# Homework Help: Help! Easy Physics - Resistivity Ratios

1. Oct 23, 2011

### I-need-help

The resistivity of Aluminium is twice that of copper. However, the density of Aluminium is one-third that of Copper.

a) For equal length and resistance, calculate the ratio:

mass of aluminium/mass of copper​

I'm thinking, the density ratio (Al:Cu) is 1:3 .....and the resistivity ratio is 2:1 ...so would the overall ratio be, 2:3 ? I have no idea...

This is easy Physics compared to what else is on this site, but my physics isn't any good, so I joined this forum in hope that someone would be able to help me!

Thank you :)

Last edited: Oct 23, 2011
2. Oct 23, 2011

### PeterO

refer to the formula of resistance - often given as R = ρL/A - to see that one of the wires will have to be thicker - thus have a greater volume of metal, thus greater mass than if it was the same thickness.

3. Oct 25, 2011

### I-need-help

Okay, thank you. I understand that, i'm just not sure how to get a ratio from that, with no other information... I'm not even sure if I need to work out a ratio actually. Oh well, thanks anyway.

4. Oct 25, 2011

### PeterO

Firstly, your original answer 2:3 was correct - but your uncertainty indicated you were not sure why.

When doing ratios, I just use the formulas and do a grand divide to produce the ratio.

In this case we want the ratio of masses.

Well I know density is mass/ volume [ σ = M/V] so

M = Vσ

Now the ratio: firstly put subscripts on the variable - I would use c for copper and a for aluminium

Ma = Vaσa
and
Mc = Vcσc

In ratio form:

Ma/Mc = Va/Vc x σac

We thus know

Ma/Mc = Va/Vc x 1/3

since we were given the ratio of the densities.

So now we need the ratio of the Volumes to complete this.
Each wire is effectively a cylinder

V =πr2h

For the wire, h = length of the wire - which is the same for both wires - so the ratio reduces to.

Va/Vc = ra2/rc2

So now we need the ratio of Radii [or diameters?]

Resistance is given by"

R = ρL/A

Since Area here is that of the circular wire,

R = ρL/∏r2

transposing

r2 = ρL/∏R

This gives

ra2 = ρaLa/∏aRa
and
rc2 = ρcLc/∏cRc

Now for these wires, Length and resistance [and of course ∏] are the same, so the ratio simplifies to

ra2/rc2 = ρac

Substituting back into:

Va/Vc = ra2/rc2

gives

Va/Vc = ρac

Then back into:

Ma/Mc = Va/Vc x 1/3

gives

Ma/Mc = ρac x 1/3

which gives

Ma/Mc = 2 x 1/3

which is 2/3 or 2:3 if you like.

While this has been lengthy to type out, when written it is much quicker.

Note: Normally the line:

ra2/rc2 = ρac

would be expressed as

ra/rc = √[ρac]

but I knew my previous formula had ra2/rc2 in it so I left it as was.

You can use this ratio technique to find the ratio of anything:

eg Ratio of two accelerations

F = ma → a = F/m

so

a1 = F1/m1
and
a2 = F2/m2

a1/a2 = F1/F2 x m2/m1

[Note that since m was in the denominator, is appears "upside down" as a ratio.]

SO once we know the ratio of the forces, and the ratio of the masses we can work out the ratio of the accelerations, without calculating/knowing the actual value of each acceleration.

5. Oct 26, 2011

### I-need-help

Ahhhh okay, thanks so much for your help. Much appreciated. :)