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1. How does the thickness and material of a wire affect long distance electricity transmission ( and power loss ) ?
2. (Original post by chowderspin)
How does the thickness and material of a wire affect long distance electricity transmission ( and power loss ) ?
Power loss in a wire depends on its resistance. The higher the resistance, the more the loss. (Other things being equal)

You should have a standard formula for resistance in terms of the cross section area, length and resistivity of a wire.
This tells you directly how resistance is affected by thickness (cross section area), length, and the material itself (resistivity).
3. The resistance, R of a material can be defined by the equation

Where is the resistivity of the material, is the length and is the cross-sectional area.

This means that as the cross-sectional area increases, the resistance decreases (we are dividing by a larger number). Alternatively, we can consider it from a non-mathematical point of view. Wires with smaller cross-sectional areas have higher resistances - they cannot move through as easily. This is like having a lot of cars trying to move down one lane of road. Wires with larger areas have less resistance - it is easier for electrons to flow through. If we consider the same number of cars as with the smaller C/S area but this time on a motorway with numerous lanes, it is easier for them to travel down the road and easier for the electrons to travel through the wire.

As for material, this is even simpler. Different materials have different resistivities; it is easier for current to flow through some materials than others. For example: copper is a reasonably good conductor of electricity - its resistivity at a given temperature is relatively low. Conversely, a lot of non-metals are poor conductors of electricity, they have high resistivities. Multiplying the length by a larger number means that resistance will be larger.

There are a number of equations for power dissipated but for this purpose, I will use . Here, let the current be constant so we can see how the power dissipated depends on resistance when current remains the same. This shouldn't be too hard now - we can see that as resistance increases, power must increase if current remains the same.

Finally, we can combine the resistivity formula with some other equations to give: and . Indeed, as we would expect, any electricity formula with resistance, can be combined with the formula we use to calculate resistance.
4. Mm. I think the aim of this site is to help people to do the work themselves, rather than answer the whole question for them.
5. (Original post by Stonebridge)
Power loss in a wire depends on its resistance. The higher the resistance, the more the loss. (Other things being equal)

You should have a standard formula for resistance in terms of the cross section area, length and resistivity of a wire.
This tells you directly how resistance is affected by thickness (cross section area), length, and the material itself (resistivity).
How could I bring the national grid into the answer?

Ive already stated how much energy is wasted each year in the National grid through heating etc.

Also, the question was, "How does the thickness and material of a wire affect long distance electricity transmission ( and power loss ) ? "

How do I bring resistivity into my answer when Im supposed to explain in terms of power loss? i.e. how can i relate the factors affecting resistivity with power loss?
6. (Original post by chowderspin)
How could I bring the national grid into the answer?

Ive already stated how much energy is wasted each year in the National grid through heating etc.

Also, the question was, "How does the thickness and material of a wire affect long distance electricity transmission ( and power loss ) ? "

How do I bring resistivity into my answer when Im supposed to explain in terms of power loss? i.e. how can i relate the factors affecting resistivity with power loss?
If you want to talk about the national grid, you have to explain that this transmits power through the cables at very high voltage. The cables are also very thick.
As power = VI, then you can achieve high power using either high voltage or high current. High voltage (and therefore lower current) is used as this gives lower energy loss.
As I said in my 1st post, power loss in the cables is due to resistance. Resistance turns the energy into heat in the cables. The higher the resistance the higher the power loss. This power loss is = I²R so you reduce power loss by reducing I and R.

Also as I said in my 1st post, the resistance of the cables depends on their cross section area (thickness), their length and their resistivity. There is a formula you should know (which dstevens gave in his post) for resistivity which gives exactly how the resistance depends on these things.
So, power loss depends on resistance. Resistance depends on cross section area, length and the resistivity of the material. This is how you answer the question.

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