36 Resistivity - 


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Unformatted text preview: 
 We
already
mentioned
that
old
theories
of
electricity
described
it
as
an
invisible
and
 incompressible
fluid,
and
considered
the
electric
voltage
as
the
pressure
difference.
 Interestingly,
Ohm’s
law
is
similar
to
Poiseuille’s
law,
which
was
formulated
about
 20
years
later,
in
1842,
for
the
laminar
flow
of
water
through
very
small
tubes:
the
 flow
rate
is
directly
proportional
to
the
pressure
drop
per
unit
length
of
the
tube.
 For
a
flow
of
liquid,
it
is
easy
to
predict
that
its
rate
would
decrease
if
the
tube
gets
 longer
and
narrower.
Similarly,
a
simple
theory
for
electric
current
maintains
that
 the
electric
resistance
R
of
a
uniform
cylindrical
sample
is
directly
proportional
to
 the
length
L
of
the
sample
and
inversely
proportional
to
its
cross‐section
area
A,
as
 sketched
in
Figure
2‐9.

 
 Ganago
/
Resistance
and
Resistivity
/
Resistance
of
Wires
/
Wiring
Your
Speakers
 Required
Reading
for
HW
03
 
 2.1.2
Resistance
and
resistivity

 
 Figure
2‐9.
For
a
uniform
cylindrical
conductor,
resistance
is
 proportional
to
its
length
L
and
inversely
proportional
to
its
cross‐ section
area
A.

 
 
 Mathematically:
 L 
 
 
 [equation
2‐6]

 R = ρ⋅ 
 A 
 The
coefficient
ρ
in
equation
[2‐6]
is
called
resistivity
and
strongly
depends
on
the
 material
of
the
sample.

 
 Learning
Objective:
Define
resistivity
and
explain
the
dependence
of
 electric
resistance
on
the
length
and
cross‐section
of
a
uniform
 cylindrical
sample.
 
 ©
2010
Alexander
Ganago

 Page
1
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Resistivity
 Ganago
/
Resistance
and
Resistivity
/
Resistance
of
Wires
/
Wiring
Your
Speakers
 Required
Reading
for
HW
03
 
 The
SI
unit
of
resistivity
is
ohm
⋅
meter
(Ω⋅m);
the
numerical
values
of
resistivity
 range
from
1.59⋅10‐8
Ω⋅m
for
silver
to
~1013
Ω⋅m
for
hard
rubber
to
~
1022
Ω⋅m
for
 Teflon
(note
the
impressive
difference
by
30
orders
of
magnitude).
Materials
with
 low
resistivity
are
called
conductors;
these
are
metals,
of
which
we
make
wires:
 copper
has
ρ
=
1.72⋅10‐8
Ω⋅m;
much
cheaper
and
lighter
aluminum
has
ρ
=
2.82⋅10‐8
 Ω⋅m.
Materials
with
high
resistivity
are
called
insulators:
we
use
them
to
wrap
wires
 in
order
to
avoid
unwanted
electric
connection,
shock
and
electrocution
of
the
users
 of
electric
appliances,
etc.


 

 Learning
Objective:
Explain
the
difference
between
conductors
and
 insulators;
provide
numerical
examples.
 
 Resistivity
of
semiconductors
is
variable:
in
pure
germanium,
it
equals
0.46
Ω⋅m;
in
 pure
silicon,
ρ
=
640
Ω⋅m,
but
these
numbers
can
decrease
by
several
orders
of
 magnitude
in
the
presence
of
impurity
atoms.
The
actual
values
of
resistivity
depend
 on
a
variety
of
factors,
including:
(a)
parameters
of
the
semiconductor
material
such
 as
density
and
type
of
impurity
atoms,
(b)
operation
conditions
such
as
the
polarity
 and
magnitude
of
applied
voltage,
(c)
environmental
parameters
such
as
 temperature,
illumination
of
the
sample,
etc.
We
will
consider
examples
later
in
the
 book.

