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Current, Resistance, and Electromotive Force
255
EXECUTE:
The ratio of the current at 20 C
°
to that at the higher temperature is
(0.860 A) (0.220 A)
3.909.
=
0
20
1
(
)
3.909
T
R
TT
R
α
=+
− =
, where
0
20 C.
T
=
°
20
0
31
/
1
3.909 1
20 C
666 C.
4.5 10 (C )
T
RR
−−
=°+
= °
×
°
EVALUATE:
As the temperature increases, the resistance increases and for constant applied voltage the current
decreases.
The resistance increases by nearly a factor of four.
25.23.
IDENTIFY:
Relate resistance to resistivity.
SET UP:
R =
ρ
L/A
EXECUTE:
(a)
R =
L/A
= (0.60
Ω
⋅
m)(0.25 m)/(0.12 m)
2
= 10.4
Ω
(b)
R =
L/A
= (0.60
Ω
⋅
m)(0.12 m)/(0.12 m)(0.25 m) = 2.4
Ω
EVALUATE:
The resistance is greater for the faces that are farther apart.
25.24.
IDENTIFY:
Apply
L
R
A
=
and
VI
R
=
.
SET UP:
2
Ar
π
=
EXECUTE:
42
7
(4.50 V) (6.54 10
m)
1.37 10
m.
(17.6 A)(2.50 m)
RA
VA
LI
L
−
−
×
===
=
×
Ω
⋅
EVALUATE:
Our result for
shows that the wire is made of a metal with resistivity greater than that of good
metallic conductors such as copper and aluminum.
25.25.
IDENTIFY
and
SET UP:
Eq. (25.5) relates the electric field that is given to the current density.
V = EL
gives the
potential difference across a length
L
of wire and Eq. (25.11) allows us to calculate
R.
EXECUTE:
(a)
Eq. (25.5):
/ so
/
EJ
J E
ρρ
==
From Table 25.1 the resistivity for gold is
8
2.44 10
m.
−
×
Ω⋅
72
8
0.49 V/m
2.008 10 A/m
m
E
J
−
=
×
×Ω
⋅
()
(
)
2
27
23
0.41 10 m
11 A
IJ
AJ
r
ππ
−
=
×
×
=
(b)
(
)
0.49 V/m 6.4 m
3.1 V
VE
L
=
(c)
We can use Ohm’s law (Eq. (25.11)):
.
R
=
3.1 V
0.28
11 A
V
R
I
=
Ω
EVALUATE:
We can also calculate
R
from the resistivity and the dimensions of the wire (Eq. 25.10):
( )
8
2
2
3
m 6.4 m
0.28 ,
0.42 10 m
LL
R
−
−
⋅
=
=
Ω
×
which checks.
25.26.
IDENTIFY
and
SET UP:
Use
V = EL
to calculate
E
and then
/
E J
=
to calculate
.
EXECUTE:
(a)
0.938 V
1.25 V/m
0.750 m
V
E
L
=
(b)
8
1.25 V/m
so
2.84 10
m
4.40 10 A/m
E
J
−
=
=
×
Ω
⋅
×
EVALUATE:
This value of
is similar to that for the good metallic conductors in Table 25.1.
25.27.
IDENTIFY:
Apply
00
1
R
RT
T
=+−
⎡⎤
⎣⎦
to calculate the resistance at the second temperature.
(a) SET UP:
1
0.0004 C
−
=°
(Table 25.1). Let
0
be 0.0 C and
be 11.5 C.
°°
EXECUTE:
( )
0
1
0
100.0
99.54
1
1+ 0.0004 C
11.5 C
R
R
−
Ω
=
Ω
+−
(b) SET UP:
1
0.0005 C
−
=−
°
(Table 25.2). Let
0
0.0 C and
25.8 C.
=
°=
°
EXECUTE:
(
)
( )
1
1
0.0160
1+
0.0005 C
25.8 C
0.0158
−
=+−=
Ω
−
°
°
=
Ω
⎢⎥
EVALUATE:
Nichrome, like most metallic conductors, has a positive
and its resistance increases with
temperature. For carbon,
is negative and its resistance decreases as
T
increases.
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Chapter 25
25.28.
IDENTIFY:
00
[1
(
)]
T
R
RT
T
α
=+−
SET UP:
0
217.3
R
=Ω
.
215.8
T
R
.
For carbon,
1
0.00050 (C ) .
−
=−
°
EXECUTE:
0
0
1
(
/
) 1
(215.8 / 217.3 ) 1
13.8 C
0.00050 (C )
T
RR
TT
−
−Ω
Ω
−
−=
=
=
−
°
°
.
13.8 C
4.0 C 17.8 C.
T
=+
=
°°
°
EVALUATE:
For carbon,
is negative so
R
decreases as
T
increases.
25.29.
IDENTIFY
and
SET UP:
Apply
L
R
A
ρ
=
to determine the effect of increasing
A
and
L
.
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 Spring '06
 Buchler
 Physics, Current, Resistance, Force

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