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Platt, David – Homework 10 – Due: Sep 30 2005, 4:00 am – Inst: Ken Shih
1
This printout should have 11 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
The due time is Central
time.
1) ±or 002, v is the average thermal speed oF
the Free electron, not the average driFt speed.
±or the same carriers the average thermal
speed only depends on Temperature .
2)
±or 003, Think about metallic material only.
Semiconductor has di²erent story.
001
(part 1 oF 1) 10 points
A length oF copper wire has a resistance 49 Ω.
The wire is cut into three pieces oF equal
length.
The pieces are then connected as
parallel lengths between points A and B.
What resistance will this new “wire” oF
length
L
0
3
have between points A and B?
Correct answer: 5
.
44444 Ω.
Explanation:
Let :
R
0
= 49 Ω
.
The new wire has length
L
=
L
0
3
and cross
section
A
= 3
A
0
, so its resistance is
R
=
ρ L
A
=
ρ
µ
L
0
3
¶
3
A
0
=
1
9
µ
ρ L
0
A
0
¶
=
R
0
9
=
49 Ω
9
=
5
.
44444 Ω
.
002
(part 1 oF 3) 10 points
Consider the application oF two potential diF
Ferences across the same metallic ohmic con
ductor where the conductor is maintained at
a constant temperature oF 300 K.
V
1
J
L
R
2
R
Case 1: The potential di²erence is
V
1
, the
current density is
J
1
, and the average thermal
speed oF the Free electron within the conductor
is
v
1
.
Case 2: The corresponding quantities are
V
2
,
J
2
, and
v
2
.
IF
V
2
= 2
V
1
, the corresponding ratio
J
2
J
1
oF
current densities is given by
1.
J
2
J
1
=
1
8
.
2.
J
2
J
1
= 1
.
3.
J
2
J
1
= 2
.
correct
4.
J
2
J
1
=
1
2
.
5.
J
2
J
1
= 16
.
6.
J
2
J
1
=
1
16
.
7.
J
2
J
1
= 8
.
8.
J
2
J
1
=
1
4
.
9.
J
2
J
1
= 4
.
Explanation:
Using the relation between current density
For an ohmic material
E
=
ρ J
, we fnd
J
2
J
1
=
E
2
ρ
E
1
ρ
=
E
2
E
1
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 Fall '08
 Turner
 Physics, Work

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