Tuesday, 1 July 2014

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

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...

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.`


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