 
 
 
 Example
2‐1

 
 
 
 

 Two
cylindrical,
round,
and
uniform
pieces
X
and
Y
of
pure
copper
wire
have
the
 same
mass
but
piece
Y
is
10
times
longer
than
piece
X.
The
resistance
of
piece
X
 equals
1
Ω.
The
resistance
of
piece
Y
equals:

 A. 1
Ω
 B. 10
Ω
 C. 100
Ω
 D. 1000
Ω
 E. Not
enough
information.

 
 Solution
 
 Suppose
that
initially
both
pieces
were
identical,
but
then
piece
Y
was
stretched
10‐ fold.
Thus
its
length
increased
10‐fold
but,
in
order
to
maintain
constant
volume
 (due
to
the
constant
mass
and
constant
density),
its
cross‐section
should
have
 decreased
10‐fold.

 
 Mathematically,

 LY=
10⋅LX
;

 AY=
0.1⋅AX
 ©
2010
Alexander
Ganago

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and
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Resistance
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Wires
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Speakers
 Required
Reading
for
HW
03
 
 
 Given
that

 RX
=
ρCu
⋅
LX
/
AX
=
1
Ω
 
 we
obtain:

 RY
=
ρCu
⋅
LY
/
AY
=
(1
Ω)
⋅
(LY/LX)
/
(AY/AX)
=
(1
Ω)
⋅
10
/
0.1
=
100
Ω.
 
 Answer:
C.
 
 2.2
Practical
perspective:
Resistances
of
wires
for
your
projects
 
 Here
is
a
straightforward
practical
application
of
what
you
just
learned:
equation
[2‐ 6]
can
help
us
choose
what
wires
to
use
in
our
project.
We
understand
that
a
thicker
 wire
has
lower
electric
resistance
and
can
probably
carry
higher
electric
currents
 (like
wider
tubes
can
pour
more
water).
As
engineers,
we
would
like
to
specify
the
 number
when
asking
for
a
thicker
wire,
and
we
expect
that
there
should
be
some
 standard
for
wire
sizes.

 
 2.2.1.
The
American
wire
gauge
(AWG)
standard

 
 Indeed,
such
standards
exist.
Since
1857,
the
American
wire
gauge
(AWG)
serves
as
 a
standardized
system,
predominantly
in
the
United
States,
to
specify
how
large
a
 wire
is,
in
terms
of
its
diameter
or
cross‐section,
and
how
much
current
it
can
safely
 transfer.
The
AWG
tables
are
compiled
for
a
single,
solid,
round
conductors;
they
are
 easy
to
find
in
reference
books
and
on
the
Internet.
Table
2‐1
provides
a
partial
 listing
for
our
discussion.
Note
that
the
left
columns
of
this
table
define
the
 diameters
of
any
wire
(including
the
needles
used
for
body
piercing)
but
the
right
 columns
list
the
electric
parameters,
which
are
valid
only
for
copper
wires
because
 they
depend
on
the
resistivity
of
the
material.


 
 Table
2‐1
 
 A
partial
listing
of
AWG
standard
for
the
size
of
wire,
as
well
as
the
resistance
and
 load
carrying
capacity
of
copper
wire

 
 AWG
 Conductor
 Conductor
 Ohms
per
 Maximum
amps
for
power
 gauge
 diameter,
 diameter,
mm
 km
 transmission
(conservative
 inches
 rule)
 0000
 0.46
 11.684
 0.16072
 302
 0
 0.3249
 8.25246
 0.322424
 150
 20
 0.032
 0.8128
 33.292
 1.5
 24
 0.0201
 0.51054
 84.1976
 0.577
 30
 0.01
 0.254
 338.496
 0.142
 36
 0.005
 0.127
 1360
 0.035
 ©
2010
Alexander
Ganago

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Resistivity
 Ganago
/
Resistance
and
Resistivity
/
Resistance
of
Wires
/
Wiring
Your
Speakers
 Required
Reading
for
HW
03
 
