Answer to Question #238934 in Electrical Engineering for Hehe

Question

A copper wire of unknown length has a diameter of 0.25 in. and a resistance of 0.28 ohm. By several successive passes through drawing dies the diameter of the wire is reduced to 0.05 in. Assuming that the resistivity of the copper remains unchanged in the drawing process, calculate the resistance of the of the reduced-size wire

Answer 1

First, find the area of the copper wire using the following equation:

$$R=\frac{\rho L}{A},\\\space\\ A=\frac{\pi D^2}{4}.$$

The volume of the wire is

$$V=AL=\frac{\pi D^2}{4}L.$$

The volume of the reduced wire:

$$V=\frac{\pi d^2}{4}l.$$

Since the volume remains unchanged,

$$\frac{\pi d^2}{4}l=\frac{\pi D^2}{4}L,\\\space\\ l=L\frac{D^2}{d^2}.$$

The resistance of the of the reduced-size wire is

$$r=\frac{\rho l}{a}=\frac{\rho LD^2}{d^2·\frac{\pi d^2}{4}}=\frac{4\rho LD^2}{\pi d^4}.$$

From the very first equation,

$$\rho L=RA=R\frac{\pi D^2}{4}.\\\space\\ r=R\frac{D^4}{d^4}=175\space\Omega.$$