# Help Easy Physics - Resistivity Ratios

• I-need-help

#### 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:

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 :)

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.

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.

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.

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.

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.

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.

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