Question

a copper wire has a diameter 0.5m and resistivity 1.6*10^-8 ohm m. (i). what will be the length of this wire to make the resistance 12 ohm? (ii). how much will be the resistance of another copper wire of same length but half the diameter?

Answer 1

Explanation:

Radius r = 0.25mm , ρ= 1.6*10^-8 Ωm, area = 6.25*10^-8 sq m

Resistance= R = ρ*L/A

=> 12*6.25*10^-8/(  1.6*10^-8 ) = L

=> L = 46.85m

(ii)

For new wire, radius is half. So area is 1/4 bcoz area proportional to square of radius.

Then Resistance = ρ*L/(A/4) = 4ρ*L/A = 4*46.85 = 187.4Ω

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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.