# Conductivity of a coated wire

## Homework Statement

A long round wire of radius a and conductivity σ is coated with a material of conductivity 0.1σ. (a) What must be the thickness of the coating so that the resistance per unit length of the uncoated wire is reduced by 50%?
2. Homework Equations [/B]
R = l/(σS)
where R is the resistance, l is the length of the wire, σ the conductivity and S the cross sectional area

## The Attempt at a Solution

regarding part a) the resistance per unit length of the uncoated wire is (1/σS) this equals 1/(σπa2
meaning that its resistance per unit length doesn't depend on the thickness of the coating.
However the soln manual
solved it by setting: resistance per unit length of coated wire = resistance per unit length of uncoated wire
Why is that so??

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gneill
Mentor
However the soln manual
solved it by setting: resistance per unit length of coated wire = resistance per unit length of uncoated wire
Why is that so??
The coating creates a conductive path that is parallel to the wire inside. How do parallel resistances add?

The coating creates a conductive path that is parallel to the wire inside. How do parallel resistances add?
as follows R1 + R2 = R1R2/(R1 + R2)

gneill
Mentor
as follows R1 + R2 = R1R2/(R1 + R2)
Better written as: R1 || R2 = R1R2/(R1 + R2)

And if R1 and R2 are equal, say R1 = R2 = R, what is the combined resistance?

• Abdulwahab Hajar
Better written as: R1 || R2 = R1R2/(R1 + R2)

And if R1 and R2 are equal, say R1 = R2 = R, what is the combined resistance?
Ok I get it, it becomes half the original
But isn't the question just asking the reduction of the resistance of the uncoated wire only whereas what you showed me would be the resistance of both.
Furthermore, how could you determine that they are in parallel??
Thank you it's much appreciated

gneill
Mentor
Ok I get it, it becomes half the original
But isn't the question just asking the reduction of the resistance of the uncoated wire only whereas what you showed me would be the resistance of both.
The question is perhaps poorly phrased, but I interpret it to mean they they want the resistance of the "new" coated wire as compared to the original uncoated wire.
Furthermore, how could you determine that they are in parallel??
Thank you it's much appreciated
Look at the way conductive paths combine. Consider a small (differential) length of the wire with the coating. The original wire will have some cross sectional area A1 while the coating will have some cross sectional area A2. Both have length dL. You could remove the coating, and form a second wire with cross section A2 in parallel with the first wire without changing the net conductivity.

• Abdulwahab Hajar
The question is perhaps poorly phrased, but I interpret it to mean they they want the resistance of the "new" coated wire as compared to the original uncoated wire.

Look at the way conductive paths combine. Consider a small (differential) length of the wire with the coating. The original wire will have some cross sectional area A1 while the coating will have some cross sectional area A2. Both have length dL. You could remove the coating, and form a second wire with cross section A2 in parallel with the first wire without changing the net conductivity.
Thank you sir, you've been very helpful