25 Lecture Lam - University of Hawaiiplam/ph272_summer/L5/25_Lecture_Lam.pdf · Title:...

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


Topics for Chapter 25

• Electric current and current density

• Resistivity and resistance

• Ohm’s Law

Intermission

• Electromotive force (emf)

• Power dissipation in resistors


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.


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).


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


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?


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


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.


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?


Current–voltage relationships

• Reminder: Ohm’s Law is linear, but current flow through other

devices may not be.


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.


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).


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.


Intermission


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)


Example: A resistor connected to a real battery

• Follow Example 25.6. Find I and terminal voltage Vab


Voltmeters and ammeters

• Follow Conceptual Example 25.7.

What happens when the

voltmeter is connected in series

with the circuit?


A source with a short circuit

• Follow Example 25.8.

• Find I and terminal voltage Vab.


Potential changes around a circuit

• The net change in potential energy must be zero for the entirecircuit.


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.

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