conductor resistance question?

You start by considering that you have a certain volume of copper wire whose physical diameter you are given. We do not know how long the wire is so lets assume you have one foot. Then you may calculate the volume of the wire. After you draw the wire through numerous dies you know have the same volume of copper wire but at a new diameter, therefore it must be longer than the original length.

So V = ((0.162inches/2)^2)*(pi)*12 inches = 0.247 Cubic inches

The new length of wire must still be this volume. Find the new length based on a diameter of 0.032 inches to get:

0.247 Cu-Ins = ((0.032 inches/2)^2)*(pi) * (New length)

New Length = 307.55 inches

Now: New Length / original length = 307.55 inches / 12 inches = 25.63 longer than the original length.

You also know that Resistance = (rho)*length / Area

So:

R1 = (rho)*length1 / Area1 solve for "roh" to get:

rho = R1*A1/(length1)

Your new resistance is given by:

R2 = (rho)*length2 / Area2 solve for "roh" to get:

rho = R2*A2/(length2)

Since both equations are equal to "rho" you can set them equal to each other:

R1*A1/(length1) = R2*A2/(length2), now solve for R2:

R2 = R1*(A1/A2)*(L2/L1) = R1*(d1^2/d2^2)*(L2/L1) = 0.4*(10.3)*(25.6) = 264 ohms.

Hope this helps,

Newton1Law

action of electrons is different in insulators and conductors. In an insulator, the electron is localized to an atom and has to "hop" from atom to atom with the intention to flow. greater potential helps the technique alongside so resistance falls with temperature. In a conductor, electrons are de-localized and propagate in a wave like way. If the atoms are in a periodic lattice with little circulation, it could provide a smaller concern to the wave. As temperature rises and the atoms flow greater, wave propagation is inhibited and resistance will improve.

The cross sectional area and length changed.

R=ρL/A

ρ = resistivity

L = length

A = area

d = 0.162 R = 0.4

d= 0.032 R =?

R = 0.8