Question

A copper wire has a diameter 0.5mm 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?

Answer 1

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Answer 2

Answer: The new resistance is 2.5 ohm.          

Explanation:

Given that,

Diameter d = 0.5 mm

Resistivity $\rho = 1.6\times10^{-8}\ ohm$

Resistance R = 10 ohm

We know that,

The resistance of the wire is

$R = \dfrac{\rho l}{A}$....(I)

$10=\dfrac{1.6\times10^{-8}\times l}{3.14\times0.25\times0.25\times10^{-6}}$

$l = \dfrac{3.14\times0.25\times0.25\times10^{-6}\times10}{1.6\times10^{-8}}$

$l = 122.6 m$

If the diameter is double then,

The new resistance is

$R' = \dfrac{\rho l}{A'}$.....(II)

Divide equation (II) by equation (I)

$\dfrac{R'}{R}=\dfrac{\rho l\times A}{A'\times\rho l}$

$\dfrac{R'}{R}=\dfrac{A}{A'}$

$\dfrac{R'}{R}=\dfrac{\pi (\dfrac{d}{2})^2}{\pi (\dfrac{d'}{2})^2}$

$\dfrac{R'}{R}=\dfrac{(\dfrac{d}{2})^2}{(\dfrac{2d}{2})^2}$

$R' = \dfrac{10}{4}$

$R' =2.5\Omega$

Hence, The new resistance is 2.5 ohm.