Question

A 0.2 cm diameter wire must carry a 20-A current. If the maximum power dissipation along the wire is 4W/m, what is the minimum allowable conductivity of the wire in Ohm-m? (a) 3.18x10 (b) 3.18x10 () 3.18x10 (d) 3.18x10

Answer 1

The conductivity of the wire is
`3.18*10^(7)\ ohm-m`.

Step-by-step explanation:

Given that,

Diameter = 0.2 cm

Current = 20 A

Power = 4 W/m

We need to calculate the conductivity

We know that,

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

Using formula of resistance

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

Where,

`\rho` = resistivity

A = area

l = length

Using formula of power

`P = i^2 R`

`R = (P)/(i^2)`

Put the value of R in equation (I)

`(P)/(i^2)=(\rho l)/(\pi r^2)`

`\rho=(P\pi r^2)/(l*i^2)`

`\sigma=(l* i^2)/(P\pi r^2)`

Put the all values into the formula

`\sigma=(1*(20)^2)/(4*3.14*(0.1*10^(-2))^2)`

`\sigma=3.18*10^(7)\ ohm-m`

Hence, The conductivity of the wire is
`3.18*10^(7)\ ohm-m`.