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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley
PowerPoint® Lectures for
University Physics, Twelfth Edition
– Hugh D. Young and Roger A. Freedman
Lectures by James Pazun
Chapter 25
Current, Resistance, and
Electromotive Force
Modified by P. Lam 7_21_2008
Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley
Topics for Chapter 25
• Electric current and current density
• Resistivity and resistance
• Ohm’s Law
Intermission
• Electromotive force (emf)
• Power dissipation in resistors
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Direction of electrical current
• The battery sets up an electric field inside the light bulb and the conductingwire which causes the electrons to move (fig. b).
• By convention, one visualizes electrical current as positive charges moving inthe direction of the electric field (fig. a), although it is really the electronswhich are moving in opposition.
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Electrical current and “drift” velocity
• In the absence of an electric field, electrons move randomly in a conductor(v~106 m/s).
• Connecting the conductor to a battery creates an electric field in the conductor,which accelerates the electrons
• However, the electrons “collide” with the atoms in the conductor.
• The combined effect of the electric field and the collision is a steady state driftof the electrons- drift velocity (vd~10-4 m/s).
Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley
Electrical current and current density
Electric current I dQdt
=Coulombsecond
= Ampere (A)
(analogous to water flow =gallons
second )
Current density J = IA
=Coulombsecond
/area =A
m2
We can relate J to the density (n) of the current carrier,
the charge of the carrier (q), and the drift velocity (vd )
J = nqvd
(q = charge of each current carrier; for electron q = -1.6x10-19C)
Technically, J is a vectorr J = nq
r v d
I = (r J • ˆ n ) dA = (
r J • ˆ n ) A if (
r J • ˆ n ) is a constant over the entire area
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Electrical current and current density-Example
Example 25.1 on P.850.
A wire carries a current of 1.67A.
The diameter of the wire is 1.02mm.
The density of free electrons in the wire is 8.5x1028 /m3
(a) Find J
(b) Find vd
JI
A and J = nqvd
Which equation do we use for part a?
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Relationship between current density and electric field
Battery E - field electron moves current
One would expect that the greater the electric field (E),
the higher the current density (J)
For many materials, the relationship between E and J are linear :
J =1
E; = resistivity of the material
IA
=1
V
L
I =A
L
V
L
A
= I
L
A
V = IR (called "Ohm's Law")
R =L
A
= resistance of the material
The unit of resistance is Ohm ( )
Unit of resistivity is Ohm
meter=m
E
L
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Resistivity and resistance - example
Resistivity is instrince to the material (analogou to density)
while resistance depends also on the size of the material
(analgous to the total mass).
Example :
Find the resistance of a copper wire whose diameter is 1.02mm
and its length is 50 meters.
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Temperature Dependence of Resistivity
• For most materials, resistivity rises withincreasing temperature.
• Table 25.2 tabulates the values of .
• One can make a thermometer based on theresistance of a material.
(T) o 1+ (T To)[ ]
Given that a piece of copper wire has a resistance (R) of 1 at 20oC.
What is its resistance at 30oC?
What is its resistance as a function of temperature?
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Current–voltage relationships
• Reminder: Ohm’s Law is linear, but current flow through other
devices may not be.
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Electromotive force and circuits
• Just like a waterfountain won’t workwithout a waterpump and acomplete curcuit forthe water to circuit,neither wouldelectrical current.
Outside the battery, positive charge
moves from higher electric potential to
low electric potential.
Inside the battery, the chemical process
acts like a “charge pump” which brings
the positive charge back up to the higher
electric potential. Fn is a non-electrostatic
force.
The electrostatic force is analogous to the
force of gravity.
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Ideal battery vs. real battery
Ideal battery :
The voltage across the two terminal
(terminal voltage = Vab) equals the
electromotive force of the battery ( )
Vab =
Real battery has internal resistance (r).
The terminal voltage is less than the
electromotive force of the battery
Vab = Ir
The terminal voltage decreases as more current (I)
is drawn from the battery.
The terminal voltage also diminishes if the
internal resistance be large (this is what happens
to old batteries; internal resistance gets too large).
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What if the circuit is “open” (not complete)?
+ -
+
+
-
-
Two external charges + and -
provides the electrostatic force.
Current flows for a brief time till
electrostatic equilibrium is
achieved inside the conductor
(E=0 inside a conductor)
Two conductors attached to the
battery terminals but not completing a
circuit. The two conductors act like a
capacitor (though very, very small
capacitance). Current flows for a
brief time until the “capacitor”
charged to the same potential
difference as the battery’s emf.
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Intermission
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Symbols for circuit diagrams
• Shorthand symbols are in use for all wiring components. Seebelow.
Ideal ammeter has zero resistance.
Ideal voltmeter doesn’t drawn any current (infinite resistance)
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Example: A resistor connected to a real battery
• Follow Example 25.6. Find I and terminal voltage Vab
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Voltmeters and ammeters
• Follow Conceptual Example 25.7.
What happens when the
voltmeter is connected in series
with the circuit?
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A source with a short circuit
• Follow Example 25.8.
• Find I and terminal voltage Vab.
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Potential changes around a circuit
• The net change in potential energy must be zero for the entirecircuit.
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Power and energy in circuits
• Compute the power dissipated by the external resistance (R) andthe internal resistance (r).
• Compute the power provided by the battery.