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Current and Resistance
79
17.19
The volume of material,
( )
2
00
L
rL
π
==
0
VA
, in the wire is constant. Thus, as the wire is
stretched to decrease its radius, the length increases such that
( )
()
2
0
ff
=
2
0
giving
2
2
0
4.0
0.25
rr
L
=
2
0
16
f
f
LL
L
=
0
L
The new resistance is then
2
2
22
0
0
16
16 4
256
4
f
RR
Ar
r
r
ρρ
ρ
ππ
== =
=
=
=
0
256 1.00
256
=
Ω
Ω
17.20
Solving
( )
0
1
TT
α
=+−
0
for the final temperature gives
()( )
3
0
0
1
-3
0
140
19
20 C
1.4
10 C
4.5
19
R
−
−
Ω
−Ω
=+
=°+
= × °
×°
Ω
17.21
From Ohm’s law,
ii
VI
R IR
∆=
=
, so the current in Antarctica is
1
3
1
3
C
58.0 C
20.0 C
1.00
1.98 A
C
88.0 C
20.0 C
i
i
fi
f
f
T
R
II
R
T
−
−
−
+−
°
°
−
°
° −
°
−
°
1
1
1
3.90
10
A
1
3.90
10
i
RT
I
−
+×
17.22
The expression for the temperature variation of resistance,
( )
1
with
, gives the temperature coefficient of resistivity of this material as
0
20 C
T
=°
( )

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