Question

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`

Step-by-step explanation:

The resistance of a wire is given by:

`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

For the wire in this problem:

`\rho = 3.3\cdot 10^(-7) \Omega m`

L = 7.0 m

r = d/2 = 0.07 cm = 0.0007 m, so the area is

`A=\pi r^2=\pi(0.0007)^2=1.54\cdot 10^(-6)m^2`

Solving for R,

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