Resistance Question

The strands are all in constant contact with each-other meaning they're not to be treated as resistors in parallel- there might be some trickery I am unaware of, though.

Per unit length, the resistance is likely to just be the sum of all the area contributions of each strand involved.

Thanks for your response -
This was the answer I got for part (a):
6 * 0.6 + 4 = 7.600 ohms
but I'm not sure if it's right, especially since the question asked for 3 decimal places...

For (b), I wondered if it had something to do with thicker conductors having a lower resistance

Alright- it might be resistors in parallel, then. It's just my conjecture but in the case of resistors in parallel, you have current flowing down each resistor and unable to transfer to the other parallel resistor- in this case you've got a bunch of aluminum strands wound around a steel core, with current able to flow through whichever way it wants. I do recall a problem similar to this where the aluminum strands were all treated as parallel to each-other too, I'll try to find it to see what explanation they offer for that, then.

In reality I expect most of the current will opt to go down the aluminum fibres around the core- current takes the path of least resistance and will go down the lower-resistance parts moreso, though some will opt for the core. See https://www.quora.com/Is-steel-cable-used-as-an-electrical-conductor-in-overhead-power-lines for some kind of illustrative explanation for this kind of thing.

For (b) you know that resistivity of the steel is going to be constant, the length constant, but you're increasing the area. $\rho = \frac{RA}{L}$ and consequently the resistance will fade for the steel component (as you said.)

I'll try to find the reasoning for why the aluminum can be treated in parallel (if that is indeed the case as this implies)

Alright. Putting aside the mental effort into deciding whether to treat the aluminum in parallel or not, we could consider just adding a bunch of cores of the wire up to give a total area for the current to flow down. Consider

$\displaystyle \rho = \frac{RA}{L} = R\frac{\sum{A}}{L}$

The resulting resistance is just equal to

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

If you treated the resistors individually...



Which looks close to what we had earlier...

$\displaystyle \sum\frac{1}{R_i} = \frac{\sum{A_i}}{\rho{L}}$

So this does imply adding them in parallel if we're just considering adding up the area to get the final area in calculating resistance.

Treat the resistors in parallel.

I recall the problem I had (identical to yours) just told us to treat them in parallel- I never actually thought about why, I guess this is why...