Calculate the resistivity of a wire of diameter 3.5 mm and length 15 cm. Resistance of the wire is 25 ohm. *
Answers
Answer:
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Step-by-step explanation:
copy from Google
Strategy
We can rearrange the equation
R
=
ρ
L
A
R=ρLA to find the cross-sectional area A of the filament from the given information. Then its diameter can be found by assuming it has a circular cross-section.
Solution
The cross-sectional area, found by rearranging the expression for the resistance of a cylinder given in
R
=
ρ
L
A
R=ρLA, is
A
=
ρ
L
R
A=ρLR.
Substituting the given values, and taking ρ from Table 1, yields
A
=
(
5.6
×
10
−
8
Ω
⋅
m
)
(
4.00
×
10
−
2
m
)
0.350
Ω
=
6.40
×
10
−
9
m
2
A=(5.6×10−8Ω⋅m)(4.00×10−2m)0.350Ω=6.40×10−9m2.
The area of a circle is related to its diameter D by
A
=
π
D
2
4
A=πD24.
Solving for the diameter D, and substituting the value found for A, gives
D
=
2
(
A
p
)
1
2
=
2
(
6.40
×
10
−
9
m
2
3.14
)
1
2
=
9.0
×
10
−
5
m
D=2(Ap)12=2(6.40×10−9m23.14)12=9.0×10−5m.
Discussion
The diameter is just under a tenth of a millimeter. It is quoted to only two digits, because ρ is known to only two digits.Strategy
We can rearrange the equation
R
=
ρ
L
A
R=ρLA to find the cross-sectional area A of the filament from the given information. Then its diameter can be found by assuming it has a circular cross-section.
Solution
The cross-sectional area, found by rearranging the expression for the resistance of a cylinder given in
R
=
ρ
L
A
R=ρLA, is
A
=
ρ
L
R
A=ρLR.
Substituting the given values, and taking ρ from Table 1, yields
A
=
(
5.6
×
10
−
8
Ω
⋅
m
)
(
4.00
×
10
−
2
m
)
0.350
Ω
=
6.40
×
10
−
9
m
2
A=(5.6×10−8Ω⋅m)(4.00×10−2m)0.350Ω=6.40×10−9m2.
The area of a circle is related to its diameter D by
A
=
π
D
2
4
A=πD24.
Solving for the diameter D, and substituting the value found for A, gives
D
=
2
(
A
p
)
1
2
=
2
(
6.40
×
10
−
9
m
2
3.14
)
1
2
=
9.0
×
10
−
5
m
D=2(Ap)12=2(6.40×10−9m23.14)12=9.0×10−5m.
Discussion
The diameter is just under a tenth of a millimeter. It is quoted to only two digits, because ρ is known to only two digits.