Question

If the gauge number of a wire is increased by $ 6 $ , the diameter is halved. If a gauge number is increased by $ 1 $ , the diameter decreases by the factor of $ {2^{\dfrac{1}{6}}} $ . Knowing this and that $ 1000ft $ of $ 10 - $ gauge Copper wire has a resistance of approximately $ 1.00\Omega $ . Estimate the resistance of $ 25ft $ of $ \;22 - $ gauge Copper wire.

Answer 1

Hint:
To solve this question, we have to use the formula which relates the resistance with the length and the area of the cross section of a wire. We can obtain the resistivity of copper in terms of the given values. Then we can use this resistivity to make further calculation for the required resistance.
Formula Used: The formula used in this solution is given as,
 $ R = \rho \dfrac{l}{A} $
Here, $ R $ is the resistance of the wire, $ \rho $ is the resistivity of the wire, $ l $ is the length of the wire, and $ A $ is the area of cross-section of the wire.

Complete step by step answer:
We know that, to calculate the resistance of a wire, we need resistivity, length and Area of cross-section of the wire.
So now, if $ \rho $ is the resistivity of $ 1000ft $ long $ 10 - $ gauge wire, and $ D $ is the diameter of such a wire, then we get Resistance of the wire as,
 $ R = \rho \dfrac{l}{A} $
Here since we are already given the resistance to be $ 1.00\Omega $ , so we can get the resistivity as,
 $ \rho = R\dfrac{A}{l} $
Now putting in the required values, we get,
 $ \rho = (1.00\Omega )\dfrac{{(\pi \dfrac{{{D^2}}}{4})}}{{1000ft}} $
 $ \rho = \dfrac{{\pi {D^2}}}{{4000}} $ ………………..(1)
Now, we shall use this value of resistivity to calculate the resistance of the $ \;22 - $ gauge copper wire. But first, let us calculate the value of Diameter for the $ \;22 - $ gauge copper wire,
From inspection we can understand that
 $ 22gauge{\text{ }} = {\text{ (}}\left( {\left( {\left( {10gauge} \right) + 6} \right)gauge} \right) + 6)gauge $
According to the question, increasing the gauge number by $ 6 $ makes the diameter halved.
So, since we increased it by $ 6 $ twice, the diameter will be halved twice. Hence, the diameter will become $ \dfrac{D}{4} $ .
So, Resistance of the 25ft of $ \;22 - $ gauge wire becomes,
 $ R = \rho \dfrac{l}{A} $
 $ R = \left( {\dfrac{{\pi {D^2}}}{{4000}}} \right)\dfrac{{25}}{{\left( {\dfrac{{\pi {{\left( {\dfrac{D}{4}} \right)}^2}}}{4}} \right)}} $ (from (1))
This can be simplified and written as,
 $
  R = \left( {\dfrac{{\pi {D^2}}}{{4000}}} \right)\dfrac{{25}}{{\left( {\dfrac{{\pi {D^2}}}{{4 \times 16}}} \right)}} \\
   \Rightarrow R = \left( {\dfrac{{\pi {D^2}}}{{4000}}} \right)\dfrac{{(25 \times 4 \times 16)}}{{\left( {\pi {D^2}} \right)}} \\
 $
This gives us,
 $
  R = \dfrac{{1600}}{{4000}} \\
   \Rightarrow R = \dfrac{4}{{10}} = 0.4\Omega. \\
 $

Note:
We should always remember that for a material at a particular temperature, its resistivity doesn’t change. Thus, we could use the same value of resistivity in both the cases. However, we could not use the same technique if the temperature of the copper wire also changed.