94
CHAPTER 17
(c) The filament is tightly coiled
to fit the required length into a small space
(d) From
( )
00
1
TT
α
=+−
LL
, where
()
1
6
4.5
10
C
−
−
=×
°
, the length at
is
0
20 C
T
=°
0
0
25 m
1
1
4.5
10
C
L
L
−
==
=
+−
+×
°
1
6
26 m
C
2 600 C
20
−
°
°
−
17.60
Each speaker has a resistance of
4.00
R
=
Ω
and can handle 60.0 W of power. From
, the maximum safe current is
2
IR
=
P
max
60.0 W
4.00
R
Ω
√
3.87 A
=
I
Thus, the system
is not adequately protected by a 4.00 A fuse.
17.61
The crosssectional area of the conducting material is
( )
22
outer
inner
Arr
π
=−
Thus,
( )( )
( )
52
3.5
10
m
4.0
10
m
1
.
21
0 m
0
.
5
01
L
R
A
ρ
−
−−
×Ω
⋅
×
Ω
×−
×
7
3.7
37 M
=×Ω
=
17.62
(a)
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This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.
 Fall '10
 STAFF
 Physics, Current, Resistance

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