P27_038.PDF
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Pobierz
Chapter 27 - 27.38
38. (a) From
P
=
V
2
/R
=
AV
2
/ρL
,wesolveforthelength:
ρP
=
(2
.
60
×
10
−
6
m
2
)(75
.
0V)
2
m)(5000W)
=5
.
85m
.
(5
.
00
×
10
−
7
Ω
·
(b) Since
L
∝
V
2
thenewlengthshouldbe
L
=
L
V
V
2
=(5
.
85m)
100V
75
.
0V
2
=10
.
4m
.
L
=
AV
2
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