1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Help! Easy Physics - Resistivity Ratios

  1. Oct 23, 2011 #1
    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. jcsd
  3. Oct 23, 2011 #2


    User Avatar
    Homework Helper

    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.
  4. Oct 25, 2011 #3
    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.
  5. Oct 25, 2011 #4


    User Avatar
    Homework Helper

    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
    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


    r2 = ρL/∏R

    This gives

    ra2 = ρaLa/∏aRa
    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


    Va/Vc = ρac

    Then back into:

    Ma/Mc = Va/Vc x 1/3


    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


    a1 = F1/m1
    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.
  6. Oct 26, 2011 #5
    Ahhhh okay, thanks so much for your help. Much appreciated. :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook