If the length of a wire of resistance `R` is uniformly stretched to `n` times its original value, what will be its new resistance?  

Expert Answers
Borys Shumyatskiy eNotes educator| Certified Educator

I suppose that the cross sectional area `A` of the original wire is constant along its length, and the same for the stretched wire (although its cross sectional area `A'` is clearly different).

It is known and is not of big surprise that the resistance of a wire from the same material in the same conditions is directly proportional to its length `L` and is inversely proportional to its cross sectional area `A.`

It is given that the new length `L'` is `n` times greater than the original, `L'=n*L.` What about the new cross sectional area `A'?` The mass of the wire and therefore its volume must remain the same, and it is equal to `L*A = L'*A' = n*L*A'.` Hence the new area `A' = A/n,`  `n` times less, and

`(L')/(A') = (nL)/(A/n) = n^2 L/A.`

Therefore the new resistance will be `n^2` times greater than the original, `R' = n^2 R.`


Access hundreds of thousands of answers with a free trial.

Start Free Trial
Ask a Question