Schoolwork123
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#1
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#1
Hi - I'm not sure how to answer this homework question. Any ideas would be much appreciated!

a) A length of electrical cable for use at high voltages consists of a steel alloy conductor surrounded by six aluminium conductors as shown in the diagram. Calculate the resistance of one kilometre of this cable given that:
Resistance of 1 km of aluminium conductor = 0. 60 Ω
Resistance of 1 km of steel alloy conductor = 4.0 Ω.
Give your answer to three decimal places.

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b) Another type of electrical cable has a much thicker conductor, of the same steel alloy surrounded by nine aluminium conductors as shown. The aluminium conductors have the same cross section area as those in part a). In what way does the resistance of 1 km of this cable differ from the resistance of 1 km of the cable in part a)? You must justify your answer.
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Callicious
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#2
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#2
You should be able to just treat it as the compound area for each segment, i.e. sum of all areas. rho = RA/L hence resistance per unit length should decrease.

Edit:
This does imply treating them in parallel. See my answer below.
Last edited by Callicious; 1 week ago
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Schoolwork123
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#3
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(Original post by Callicious)
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
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Callicious
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#4
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(Original post by Schoolwork123)
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
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. (I expect one way to consider this would be that you'd expect energy dissipation to be minimized and would thus expect more current to go down the less resistive route, but I digress...)

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...ad-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)
Last edited by Callicious; 1 week ago
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Schoolwork123
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#5
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(Original post by Callicious)
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...ad-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)
Thank you so much for your help - I agree with your idea, and I wondered about whether the aluminium would be treated in parallel too. The question seemed suspiciously straightforward otherwise.
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Callicious
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#6
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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_{tot} = \displaystyle \frac{\rho{L}}{\sum{A}}

If you treated the resistors individually...

R_i = \frac{\rho{L}}{A_i}

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

Which looks close to what we had earlier...

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

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...
Last edited by Callicious; 1 week ago
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Schoolwork123
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#7
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#7
(Original post by Callicious)
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...

R_i = \frac{\rho{L}}{A_i}

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

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...
Thanks so much - I'll try that.
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Schoolwork123
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#8
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#8
You were right - treating them as resistors in parallel gave me an answer of 0.098 ohms, which was correct. Thanks so much for your explanation!
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Joinedup
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#9
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#9
If the wires were all insulated from each other it'd be straightforward to think of it as resistances in parallel...

I would reason for uniform wires that since they're connected together at each end they've all got the same PD across their length... and also at any point along the cable the wires will also all be at the same potential - so you don't need to complicate it by worrying about any current flowing between the wires.
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