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    Hello, I'm stuck on these questions

    Q1) What length of copper wire of diameter 0.38mm would be need to limit the current in it to 0.5A when connected to a 2V battery?

    A=\pi (0.19 \times10^{-3})^2

    A=1.13 \times 10^{-7}

    V=IR

    2=0.5R

    R=4

    R=\frac{pL}{A}

    4=\frac{1.7\times10^{-8}\times L}{1.13\times 10^{-7}}

    \frac{4(1.13 \times10^{-7})}{1.7 \times10^{-8}}=L, L=26.7

    Is that correct?

    Q2)A 0.28mm diameter copper wire, 15.0m long, is connected to a 3.0V battery. Find its resistance and the current flowing.

    A=\pi(0.14 \times10^{-3})^2

    A=6.18 \times 10^{-8}

    ...What do I do next to answer the question?

    Thanks a lot.





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    (Original post by student473)
    Hello, I'm stuck on these questions

    Q1) What length of copper wire of diameter 0.38mm would be need to limit the current in it to 0.5A when connected to a 2V battery?

    A=\pi (0.19 \times10^{-3})^2

    A=1.13 \times 10^{-7}

    V=IR

    2=0.5R

    R=4

    R=\frac{pL}{A}

    4=\frac{1.7\times10^{-8}\times L}{1.13\times 10^{-7}}

    \frac{4(1.13 \times10^{-7})}{1.7 \times10^{-8}}=L, L=26.7

    Is that correct?
    Yes, but don't forget the units!

    Q2)A 0.28mm diameter copper wire, 15.0m long, is connected to a 3.0V battery. Find its resistance and the current flowing.

    A=\pi(0.14 \times10^{-3})^2

    A=6.18 \times 10^{-8}

    ...What do I do next to answer the question?

    Thanks a lot.
    Use R=ρL/A to find resistance (as you did in the other question)
    then Ohm's Law to find current.


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    (Original post by Stonebridge)
    Yes, but don't forget the units!

    Use R=ρL/A to find resistance (as you did in the other question)
    then Ohm's Law to find current.

    [/QUOTE]

    Stupid question but how do I use R=ρL/A when I only have 2 values to input? I use V=IR but I again only have 1 value, which is voltage.
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    You have L and A (via the diameter) given in the question, and use the same value for ρ as in the other part. This gives you R to use with V to find the current.
 
 
 
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