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78. (a) Let
∆
T
be the change in temperature and
κ
be the coefficient of linear expansion
for copper. Then
∆
L =
L
∆
T
and
∆
∆
L
L
T
==
×
°
=
×
−−
(.
/
.
.
17 10
17 10
55
K)(1.0 C)
This is equivalent to 0.0017%. Since a change in Celsius is equivalent to a change on the
Kelvin temperature scale, the value of
used in this calculation is not inconsistent with
the other units involved. Incorporating a factor of 2 for the twodimensional nature of
A
,
the fractional change in area is
∆
∆
A
A
T
×
°
=
×
2
2 17 10
34 10
/
.
K)(1.0 C)
which is 0.0034%. For small changes in the resistivity
ρ
, length
L
, and area
A
of a wire,
the change in the resistance is given by
∆∆
∆
∆
R
RR
L
L
R
A
A
=
∂
∂
+
∂
∂
+
∂
∂
.
Since
R =
L
/
A
,
∂
R
/
∂
= L
/
A = R
/
,
∂
R
/
∂
L
=
/
A = R
/
L
, and
∂
R
/
∂
A = –
L
/
A
2
–R
/
A
.
Furthermore,
∆
/
=
α
∆
T
, where
is the temperature coefficient of resistivity for copper
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 Spring '10
 Rebello,NobelS

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