if a wire of initial length, and radius has a measured resistance of 1.0 ohms. the wire is drawn under tensile stress to a new uniform radius of 0.25r. what is the new resistance of the wire?
assume the volume of the wire is constant cross sectional area = pi*R^2 A1 = PI * r^2 A2= PI * (.25*r)^2 = PI * .0625*(r)^2 = 16* A2 If volume stays constant, Length2 = 16* Length1 since resistance is proportional to length/area R2/R1= L2L1*A1/A2=16*16 = 256 The resistance now 256 times as great
Wire resistance =( length/area) x a constant When drawn, the wire radius is 1/4 of original, and it's length is 16 times of original (because volume is unchanged). Solve for new resistance/
