Resistivity Isaac physics

I get a tiny resistance using $R=\frac{\rho l}{A}$ of $\frac {0.5 \times 10^{-5}}{\pi}$ Ohm

Website says your answer is wrong!!

What is the website answer? I will work agaain I forgot to square radii and alos made a unit erro

https://isaacphysics.org/questions/ch_c_p1?board=physicsskills_book_ch_c1
Scroll to the bottom and click c1.12

https://isaacphysics.org/questions/ch_c_p1?board=physicsskills_book_ch_c1
Scroll to the bottom and click c1.12

Once I fixed my units it works out. Here is the working for Aluminium.
$R=\frac{\rho l}{A}=\frac{2.5 \times 10^{-8}\times 20 \times 10^3}{\pi \times (2 \times 10^{-3})^2}$ Which all cancels down to $\frac{125}{\pi}$. Divide this by 10 to get the Aluminium resistance.

A similar calculation for Copper gives $\frac{75}{\pi}$. 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.
$R=\frac{\rho l}{A}=\frac{2.5 \times 10^{-8}\times 20 \times 10^3}{\pi \times (2 \times 10^{-3})^2}$ Which all cancels down to $\frac{125}{\pi}$. Divide this by 10 to get the Aluminium resistance.

A similar calculation for Copper gives $\frac{75}{\pi}$. 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

I think I get it, I get the correct answer doing this aswell and made sense to me.

So I've deduced this:

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
$\frac{1}{R_t}=\frac{1}{R_1}+ \frac{1}{R_2} =\frac{R_2+R_1}{R_1R_2}$

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

${R_t}=\frac{R_1+R_1}{R_1+R_2}$

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!