Answer:
The resistance [tex]R[/tex] of a wire is calculated by the following formula:
[tex]R=\rho\frac{l}{s}[/tex] (1)
Where:
[tex]\rho[/tex] is the resistivity of the material the wire is made of. In this case we are talking about copper, and its resistivity at [tex]0\ºC[/tex] is:
[tex]\rho=1.72(10)^{-8}m\Omega[/tex]
[tex]l[/tex] is the length of the wire, which in this case is [tex]1000ft[/tex], but we will make the conversion to meters, in order to work with the same units:
[tex]l=1000ft\frac{0.3048m}{1ft}=304.8m[/tex]
[tex]s[/tex] is the transversal area of the wire. In this case is a circumference, so we will use the formula of the area of the circumference:
[tex]s=\pi{r}^{2}[/tex]
Now, in this problem we have two transversal areas:
[tex]s_{1}=10380cm=103.8{m}^{2}[/tex]
[tex]s_{2}=5178cm=51.78{m}^{2}[/tex]
And two resistances:
[tex]R_{1}=0.9989\Omega[/tex] and [tex]R_{2}[/tex] which is the one we are asked to find.
Taking into account we are working with the same material (copper), the [tex]\rho[/tex] will be the same. In addition, the length [tex]l[/tex] of the wire is the same, the only thing that changes is the transversal area.
According to this given data, we have a system of two equations, as follows:
[tex]R_{1}=\rho\frac{l}{s_{1}}[/tex] (2)
[tex]R_{2}=\rho\frac{l}{s_{2}}[/tex] (3)
We know all the values, except [tex]R_{2}[/tex].
At this point of the problem, we will approach it in the following way:
From (2) let’s isolate [tex]l\rho[/tex]:
[tex]l\rho=R_{1}S_{1}[/tex] (4)
From (3) let’s also isolate [tex]l\rho[/tex]:
[tex]l\rho=R_{2}S_{2}[/tex] (5)
If equation (4)=equation (5)
[tex]R_{1}S_{1}=R_{2}S_{2} [/tex] (6)
Now, we have to find the value of [tex]R_{2}[/tex]:
[tex]R_{2}=\frac{R_{1}S_{1}}{S_{2}}[/tex] (7)
Finally:
[tex]R_{2}=2.002\Omega[/tex]
Rounding:
[tex]R_{2}=2\Omega[/tex]>>>>>This is the result
It is important to note that as the resistance of the wire is inversely proportional to its cross-sectional area, when the area decreases the resistance increases.
