Question

a wire of resistance r is drawn so that its radius get halfed find new resistance

Answer 1

If you stretch it uniformly (maybe by drawing the wire though a die) until the diameter is half the original value, so the cross-sectional area is 1/4, then the length will have increased by a factor of 4 (assuming the drawing process keeps the volume fixed, which is not quite right, but close enough for the concept here), then the resistance will increase by 4 x 4 = 16.  One factor of 4 due to the reduced area and another factor of 4 due to the increased length.  And we are talking about DC resistance here since someone talked about high frequency behaving differently.

It is also possible that stretching the wire could change the resistivity a little too, so then the factor of 16 would be a little different. 

HOPE IT HELPS YOU.... PLS MARK ME AS BRAINLIEST....

Answer 2

$<b><i><u>$

A given resistance cable R is stretched to reduce its diameter to half its previous value. What will be its new resistance?

If a metallic wire has length L, diameter d and material of electrical resistivity rho, the resistance R of a wire is inversely proportional to the cross section A of the wire,

R = rho L / A

being

A = π d^2 / 4

Assuming that the wire is always used with the same length (L), the variation of the diameter (d) will cause a variation in the electrical resistance (R). So we are talking about its electrical resistance per unit length.

IN THIS CASE it could be said that if the diameter is reduced by half (d/2) , the area of ​​the cross section is reduced four times (A/4) and then, the resistance would increase four times, i.e., the new resistance will be 4R . . . right?

WRONG ! There is something more, something very important.

The process to reduce the diameter is called "drawing" (“trefilado”, in Spanish), which in this case corresponds to a cold work of 75% (% CW = 75), which is a huge change which produces a lot deformation in the grains of the polycrystalline material that composes the conductor. This large number of deformation defects greatly increases the electrical resistivity rho of the material, and therefore, the answer to the question is

THE NEW RESISTANCE WILL BE GREATER THAN 4R,

and we can´t know how much greater the resistance can be, without having more detailed information about the microstructure and composition of the material.

After drawing we can make an annealing of “recovery” to the wire, which is a heat treatment in a furnace at a moderate temperature (which could be, depending the melting point of the conductor material, between 100 and 400 celsiuls degrees). This “low” temperature is high enough to activate the diffusion of atoms in the metallic net. The rearrangement of the crystal structure is a way to recover almost all the initial properties of the material, which in this case means reducing the value of rho.

Then, assuming a "perfect" recovery treatment, we could say that the new resistance will be exactly 4R. However, in practice it will always be at least a little higher (say between 2 and 8%).