Help Easy Physics - Resistivity Ratios

[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 :)

 

[PeterO]

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.

 

[I-need-help]

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.

 

[PeterO]

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

M_a = V_aσ_a
and
M_c = V_cσ_c

In ratio form:

M_a/M_c = V_a/V_c x σ_a/σ_c

We thus know

M_a/M_c = V_a/V_c 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.

V_a/V_c = r_a²/r_c²

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

r_a² = ρ_aL_a/∏_aR_a
and
r_c² = ρ_cL_c/∏_cR_c


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

r_a²/r_c² = ρ_a/ρ_c

Substituting back into:

V_a/V_c = r_a²/r_c²

gives

V_a/V_c = ρ_a/ρ_c

Then back into:

M_a/M_c = V_a/V_c x 1/3

gives

M_a/M_c = ρ_a/ρ_c x 1/3

which gives

M_a/M_c = 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:

r_a²/r_c² = ρ_a/ρ_c

would be expressed as

r_a/r_c = √[ρ_a/ρ_c]

but I knew my previous formula had r_a²/r_c² 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.

 

[I-need-help]

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. :)