695
19.13.
19.13 View the pipeline as a big long resistor. Its resistance is given by Equation 19.13 on page 842 of
the text,
R=
ρ`
,
A
where ρ is the resistivity of steel, ` is the length of the pipeline, and A is its cross-sectional area (but just
the steel part of the cross section — don’t include the hollow part that ordinarily carries the oil).
We’re given the resistivity of steel as ρ = 1.80 × 10−7 Ω·m , and we’re given ` = 1.27 × 106 m .
It remains to find A. Consider a cross section of the pipe. The steel portion of the cross section looks
like a washer. Cut the washer at one point, and straighten it out. You’ll get a strip of steel whose length
is the circumference of the pipe, π(1.20 m ), and whose width is the thickness of the pipe’s wall, 0.010 m .
Hence, A = π(1.20 m )(0.010 m ) = 0.038 m 2 .
Thus,
R=
(1.80 × 10−7 Ω·m )(1.27 × 106 m )
= 6.0 Ω .
0.038 m 2
19.14 We are told that the old and new wires have the same volume. Let A be the old and A0 be the new
cross sectional area of the wire. Then
A` = A0 (10`) =⇒ A0 =
A
.
10
The resistance of the original wire is given by Equation 19.13 on page 842 of the text,
R=
ρ`
,
A
so the resistance of the new wire is
R0 =
ρ(10`)
ρ(10`)
=
= 100
0
A
(A/10)
ρ`
A
= 100R.
Therefore the resistance increases by a factor of 100.
19.15
a) The resistance of the cube is given by Equation 19.13 on page 842 of the text,
R=
b)
ρ
ρ`
2.0 × 10−8 Ω·m
ρ`
= 2 =⇒ ` =
=
= 2.0 × 10−2 m .
A
`
R
1.0 × 10−6 Ω
Let a be the length of a side of the square plate. Since the volume of material remains the same,
`3 = a2
√
p
`
=⇒ a = 10`2 = 10(2.0 × 10−2 m )2 = 6.3 × 10−2 m .
10
c) The resistance is
R=
19.16
(2.0 × 10−8 Ω·m )(2.0 × 10−2 m )/10)
ρ(`/10)
=
= 1.0 × 10−8 Ω .
a2
(6.3 × 10−2 m )2
Consider a length ` of wire with resistivity ρ . If the wire is gauge 4, then its resistance is
R=
ρ`
.
21.1 × 10−6 m 2
If you increase the gauge to gauge 6, the resistance is then
R0 =
ρ`
.
13.3 × 10−6 m 2
696
CHAPTER 19.
ELECTRIC CURRENT, RESISTANCE, AND DC CIRCUIT ANALYSIS
To find the factor by which the resistance increases, evaluate their ratio
21.1 × 10−6 m 2
R0
=
≈ 1.6.
R
13.3 × 10−6 m 2
A similar result is obtained if you increase the gauge number by 2 with other initial gauge numbers. For
example, if you increase the gauge from 6 to 8, the ratio of the resistances is
13.7 × 10−6 m 2
R0
≈ 1.6,
=
R
8.37 × 10−6 m 2
so the result does not depend upon the gauge of the initial wire.
19.17
From Equation 19.3 on page 840 of the text, the current in the big wire is
I = nqAhvi =⇒ hvi =
I
,
nqA
where n is the number of charge carriers per unit volume, q is the amount of charge per carrier, hvi is the
2
d
, so
drift speed, and A is the cross-sectional area. Let d be its diameter. Then A = π
2
hvi =
4I
I
.
2 =
2
nqπd
d
nqπ
2
d
For the small wire, I, n, and q are the same, but the diameter is , so the cross-sectional area is A0 = π
2
Therefore its drift speed will be
hv0 i =
2
d
.
4
I
16I
.
2 =
nqπd2
d
nqπ
4
The drift speed in the small wire is greater than that in the big wire by a factor of 4.
19.18
a)
Use Equation 19.15 on page 845 of the text,
ρ = ρ0 [1 + α(T − T0 )]
=⇒ 0 Ω · m = ρ0 [1 + α(T − T0 )]
=⇒ (T − T0 ) = −
1
1
= 2 × 103 K .
=−
α
−0.5 × 10−3 K −1
If we take the reference temperature to be 20◦C= 293 K = T0 , then
T = 2 × 103 K + 293 K = 2 × 103 K .
b) The result of part a) is cooler than the melting temperature. It is unlikely that the resistance of diamond
vanishes when the temperature is 2×103 K , which calls into question the assumption about the temperature
independence of α.