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# Resistivity Isaac physics

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Why bother with a post grad? Are they even worth it? Have your say! 26-10-2016
1. Hey guy, im attempting the questions in the isaac physics book. One of the questions is C1.12:

The answer i get is 76000, but this seems to be incorrect! The isaac physics website also doesnt give the correct answer to you so i dont know whats right.
2. I get a tiny resistance using
3. I get a tiny resistance using of Ohm

Some detail. n identical strands in parallel will have 1/n times the single strand resistance. One you have R_al and R_cu then use
4. I seem to get an answer of 756 ohms.
5. (Original post by nerak99)
I get a tiny resistance using of Ohm
6. (Original post by FusionNetworks)
What is the website answer? I will work agaain I forgot to square radii and alos made a unit erro
7. (Original post by nerak99)
What is the website answer? I will work agaain I forgot to square radii and alos made a unit erro
https://isaacphysics.org/questions/c...lls_book_ch_c1
Scroll to the bottom and click c1.12
8. (Original post by FusionNetworks)
https://isaacphysics.org/questions/c...lls_book_ch_c1
Scroll to the bottom and click c1.12
Thanks, Seems my answer is wrong. I'll keep on trying though
9. (Original post by FusionNetworks)
https://isaacphysics.org/questions/c...lls_book_ch_c1
Scroll to the bottom and click c1.12
10. Once I fixed my units it works out. Here is the working for Aluminium.
Which all cancels down to . Divide this by 10 to get the Aluminium resistance.

A similar calculation for Copper gives . Divide this by 15.

Then use the formula as above to get to the correct answer. Which FusionWorks can supply. However if you need the answer from me later then just shout.
11. (Original post by nerak99)
Once I fixed my units it works out. Here is the working for Aluminium.
Which all cancels down to . Divide this by 10 to get the Aluminium resistance.

A similar calculation for Copper gives . Divide this by 15.

Then use the formula as above to get to the correct answer. Which FusionWorks can supply. However if you need the answer from me later then just shout.
Seems to be correct, I don't understand why we divide by 10 & 15 rather then multiplying with it?
12. (Original post by nerak99)
Once I fixed my units it works out. Here is the working for Aluminium.
Which all cancels down to . Divide this by 10 to get the Aluminium resistance.

A similar calculation for Copper gives . Divide this by 15.

Then use the formula as above to get to the correct answer. Which FusionWorks can supply. However if you need the answer from me later then just shout.
Hi thabks for your answer! Your right the final answer is 1.1ohm, as you then have to recognise that the wires are parallel. However, im not sure why u divide by 10 and 15? I thought you had to multiply
13. I think I get it, I get the correct answer doing this aswell and made sense to me.

So I've deduced this:
14. (Original post by FusionNetworks)
Hi thabks for your answer! Your right the final answer is 1.1ohm, as you then have to recognise that the wires are parallel. However, im not sure why u divide by 10 and 15? I thought you had to multiply
In the case of parallel wires, the Resistance goes down as you have more of them. If you had two wires of 12 ohms each in parallel then the total would be 6 Ohm. This is because the wires are identical and being in parallel you get twice the current for the same potential difference. So, in the case of parallel wires if you have n of them and they are all identical, you can just divide by n for the Total Resistance.

In the case of wires that are not identical then you have to use 1/(R total) =1/R1+1/R2+1/R3 ...

Because the case for two resistors is so common we can say

and Hence (because we have a single fraction on either side of the equation)

Which is what I use here. As an exercise you can do something similar for three but it is more complex and not so useful.
I think I get it, I get the correct answer doing this aswell and made sense to me.

So I've deduced this:

Yes, the times 10 in your example is because 10 identical wires end up dividing the resistance by 10 but since you are dealing with 1/R in your formula, you multiply by 10.
16. (Original post by nerak99)
In the case of parallel wires, the Resistance goes down as you have more of them. If you had two wires of 12 ohms each in parallel then the total would be 6 Ohm. This is because the wires are identical and being in parallel you get twice the current for the same potential difference. So, in the case of parallel wires if you have n of them and they are all identical, you can just divide by n for the Total Resistance.

In the case of wires that are not identical then you have to use 1/(R total) =1/R1+1/R2+1/R3 ...

Because the case for two resistors is so common we can say

and Hence (because we have a single fraction on either side of the equation)

Which is what I use here. As an exercise you can do something similar for three but it is more complex and not so useful.
Thanks so much that's a neat trick to use!
17. (Original post by FusionNetworks)
Thanks so much that's a neat trick to use!
Thanks and I have just noticed a typo, which is important in the circumstances.
For two resistors in parallel.

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