Resistivity

Okay, so here is a question:
The resistivity of aluminium is 2.6*10^-8 ohm metres. Calculate the cross sectional area A of a single strand of cable
the resistance is 2.0 ohm per KM
and length is 1000metres.
So I thought it'd be right to do:
(2.6*10^-8 * 1000m)/2*1000
As it is resistivity per km, so we need to get it into metres, however this is wrong, and should just be 2 on the bottom, could someone please explain why?

The question tells us the resistance of the cable is 2 ohms per km.

i.e. A 1km length of the cable has a total resistance of 2 ohms.

$\rho = \frac{RA}{L}$

$A = \frac{\rho L}{R}$

$A = \frac{2.6 \mathrm{x}10^{-8} \mathrm{\ x \ }1000}{2} = 1.3 \mathrm{x}10^{-5} \mathrm{\ m}^2$

so it's like the resistance is 0.002 ohms per metre, and they've already scaled it up for us to 1000m?
I am still struggling to understand why we scale up.
In every question I have done up to this point it is simply in metres, not Kilometres.
It is really frustrating me.
it feels like we have left it in KM as there is 1000m so it is just 2 ohms per km * 1
okay, I guess.
But now I am confused as we have done it in metres on the top then m^3 on the bottom, so we're gonna end up with m^-2.

Another floor in my logic is that it seems that we leave it in km because it happens to be our exact length, but then when we are doing it in metres for other things, e.g. if the length was 600 metres and everything was in metres, and we wanted the area we wouldnt then times our resistance by 600 so it equals our metres value, do you see what I mean?

I am tying myself in knots here, so I'm afraid you're gonna have to delve a little bit deeper in order for me to understand. I keep getting moments where it makes sense, then it drifts away as I look at it and say "hold on but ...means..."
the km and metres has really messed with me, and to be honest I honestly dont see how we can say:
our pl will be in metres our R is gonna be in km.
You cant just mix scales, or at least in my head it doesnt work.
It doesnt seem right to say "here is our resistivity in 10000m" then divide it by 10km
surely we have to stick to the same scale or we're gonna be a factor of 10^3 out?
Further help would be really appreciated, thanks so far though.

Okay, so here is a question:
The resistivity of aluminium is 2.6*10^-8 ohm metres. Calculate the cross sectional area A of a single strand of cable
the resistance is 2.0 ohm per KM
and length is 1000metres.
So I thought it'd be right to do:
(2.6*10^-8 * 1000m)/2*1000
As it is resistivity per km, so we need to get it into metres, however this is wrong, and should just be 2 on the bottom, could someone please explain why?

Be very careful with the wording.

The question clearly states ohms per km. i.e. 2 ohms per km is a value of resistance not resistivity.

But again I come back to the fact that we have left it as 2 as it is the same distance as our cable.
Does that mean i should make my ohms per metre into ohms per 624 metres if a question gave me that, that was the length of a given wire that I would have to calculate the resistivity of? No, we wouldn't, so I dont understand why on earth we make the exception here?!

It's not an exception - it's the application of the rules!

Please read through the whole of this post very carefully. I have written it in the fullest way possible. All of the information you need to understand the concept of resistivity is in here.

The materials property of resistivity is used to calculate the actual total resistance of any given volume of that material:

Resistivity is an intrinsic property of the material to inhibit current flow. It relates the electrical properties of a material to it's physical dimensions of cross sectional area and length.

The unit of resistivity is $\mathrm{ohm \ metres} = \Omega \mathrm{m}$

$\rho = \frac{RA}{L}$

equating units:

$\frac{RA}{L} = \frac{\Omega \mathrm{\ x \ m^2}}{\mathrm{m}} = \Omega \mathrm{m}$

Resistance on its own is purely a ratio between voltage and current and no information can be deduced about the physical dimensions of the material giving rise to that resistance.

Ohms law:

$R = \frac{V}{I}$

The unit of resistance is $\mathrm{ohms} = \Omega$

The analogy to resistivity would be the density of a material for instance. i.e 20 kg/m³ is not the same as saying 20 kg. The latter of which says nothing about the dimensions of that 20kg whereas the former can be applied to the physical dimensions in order to calculate the total mass. But even this is not accurate enough because the direct equivalent for resistance and resistivity would be mass and densitivity respectively with densitivity units of kg metres!

Going back to the original question

Resistance:

The question tells is the resistance of the cable is 2 ohms per km. That is, the total resistance of the cable is increasing by 2 ohms for every 1km length.

i.e. more information than just pure resistance is given, which therefore relates the length of the cable to resistance. But, it's still only in one dimension (length). It's still not resistivity.

Thus if the cable was 10 km long, then its total resistance would be $R = 2 \mathrm{\ x \ }\frac{10km}{1km} = 20\Omega$

If the cable was 400m long, then it's total resistance would be $R = 2 \mathrm{\ x \ }\frac{0.4km}{1km} = 0.8\Omega$

In our question, we are also told the length if the cable is 1km, therefore the total resistance for this 1km length of cable is: $R = 2 \mathrm{\ x \ }\frac{1km}{1km} = 2\Omega$

If the question stated the resistance of the cable was 27 ohms per 624 metres and the length was 2km then the total resistance would be calculated as:

$R = 27 \mathrm{\ x \ }\frac{2km}{0.624km} = 86.54\Omega$


Ohms/m or ohms/km is a standard unit of measure relating a cables resistance to a standard unit of length.

Resistivity:

The question asks for the cross sectional area and gives us the resistivity of the cable material as $\rho = 2.6 \mathrm{x}10^{-8} \Omega m$

Resistivity is defined as

$\rho = \frac{RA}{L}$

therefore

$A = \frac{\rho L}{R}$

where

$R = 2\Omega$ (previously calculated)

$L = 1 \mathrm{km}$

$\rho = 2.6 \mathrm{x}10^{-8} \Omega m$

substituting values:

$A = \frac{\rho L}{R} = \frac{2.6 \mathrm{x}10^{-8} \mathrm{\ x \ }1 \mathrm{x}10^3}{2} = 1.3 \mathrm{x}10^{-5} \mathrm{m}^2$