Answer:
a) 2.037 ohm/meter
b) 1487.9 Ω
Explanation:
a)
We all know that:
[tex]R = \rho \frac{l}{A}[/tex]
where:
[tex]\rho[/tex] = resistivity
and which can be written as:
[tex]\rho = \frac{1}{ \sigma}[/tex]
where [tex]\sigma[/tex] = conductivity
l = length = 2 cm = [tex]2*10^{-2}m[/tex]
A = area = [tex]\pi r ^2[/tex]
where
r = radius of the cylinder
r = [tex]\frac{D}{2}[/tex]
= [tex]\frac{0.5}{2}[/tex]
= 0.25 cm
Also by ohm's law
V = I R
[tex]R = \frac{V}{I}[/tex]
= [tex]\frac{1 V }{2*10^{-3} A}[/tex]
= 500 Ω
[tex]R = \rho \frac{l}{A}[/tex]
[tex]R = \frac{l}{\sigma A}[/tex]
[tex]\sigma = \frac{l}{R A}[/tex]
[tex]\sigma = \frac{2*10^{-2}m}{500* \pi * (0.25)^2 * 10^{-4}}[/tex]
[tex]\sigma = 2.037 ohm/ meter[/tex]
b)
Since [tex]R = \frac{\rho * l}{A *l}[/tex]
[tex]R = \frac{\rho *l}{V}[/tex]
where ; V = [tex]\pi r^2 l[/tex]
[tex]\pi r^2 l = \pi r_1^2 l_1+ \pi r_2^2 l_2[/tex]
r = 0.25 cm, l = 2cm , [tex]r_1 = 0.25cm[/tex] , [tex]r_2 = 0.1 cm[/tex]
So ; we are tasked to determine [tex]l_1[/tex] and [tex]l_2[/tex]
[tex]l' = l_1 + l_2[/tex]
[tex]l_1 = \frac{l'}{2} =l_2[/tex]
[tex]l_1 = l_2[/tex]
[tex]l_1 = \frac{\pi r^2 l}{\pi r^2_1+ \pi r_2^2}[/tex]
[tex]l_1 = \frac{\pi 0.25^2 *2}{\pi 0.25^2_1+ \pi 0.1_2^2}[/tex]
= 1.724 cm
[tex]l' = l_1 + l_2[/tex]
[tex]l' = 1.724+1.724[/tex]
[tex]l'[/tex] ≅ 3.45 cm
Total Resistance R' is calculated as follows:
[tex]R' = \frac{\rho l'^2}{V}[/tex]
= [tex]\frac{1}{2.037} *\frac{3.45^2*10^{-4}}{\pi * (0.25)^2*2*10^{-6}}[/tex]
R' = 1487.9 Ω
