Question

PLEASE HELP!! PHYSICS QUESTION-DOES ANYONE KNOW HOW TO DO THIS WITH THE WORKING OUT?!?

Taking the resistivity of platinoid as 3.3 x 10-7 m, find the resistance of 7.0 m of platinoid wire of average diameter 0.14 cm.

Answer 1

`1.5 \Omega`

Explanation:

The resistance of a wire is given by the equation

`R=(\rho L)/(A)`

where :

`\rho` is the resistivity of the material

L is the length of the wire

A is the cross-sectional area of the wire

Here we have the following data:

`\rho = 3.3\cdot 10^(7) m` is the resistivity

L = 7.0 m is the length of the wire

`d=0.14 cm` is the diameter of the wire, so the radius is

`r=(d)/(2)=(0.14)/(2)=0.07 cm = 7\cdot 10^(-4)m`

So the cross-sectional area is

`A=\pi r^2 = \pi (7\cdot 10^(-4))^2=1.54\cdot 10^(-6) m^2`

Substituting, the resistance of the wire is:

`R=((3.3\cdot 10^(-7))(7.0))/((1.54\cdot 10^(-6)))=1.5 \Omega`