drift to the left.
C. Move upward.
D. Move downward.
Electric field from
nonuniform surface
charges is to the right.
Force on negative
electrons is to the left.
E. On average, remain at
rest.
Slide 30-36
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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
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A Model of Conduction
 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.
Slide 30-38
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A Model of Conduction
 The graph
shows the
speed of an
electron during
multiple
collisions.
 The average
velocity is the
electron drift
speed
Slide 30-39
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The Current Density in a Wire
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/m2.
Slide 30-48
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QuickCheck 30.4
The current density in this wire is
A. 4 x 106 A/m2.
B. 2 x 106 A/m2.
C. 4 x 103 A/m2.
D. 2 x 103 A/m2.
E. Some other value.
Slide 30-49
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QuickCheck 30.4
The current density in this wire is
A. 4 x 106 A/m2.
B. 2 x 106 A/m2.
C. 4 x 103 A/m2.
D. 2 x 103 A/m2.
E. Some other value.
Slide 30-50
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Conservation of Current
Slide 30-53
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Conservation of Current
å Iin = å Iout
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Slide 22-15
QuickCheck 30.6
The current in the fourth wire is
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
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QuickCheck 30.6
The current in the fourth wire is
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
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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
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Current Density Activity
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Slide 21-16
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
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Conductivity and Resistivity
Slide 30-67
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 A battery and
is a source
of
Batteries
Current
potential difference Vbat.
 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 = Vwire/R
Graph V for circuit
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Slide 30-74
Ohm’s Law
 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.
Slide 30-75
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QuickCheck 30.11
The current through a
wire is measured as the
potential difference ΔV is
varied. What is the
wire’s resistance?
A.
B.
C.
D.
E.
0.01 .
0.02 .
50 .
100 .
Some other value.
Slide 30-76
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QuickCheck 30.11
The current through a
wire is measured as
the potential difference
V is varied. What is
the wire’s resistance?
A.
B.
C.
D.
E.
0.01 .
0.02 .
50 .
100 .
Some other value.
Slide 30-77
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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
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Resistance Activity
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Slide 21-16
Resistivity
The resistance of a wire
depends on its dimensions
and the resistivity of its
material:
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Slide 22-22
QuickCheck 30.10
Wire 2 is twice the length
and twice the diameter of
wire 1. What is the ratio
R2/R1 of their resistances?
A. 1/4.
B. 1/2.
C. 1.
D. 2.
E. 4.
Slide 30-72
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QuickCheck 30.10
Wire 2 is twice the length
and twice the diameter of
wire 1. What is the ratio R2/R1
of their resistances?
A. 1/4.
B. 1/2.
C. 1.
D. 2.
E. 4.
Slide 30-73
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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?
A.
B.
C.
D.
E.
iii. Which would cause no change?
Increasing the length of the wire
Keeping the wire the same length, but making it thicker
Using a battery with a higher rated voltage
Making the wire into a coil, but keeping its dimensions the
same
Increasing the temperature of the wire
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Slide 22-15
Example Problem
The filament of a 100-W bulb carries a current of 0.83 A at the
normal operating voltage of 120 V.
A.
B.
What is the resistance of the filament?
If the filament is made of tungsten wire of diameter 0.035
mm, how long is the filament?
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Slide 22-28
Conceptual Example Problem
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?)
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Slide 22-29
Power in Circuits
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Slide 22-30
Energy and Power in Resistors
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Slide 22-31
Checking Understanding
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.
B.
C.
D.
1.0 W
2.0 W
3.0 W
4.0 W
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Slide 22-32
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.
B.
C.
D.
1.0 W
2.0 W
3.0 W
4.0 W
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Slide 22-33
Example Problem
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?
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Slide 22-34
Conceptual Example Problem: Electric Blankets
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?
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Slide 22-35