Question

A cylindrical resistor of length 2 cm and diameter of 0.5 cm has a voltage of 1 V applied between the two ends. This voltage results in a current of 2 mA.

a. Find the conductivity of the material from which the resistor is made.
b. Next, the resistor is reshaped so that one half of its length remains the same while the other half has its diameter reduced to 0.2 cm. Find the total resistance.

Answer 1

a) 2.037 ohm/meter

b) 1487.9 Ω

Step-by-step explanation:

a)

We all know that:

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

where:

`\rho` = resistivity

and which can be written as:

`\rho = (1)/( \sigma)`

where
`\sigma` = conductivity

l = length = 2 cm =
`2*10^(-2)m`

A = area =
`\pi r ^2`

where

r = radius of the cylinder

r =
`(D)/(2)`

=
`(0.5)/(2)`

= 0.25 cm

Also by ohm's law

V = I R

`R = (V)/(I)`

=
`(1 V )/(2*10^(-3) A)`

= 500 Ω

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

`R = (l)/(\sigma A)`

`\sigma = (l)/(R A)`

`\sigma = (2*10^(-2)m)/(500* \pi * (0.25)^2 * 10^(-4))`

`\sigma = 2.037 ohm/ meter`

b)

Since
`R = (\rho * l)/(A *l)`

`R = (\rho *l)/(V)`

where ; V =
`\pi r^2 l`

`\pi r^2 l = \pi r_1^2 l_1+ \pi r_2^2 l_2`

r = 0.25 cm, l = 2cm ,
`r_1 = 0.25cm` ,
`r_2 = 0.1 cm`

So ; we are tasked to determine
`l_1` and
`l_2`

`l' = l_1 + l_2`

`l_1 = (l')/(2) =l_2`

`l_1 = l_2`

`l_1 = (\pi r^2 l)/(\pi r^2_1+ \pi r_2^2)`

`l_1 = (\pi 0.25^2 *2)/(\pi 0.25^2_1+ \pi 0.1_2^2)`

= 1.724 cm

`l' = l_1 + l_2`

`l' = 1.724+1.724`

`l'` ≅ 3.45 cm

Total Resistance R' is calculated as follows:

`R' = (\rho l'^2)/(V)`

=
`(1)/(2.037) *(3.45^2*10^(-4))/(\pi * (0.25)^2*2*10^(-6))`

R' = 1487.9 Ω