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Nov 15, 2013 (3 years and 6 months ago)

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Resistance in a Conductor

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Resistance

to electric current flow


electrical resistance:

is anything that impedes the flow charges (I) through a conductor


resistor:

is anything that provides a significant amount of resistance within a conductor



The flow of charges in a conductor is dependent on

1)

The voltage (V)

2)

Resistance (R)

Electrical Resistance

(R) is measured as R=











= 1 ohm (Ω)


Examples


1

Ω means that 1 volt is required to produce 1 amp of current


2

Ω means that
2 volt is required to produce 1 amp of current


3

Ω means that 3 volt is required to produce 1 amp of current


More voltage is required to maintain the same flow of charges as the resistance of the conductor
increases.


Ohm’s Law

V=IR


When a conduct
or has a constant resistance over a wide range of voltage Ohm’s Law is applied.


As the voltage increases through a conductor and the resistance remains constant the current
increases. ( that is I


V when R is a constant)



4
FACTORS THAT INFLUENCE RESI
STANCE

of a conductor

1)

Length (L
) of the resistor (conductor)

The resistance is directly proportional to the length of the resistor


If the length is doubled the resistance is doubled


If the length is reduced to a third the resistance is reduced to a third

2)

Cross sectional area (A)

of the resistor (conductor)

The resistance is inversely proportional to the cross sectional area of the resistor


If the cross sectional area is double the resistance will be reduced by ½


If the cross sectional are is reduced to

1/3 the resistance will be increased by 3














3)

The material

that the resistor is made of. Resistance varies due to the different atomic properties
of each substance.

This resistance due to atomic structure is called
resistivity (ρ)

r
esisti
vity (ρ)

is measured in ____Ω∙m

The
resistivity

is found on a chart. The chart is normally based on the resistor having a
temperature of 20
o
C (68
o
F)

Example of resistivity table




R =

(


)



= resistance at 20
o
C


Example

A)

Find the resistance of a

8.50 m length of copper wire that has a cross sectional area
of 50.3 mm
2
. The temperature of the resistor (copper wire) is 20
o
C


R =?


L = 8.50m


A =50.3 mm
2

= 50.3 x 10
-
6
m
2


ρ = 1.68 x 10
-
8

Ω∙m



R =ρ
(


)

=(1.68 x 10
-
8

Ω∙m)
(















)

= 2
.84 x 10
-
3

Ω∙m

B)

Find the electric current flow through this copper wire if the voltage is 12.0V

I = ?

V = 12.0V




V = IR

R = 2.84 x 10
-
3

Ω∙m






I = V/R = 12.0V/2.84 x 10
-
3

Ω∙m = 4,225 A = 4,230A





4)

The
temperature of the resistor

The
resistivity

(ρ)

of most resistors changes as their temperature changes

ρ =


(1 + αΔT) ρ
o

= resistivity at 20
o
C




α is a constant over a certain temperature range




α is called the
temperature coefficient of resistivity





ΔT = T
F



T
o

T
0

= 20
o
C

Example

A)

Find the resistance of a 8.50 m length of copper wire that has a cross sectional area
of 50.3 mm
2
. The temperature of the resistor (copper wire) is 38
o

(100F)


R =?


L = 8.50m


A =50.3 mm
2

= 50.3 x 10
-
6
m
2


ρ
o

= 1.68 x 10
-
8

Ω∙m


α = 0.0068/C


T = 38
o

C


Δ
T = 38.0
o
C
-
20.0
o
C = 18
o
C



ρ =


(1 + αΔT)

ρ
o

= resistivity at 20
o
C



ρ = (1.68 x 10
-
8

Ω∙m)(1 + 0.0068/C ∙18.0C) = 1.89 x 10
-
8

Ω∙m



R =

(


)

=(1.89 x 10
-
8

Ω∙m)
(














)

= 3.19 x 10
-
3

Ω

B)

Find the electric current flow through this copper wire if the voltage is 12.0 v.

I = ?

V = 12.0V




V = IR

R = 3.19 x 10
-
3

Ω∙m






I = V/R = 12.0V/
3.19

x 10
-
3

Ω∙m =
3,762

A =
3,760
A



[some

resistors have a negative temperature coefficient(α). This means that their resistivity decreases as
temperature rises]


Superconductors
:

are electrical conductors that have
exactly zero
electrical resistance (R= 0.000000Ω)

Read chapter 17 sect 3 on supe
r conductors




Electric Power
& Cost

Word Press


Electric Power:


Power is the amount of work performed in a given time. P =







=



= watts



Electric power P =







=




=
IV

=


(


)

=





= watts







V= IR

Ohms Law I= V/R also


P = IV =
(


)

V =






P = IV=
I

(IR) = I
2
R





Three formulas for electric power


P =IV= I
2
R =





Joule Heat
: conversion of electric potential energy into thermal energy (heat)


This can be useful or it can be wasteful side effect


If the thermal energy is required as part of an operation it is good



Electric clothes dryer



Electric hot water heater



Heating home



Electric burners or coils on stove



Industrial drying apparat
us

If the thermal energy is produced as a by product of some other electric operation it may be
thought of as wasted energy.

Light bulbs give off thermal energy (some give off 95% of electric energy as heat not
light



Electric motors produce thermal ener
gy



The wires conducting electric current produce thermal energy.

All of this thermal energy is produced due to the resistance in the conductor.


To reduce this thermal energy loss the resistance must be reduced or eliminated


The equation used to determine thermal energy production by a
conductor is


Joule heat = I
2
R



Electric
E
nergy
C
ost



Power = energy /time So energy = (power) (time)




Electric energy is sold in units of
power X time

=
kilowatt
-
hours



A kilowatt h
our = 1000 watts x 1hour = 1000J/s X 3600 sec = 3.6 x 10
6

J



EXAMPLE

A) A light bulb of 60.0 watts, a refrigerator of 850 watts and a desk top computer1200 watts are all
running for 2.00 hours.


A) how many kilowatt hours of energy were used?



60.0 w +
850.0 w + 1200.0 w = 2110 w
(





)

= 2.11 kwatts




2,11 x 2.00 hrs = 4.22 kw hr

B) The cost for 1 kwatt∙hr is 11.5 cents. Determine the cost in dollars and cents to operate the 3
items?



4.22 kwatt∙hrs
(







)

= $0.485