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JulijaS [17]
3 years ago
5

A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w

ire to that of the tungsten wire.
Physics
1 answer:
spayn [35]3 years ago
8 0

Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

  1. directly proportional to its length i.eR\propto l
  2. inversely proportional to its cross section area i.eR\propto \frac{1}{A}

Therefore

R=\rho\frac{l}{A}

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2

R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }

\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }......(1)

Again for tungsten:

R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }

\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }........(2)

Given that R_1=R_2   and    l_1=l_2

Dividing the equation (1) and (2)

\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}

\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}   [since R_1=R_2   and    l_1=l_2]

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}

\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

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