 
 First
of
all,
note
that
the
larger
AWG
number
corresponds
to
thinner
wire,
 which
has
higher
resistance
and
can
carry
smaller
currents.
Electricians
often
 use
this
table:
forexample,
No.
0
AWG
(usually,
pronounced
“one
aught”)
wire
can
 carry
up
to
150
A,
typical
for
a
main
household
input,
while
inside
the
house
 individual
connections
can
be
made
with
thinner
wire.
The
rule
of
thumb
is
that
 when
the
diameter
of
wire
is
doubled,
the
AWG
will
decrease
by
6;
thus
No.
24
AWG
 (usually,
pronounced
“24
gauge”)
is
two‐fold
thicker
than
No.
30
AWG
and
4‐fold
 thicker
than
No.36
AWG.
This
formula
is
approximate
but
quite
accurate
(check
it
 with
Table
2‐1).
Additionally,
a
decrease
of
10
AWG
numbers
corresponds
to
an
 increase
of
the
wire
cross‐section
and
weight
by
a
factor
of
~10
and
to
a
decrease
of
 resistance
(for
the
same
material)
by
about
a
factor
of
10
(check
Table
2‐1
to
see
 how
accurate
these
estimates
are).

 
 
 Example
2‐2

 
 A
piece
of
AWG
#20
wire
is
made
of
aluminum
(ρ
=
2.82·10‐8
Ω·m;
density
=
2.70
 g/cm3)
and
has
mass
m=1
kg.
Calculate
the
length
of
this
wire
in
meters
and
its
 electric
resistance
in
Ω.
Obtain
the
diameter
of
wire
from
Table
2‐1.
Compare
your
 result
for
electrical
resistance
with
the
data
in
Table
2‐1.
 
 Solution
 
 mass Density
=
 .




From
the
given
values,
the
volume
of
this
piece
of
wire
equals:
 volume 1000 g 
=
370.4
cm3
 2.70 g / cm 3 According
to
Table
2‐1,
diameter
of
AWG
#20
wire
is
0.8128
mm.

 
 Thus
we
can
calculate
its
cross‐section,
expressing
it
in
cm2:
 ¼
π
⋅
(0.8128
mm)2
=
0.0519
mm2
=
5.189
⋅
10‐3
cm2
 
 The
length
of
the
wire
equals
 370.4 cm 3 
 
=
7.138
·
104
cm
 5.189 ⋅ 10 −3 cm 2 
 The
length
of
wire
equals
713.8
m.

 
 
We
can
find
the
electric
resistance:
 L 713.8 m R
=
ρ
·
 =
2.82
·
10‐8
Ω·m· 
 A 5.189 ⋅ 10 −7 m 2 R
=
38.79
Ω
 ©
2010
Alexander
Ganago

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 Ganago
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Resistance
and
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Resistance
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Wires
/
Wiring
Your
Speakers
 Required
Reading
for
HW
03
 
 
 From
Table
2‐1,
we
obtain
that
the
resistance
of
1
km
of
AWG
#20
copper
wire
 equals
R1
=
33.292
Ω
 
 From
this
value,
we
can
find
the
resistance
of
AWG
#24
aluminum
wire
that
is
 0.7138
km
long:
 ρ 0.7138 RAl
=
R1
·
 Al ⋅ =
38.96
Ω
 ρCu 1 
 This
result
is
in
perfect
agreement
with
our
calculated
resistance
R.
 
 
 
 2.2.2
Resistance
of
the
wire
and
the
load
resistance

 
 The
equations
relating
electric
resistance
to
resistivity
of
materials
and
the
size
of
 conductors
have
many
practical
applications.
One
of
the
big
ideas
is
that
the
 resistance
of
wire
should
be
much
lower
than
the
resistance
of
the
load,
to
 which
the
electric
power
has
to
be
transferred
through
this
wire.
We
will
learn
 the
theory
behind
it
later
in
this
chapter.
Since
it
is
the
ratio
of
resistances
 (wire/load)
that
matters,
the
choice
of
wire
is
dictated
by
the
particular
 applications.
Here
are
two
examples.
 
 Application
2‐1.

Can
we
neglect
the
resistance
of
the
wire?

 The
concept
of
an
ideal
conductor.
 
