Question

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 wire to that of the tungsten wire.

Answer 1

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

`√(3) :√(10)`

Step-by-step 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.e
   `R\propto l`
2. inversely proportional to its cross section area i.e
   `R\propto (1)/(A)`

Therefore

`R=\rho(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 ((d_1)/(2) )^2`

`R_1=\rho_1(l_1)/(\pi((d_1)/(2))^2 )`

`\Rightarrow ((d_1)/(2))^2=\rho_1(l_1)/(\pi R_1 )`......(1)

Again for tungsten:

`R_2=\rho_2(l_2)/(\pi((d_2)/(2))^2 )`

`\Rightarrow ((d_2)/(2))^2=\rho_2(l_2)/(\pi R_2 )`........(2)

Given that
`R_1=R_2` and
`l_1=l_2`

Dividing the equation (1) and (2)

`\Rightarrow( ((d_1)/(2))^2)/( ((d_2)/(2))^2)=(\rho_1(l_1)/(\pi R_1 ))/(\rho_2(l_2)/(\pi R_2 ))`

`\Rightarrow( (d_1)/(d_2) )^2=(1.68* 10^(-8))/(5.6* 10^(-8))` [since
`R_1=R_2` and
`l_1=l_2`]

`\Rightarrow( (d_1)/(d_2) )=\sqrt{(1.68* 10^(-8))/(5.6* 10^(-8))}`

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

`\Rightarrow d_1:d_2=√(3) :√(10)`

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

`√(3) :√(10)`