Question

If a copper wire is stretched to make its radius decrease by 0.1%, then the % increase in its resistance is approximately?

Answer 1

`0.4 " %"`

Explanation:

A thinner longer wire will have greater resistance.

As the wire has circular cross section:

• `R(r, L , rho) = (rho L)/(pi r^2)`

Wire and elongated wire will have same volume:

`V_o = pi r^2 L implies L(r) = V_o/(pi r^2)`

`implies R(r, rho) = (rho V_o)/(pi^2 r^4)`

To first order:

`implies dR = - 4(rho V_o)/(pi^2 r^5) dr`

`implies (dR)/R = -( 4(rho V_o)/(pi^2 r^5) dr)/( (rho V_o)/(pi^2 r^4))`

`= - 4(dr)/r = -4 (-0.1) = 0.4" %"`