P27_050.PDF
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)
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Chapter 27 - 27.50
50. (a) We denote the copper wire with subscript
c
and the aluminum wire with subscript
a
.
A
=
(2
.
75
×
10
−
8
Ω
·
m)(1
.
3m)
=1
.
3
×
10
−
3
Ω
.
(5
.
2
×
10
−
3
m)
2
(b) Let
R
=
ρ
c
L/
(
πd
2
/
4) and solve for the diameter
d
of the copper wire:
d
=
4
ρ
c
L
4(1
.
69
×
10
−
8
Ω
·
m)(1
.
3m)
=4
.
6
×
10
−
3
m
.
π
(1
.
3
×
10
−
3
Ω)
R
=
ρ
a
L
πR
=
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chap04
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