Practice portal › Current Electricity › Resistance and Resistivity
Asked in GSEB Board March 2018 · Resistance of a wire and stretching
Given: stretching reduces the area from A to A/n.
Idea: the amount of metal does not change, so the volume AL stays the same.
New length: A/n L'=AL, so L'=nL.
R'=ρ(L')/(A')=ρ(nL)/(A/n)=n² ρL/A.
So R'=n²R: the wire is n times longer and n times thinner, and each change multiplies the resistance by n.
Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.
Practise this with a timer