A copper wire has a diameter 0.2mm and resistivity of 1.6×10^-8ohm m.what will be the length of the wire to make its resistance 10 ohm? How much does the resistance change if the diameter is doubled?
Answers
Answer: The new resistance is 2.5 ohm.
Explanation:
Given that,
Diameter d = 0.5 mm
Resistivity \rho = 1.6\times10^{-8}\ ohmρ=1.6×10
−8
ohm
Resistance R = 10 ohm
We know that,
The resistance of the wire is
R = \dfrac{\rho l}{A}R=
A
ρl
....(I)
10=\dfrac{1.6\times10^{-8}\times l}{3.14\times0.25\times0.25\times10^{-6}}10=
3.14×0.25×0.25×10
−6
1.6×10
−8
×l
l = \dfrac{3.14\times0.25\times0.25\times10^{-6}\times10}{1.6\times10^{-8}}l=
1.6×10
−8
3.14×0.25×0.25×10
−6
×10
l = 122.6 ml=122.6m
If the diameter is double then,
The new resistance is
R' = \dfrac{\rho l}{A'}R
′
=
A
′
ρl
.....(II)
Divide equation (II) by equation (I)
\dfrac{R'}{R}=\dfrac{\rho l\times A}{A'\times\rho l}
R
R
′
=
A
′
×ρl
ρl×A
\dfrac{R'}{R}=\dfrac{A}{A'}
R
R
′
=
A
′
A
\dfrac{R'}{R}=\dfrac{\pi (\dfrac{d}{2})^2}{\pi (\dfrac{d'}{2})^2}
R
R
′
=
π(
2
d
′
)
2
π(
2
d
)
2
\dfrac{R'}{R}=\dfrac{(\dfrac{d}{2})^2}{(\dfrac{2d}{2})^2}
R
R
′
=
(
2
2d
)
2
(
2
d
)
2
R' = \dfrac{10}{4}R
′
=
4
10
R' =2.5\text{\O}megaR
′
=2.5Ømega
Hence, The new resistance is 2.5 ohm.
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