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.
I get a tiny resistance using R=Aρl of π0.5×10−5 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 Rtot=Ra+RcRal×Rcu
Once I fixed my units it works out. Here is the working for Aluminium. R=Aρl=π×(2×10−3)22.5×10−8×20×103 Which all cancels down to π125. Divide this by 10 to get the Aluminium resistance.
A similar calculation for Copper gives π75. 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.
Once I fixed my units it works out. Here is the working for Aluminium. R=Aρl=π×(2×10−3)22.5×10−8×20×103 Which all cancels down to π125. Divide this by 10 to get the Aluminium resistance.
A similar calculation for Copper gives π75. 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?
Once I fixed my units it works out. Here is the working for Aluminium. R=Aρl=π×(2×10−3)22.5×10−8×20×103 Which all cancels down to π125. Divide this by 10 to get the Aluminium resistance.
A similar calculation for Copper gives π75. 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
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 Rt1=R11+R21=R1R2R2+R1
and Hence (because we have a single fraction on either side of the equation)
Rt=R1+R2R1+R1
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.
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 Rt1=R11+R21=R1R2R2+R1
and Hence (because we have a single fraction on either side of the equation)
Rt=R1+R2R1+R1
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.