 When
building
circuits
for
a
lab
project,
we
use
prototyping
boards
several
inches
 long
and
wide,
and
we
often
choose
resistors
from
10
Ω
to
10
MΩ.
To
connect
 resistors
to
each
other
and
to
the
source
of
electric
power,
we
use
copper
wire
No.
 24
AWG
(see
Table
2‐1
for
its
specifications).
Assuming
that
a
piece
of
such
wire
is
 10‐cm
(~4
in)
long,
let
us
calculate
its
resistance
from
equation
[2‐6].
As
mentioned
 before,
the
resistivity
of
copper
is
ρ
=
1.72⋅10‐8
Ω⋅m;
the
cross‐section
area
for
No.
 24
AWG
is:
 5.11 ⋅ 10 −4 m D2 π⋅ = 3.142 ⋅ = 2.05 ⋅ 10 −7 m 2 
 4 4 
 Substitute
the
length
of
our
piece
of
wire
0.1
m,
and
obtain:

 R
=
8.4
mΩ
=
8.4
⋅
10‐3
Ω.

 
 We
found
that
the
resistance
of
such
a
piece
of
wire
is
smaller
by
a
factor
of
1,000
or
 more
than
any
other
resistance
in
our
lab
circuit;
therefore,
for
nearly
all
practical
 purposes,
we
can
safely
neglect
the
wire
resistance
in
our
calculations.
This
leads
us
 to
the
important
concept
of
an
ideal
conductor
that
has
zero
resistance.

 
 ( ) 2 ©
2010
Alexander
Ganago

 Page
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Resistivity
 Ganago
/
Resistance
and
Resistivity
/
Resistance
of
Wires
/
Wiring
Your
Speakers
 Required
Reading
for
HW
03
 
 Learning
Objective:
Explain
and
apply
the
concept
of
an
ideal
 conductor.
 
 By
the
way,
Table
2‐1
lists
the
resistance
of
No.
24
AWG
copper
wire
as
84
Ω
per
 kilometer;
since
1
km
is
104
times
longer
than
our
10‐cm
piece
of
wire,
the
result
of
 our
calculation
is
in
perfect
agreement
with
the
standard
table.
 
 Learning
objective:
Calculate
the
resistance
of
copper
wire
of
the
 given
AWG
number
and
length.
Explain
and
apply
the
concept
of
an
 ideal
conductor.

 
 
 Example
2‐3

 
 
 
 
 Assume
that:
(1)
all
wiring
in
a
prototype
circuit
is
made
with
No.
36
AWG
copper
 wire
and
has
the
total
length
of
25
cm,
and
(2)
the
total
wire
resistance
should
not
 exceed
0.5%
of
the
smallest
resistance
in
the
circuit.
Determine
the
smallest
 resistance
acceptable
in
this
project.


 
 Solution
 
 From
Table
2‐1,
obtain
the
resistance
of
No.
36
AWG
copper
wire
equal
to
1360
Ω
 per
km,
or
1.36
Ω
per
m;
thus
the
resistance
of
25‐cm
wire
equals
0.34
Ω.
In
order
to
 keep
this
value
below
0.5%
of
the
smallest
resistance
in
the
circuit,
that
resistance
 should
be
at
least
(0.34
Ω)
·
(200)
=
68
Ω.

 
 Answer:
68
Ω.
 
 
Application
2‐2.
How
to
choose
the
speaker
wires
for
your
audio
system?
 
 When
you
build
an
audio
system
or
home
theater,
you
probably
wire
the
speakers
to
 the
amplifier
or
A/V
receiver
yourself,
without
hiring
an
expert.
Here
you
can
learn
 how
to
choose
the
wire
as
the
expert
does.
A
conservative
strategy
recommends
 that
the
wire
resistance
be
kept
below
5%
of
the
speaker
resistance,
which
is
 usually
from
2
to
8
Ω.
Note
that
the
length
of
the
speaker
wire,
which
you
should
use
 in
your
calculations,
equals
double
the
distance
between
the
speaker
and
the
 amplifier,
because
you
need
two
wires
(from
the
amplifier
to
the
speaker
and
back
 from
the
speaker
to
the
amplifier)
to
build
a
closed
circuit.
Thus,
in
the
most
 demanding
case
of
a
2‐Ωspeaker,
the
wire
resistance
should
not
exceed
100
mΩ;
 for
No.
24
AWG
wire,
it
limits
the
distance
between
the
speaker
and
the
amplifier
to
 below
60
cm
(below
2
ft).
Using
No.
20
AWG
wire,
you
can
position
a
2‐Ω
speaker
as
 far
as
1.5
m
(~5
ft)
from
the
amplifier
(see
Table
2‐1);
with
No.14
AWG
wire,
you
 can
move
this
speaker
as
far
as
6
m
(~20
ft)
from
the
amplifier
(use
the
 approximate
formula
explained
above),
etc.
Of
course,
for
speakers
with
higher
 ©
2010
Alexander
Ganago

