fierro (jmf2547) – HW14 – berk – (60290)
1
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printout
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have
7
questions.
Multiplechoice questions may continue on
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before answering.
001
10.0 points
An electric field of 1824 V
/
m is applied to a
section of silver of uniform cross section.
Assume:
The
resistivity
ρ
of
silver
is
1
.
59
×
10
−
8
Ω
·
m at 20
◦
C.
The tempera
ture coefficient of silver is 0
.
0038(
◦
C)
−
1
.
Calculate the resulting current density if
the specimen is at a temperature of 6
◦
C.
Correct answer: 1
.
21163
×
10
11
A
/
m
2
.
Explanation:
Let :
ρ
20
= 1
.
59
×
10
−
8
Ω
·
m
,
T
= 20
◦
C
,
α
= 0
.
0038(
◦
C)
−
1
,
and
E
= 1824 V
/
m
.
By Ohm’s Law,
J
=
σE
ρ
=
ρ
20
[1 +
α
(
T
−
T
20
)]
= (1
.
59
×
10
−
8
Ω
·
m)
× {
1 +
bracketleftbig
0
.
0038(
◦
C)
−
1
bracketrightbig
(6
◦
C
−
20
◦
C)
}
= 1
.
50541
×
10
−
8
Ω
·
m
.
Thus,
J
=
σE
=
E
ρ
=
1824 V
/
m
1
.
50541
×
10
−
8
Ω
·
m
=
1
.
21163
×
10
11
A
/
m
2
.
002
10.0 points
The resistance to the right of A
′
B
′
is the
same as the resistance to the right of AB; that
is,
R
AB
=
R
→
A
′
B
′
.
r
r
r
2
r
2
r
2
r
A
A
′
B
B
′
What is the resistance
R
AB
between the
terminals
A
and
B
of this infinitely repeating
chain of resistors.
1.
R
AB
=
r
+
2
r R
AB
r
+
R
AB
,
therefore
R
AB
=
5
2
r
2.
R
AB
=
r
+ 2
r
−
R
AB
,
therefore
R
AB
=
3
2
r
3.
R
AB
=
r
+
2
r R
AB
2
r
+
R
AB
,
therefore
R
AB
= 2
r
correct
Explanation:
R
series
=
R
1
+
R
2
+
R
3
+
· · ·
1
R
parallel
=
1
R
1
+
1
R
2
+
1
R
3
+
· · ·
The infinite chain can be redrawn as follows
r
2
r
R
AB
A
A
′
B
B
′
R
AB
=
r
+
1
1
2
r
+
1
R
AB
=
r
+
2
r R
AB
R
AB
+ 2
r
(
R
AB
−
r
) (
R
AB
+ 2
r
) = 2
r R
AB
R
2
AB
−
r R
AB
−
2
r
2
= 0
R
AB
=
r
±
radicalbig
r
2
+ 4 (2
r
2
)
2
=
r
+ 3
r
2
=
2
r
.
003
10.0 points
Calculate the average drift speed of electrons
traveling through a copper wire with a cross
sectional area of 80 mm
2
when carrying a
current of 60 A (values similar to those for the
electric wire to your study lamp).
Assume
one electron per atom of copper contributes
to the current.
The atomic mass of copper
is 63
.
5 g
/
mol and its density is 8
.
93 g
/
cm
3
.
Avogadro’s number
N
A
is 6
.
02
×
10
23
.
Correct answer: 5
.
5369
×
10
−
5
m
/
s.
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fierro (jmf2547) – HW14 – berk – (60290)
2
Explanation:
Let :
N
= 1
,
M
= 63
.
5 g
/
mol
,
ρ
= 8
.
93 g
/
cm
3
,
A
= 80 mm
2
= 8
×
10
−
5
m
2
,
I
= 60 A
,
and
q
e
= 1
.
6
×
10
−
19
C
/
electron
.
We first calculate
n
, the number of current
carrying electrons per unit volume in copper.
Assuming one free conduction electron per
atom,
n
=
N
A
ρ
M
, where
N
A
is Avogodro’s
number and
ρ
and
M
are the density and the
atomic weight of copper, respectively
n
≡
parenleftbigg
1
electron
atom
parenrightbigg
N
A
ρ
M
.
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