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Resistivity
 Ganago
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Resistance
and
Resistivity
/
Resistance
of
Wires
/
Wiring
Your
Speakers
 Required
Reading
for
HW
03
 
 resistances,
the
length
of
the
wires
can
be
proportionally
increased,
for
example,
up
 to
6
m
(~20
ft)
with
No.20
AWG
wire
for
an
8‐Ω
speaker.


 
 Learning
Objective:
Determine
the
size
of
wires
for
connection
of
 speakers
to
the
output
of
an
audio
system.

 
 Note
that
in
these
two
Applications
we
reached
exactly
opposite
conclusions
on
 whether
the
resistance
of
a
piece
of
copper
wire
(such
as
No.24
AWG)
is
negligible
–
 or
should
serve
as
the
most
important
constraint
in
the
design
of
your
system.
The
 take‐home
message
is
that
decisions
in
Electrical
Engineering,
as
in
other
fields,
are
 determined
by
the
applications,
and
old
recipes
should
not
be
blindly
applied
to
new
 projects.



 
 
 Example
2‐3

 
 A
friend
of
yours
bought
new
4‐Ω
speakers
for
his
audio
system
and
asked
you
to
 determine
how
far
from
the
output
amplifier
he
could
put
them
using
No.
20
AWG
 copper
wire.
Provide
two
answers
–
based
on
your
calculations
using
equation
[2‐6]
 and
based
on
the
resistance
per
km
data
listed
in
Table
2‐1,
and
discuss
whether
 these
answers
agree.
In
both
cases,
assume
that
the
wire
resistance
should
be
kept
 below
5%
of
the
speaker
resistance.

 
 
 Solution

 
 Assuming
the
wire
resistance
equal
5%
of
the
speaker
resistance
(4
Ω),
we
find
the
 maximal
wire
resistance
of
200
mΩ.
From
equation
[2‐6],
substituting
the
wire
 diameter
from
table
2‐1,
we
determine
the
maximal
length
of
wire:
 2 π ⋅ 8.128 ⋅ 10 −4 m A L = R ⋅ = ( 0.2 Ω ) ⋅ 4 = 6.03 m 
 ρ 1.72 ⋅ 10 -8 Ω ⋅m ( ( ) ) 
 Thus
the
maximal
distance
between
the
speaker
and
the
amplifier
is
3
m
(~10
ft).
 
 Look
up
Table
2‐1
and
see
that
No.
20
AWG
wire
has
resistance
of
33.292
Ω
per
 kilometer,
or
33.292
mΩ
per
meter.
Since
we
have
already
found
that
the
maximal
 wire
resistance
equals
200
mΩ,
we
calculate
that
the
maximal
wire
length
is:


 200 mΩ = 6.01 m 
 33.292 mΩ /m 
 This
result
is
in
excellent
agreement
with
the
maximal
wire
length
that
we
 determined
above
from
equation
[2‐6].

 ©
2010
Alexander
Ganago

 Page
7
of
8
 File:
Resistivity
 Ganago
/
Resistance
and
Resistivity
/
Resistance
of
Wires
/
Wiring
Your
Speakers
 Required
Reading
for
HW
03
 
 
 Answer:
3
m
(~10
ft)
max.

 
 ©
2010
Alexander
Ganago

 Page
8
of
8
 File:
Resistivity
 ...
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This note was uploaded on 12/06/2010 for the course EECS 314 taught by Professor Ganago during the Spring '07 term at University of Michigan.

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