91Ó°ÊÓ

The radius of copper wire,$\mathrm{a}=0.250\mathrm{mm}$

The outer radius of the aluminum jacket,$\mathrm{b}=0.380\mathrm{mm}$

Current in the composite wire,$\mathrm{i}=2.00\mathrm{A}$

Potential differences maintained between the ends,

The resistivity of copper,${\mathrm{ÒÏ}}_{\mathrm{C}}=1.69×{10}^{-8}\mathrm{Î\copyright }.\mathrm{m}$

The resistivity of aluminum,${\mathrm{ÒÏ}}_{\mathrm{Al}}=2.75×{10}^{-8}\mathrm{Î\copyright }.\mathrm{m}$

The copper wire and the aluminum sheath are connected in parallel, so the potential difference is the same for them. Using Ohm's law, we can get the relation between the current and resistance values of copper and aluminum. But according to the concept of resistivity of a material, it can be said that the resistance is directly proportional to the product of the resistivity, length of material and inversely proportional to the area of the material can be given. This gives us the further current relation with the resistivity and the area value of both materials for similar length values.

Formulae:

The voltage equation using Ohm’s law,$\mathrm{V}=\mathrm{IR}$ (1)

The resistance of a material,

$\mathrm{R}=\frac{\mathrm{ÒÏL}}{\mathrm{A}}$ (2)

Since the copper wire and the aluminum, jacket are placed parallel in connection; the potential across both the material is the same. Thus, we can get that voltage across the copper wire is equal to the voltage across aluminum material. That is given using equation (1) as:

${\mathrm{I}}_{\mathrm{C}}{\mathrm{R}}_{\mathrm{C}}={\mathrm{I}}_{\mathrm{Al}}{\mathrm{R}}_{\mathrm{Al}}................\left(3\right)$

Now, using equation (2) and the given data, the resistance of the copper wire can be given as follows:

${\mathrm{R}}_{\mathrm{C}}=\frac{{\mathrm{ÒÏ}}_{\mathrm{C}}\mathrm{L}}{{\mathrm{Ï€²¹}}^2}....................\left(4\right)\left(âˆ\mu \mathrm{Cross}-\mathrm{sectional}\mathrm{area},{\mathrm{A}}_{\mathrm{C}}={\mathrm{Ï€²¹}}^2\right)$

Now, using equation (2) and the given data, the resistance of the aluminum sheath can be given as follows:

${\mathrm{R}}_{\mathrm{Al}}=\frac{{\mathrm{ÒÏ}}_{\mathrm{Al}}\mathrm{L}}{\mathrm{Ï€}\left({\mathrm{b}}^2-{\mathrm{a}}^2\right)}.........\left(5\right)\left(âˆ\mu \mathrm{Cross}-\mathrm{sectional}\mathrm{area},{\mathrm{A}}_{\mathrm{Al}}=\mathrm{Ï€}\left({\mathrm{b}}^2-{\mathrm{a}}^2\right)\right)$

Again, the total current flowing through the combination can be given as:

data-custom-editor="chemistry" $\mathrm{i}={\mathrm{I}}_{\mathrm{C}}+{\mathrm{I}}_{\mathrm{Al}}.........................\left(6\right)$

Now, substituting equations (4), (5), and (6) in equation (3), the current flowing through the copper wire can be given as:

$\begin{array}{rcl}{\mathrm{I}}_{\mathrm{C}}\left(\frac{{\mathrm{ÒÏ}}_{\mathrm{C}}\mathrm{L}}{{\mathrm{Ï€²¹}}^2}\right)& =& \left(\mathrm{i}-{\mathrm{I}}_{\mathrm{c}}\right)\left(\frac{{\mathrm{ÒÏ}}_{\mathrm{Al}}\mathrm{L}}{\mathrm{Ï€}\left({\mathrm{b}}^2-{\mathrm{a}}^2\right)}\right)\\ {\mathrm{I}}_{\mathrm{C}}\left(\frac{{\mathrm{ÒÏ}}_{\mathrm{C}}\mathrm{L}}{{\mathrm{Ï€²¹}}^2}+\left(\frac{{\mathrm{ÒÏ}}_{\mathrm{Al}}\mathrm{L}}{\mathrm{Ï€}\left({\mathrm{b}}^2-{\mathrm{a}}^2\right)}\right)\right)& =& \mathrm{i}\left(\frac{{\mathrm{ÒÏ}}_{\mathrm{Al}}\mathrm{L}}{\mathrm{Ï€}\left({\mathrm{b}}^2-{\mathrm{a}}^2\right)}\right)\\ {\mathrm{I}}_{\mathrm{C}}& =& \frac{\mathrm{i}\left({\displaystyle \frac{{\mathrm{ÒÏ}}_{\mathrm{Al}}\mathrm{L}}{\mathrm{Ï€}\left({\mathrm{b}}^2-{\mathrm{a}}^2\right)}}\right)}{\left({\displaystyle \frac{{\mathrm{ÒÏ}}_{\mathrm{C}}\mathrm{L}}{{\mathrm{Ï€²¹}}^2}}+\left({\displaystyle \frac{{\mathrm{ÒÏ}}_{\mathrm{Al}}\mathrm{L}}{\mathrm{Ï€}\left({\mathrm{b}}^2-{\mathrm{a}}^2\right)}}\right)\right)}\\ {\mathrm{I}}_{\mathrm{C}}& =& \frac{\mathrm{i}\left({\displaystyle \frac{{\mathrm{ÒÏ}}_{\mathrm{Al}}\mathrm{L}}{\mathrm{Ï€}\left({\mathrm{b}}^2-{\mathrm{a}}^2\right)}}\right)}{\left({\displaystyle \frac{{\mathrm{ÒÏ}}_{\mathrm{C}}\mathrm{L}\left(\mathrm{Ï€}\left({\mathrm{b}}^2-{\mathrm{a}}^2\right)\right)+{\mathrm{ÒÏ}}_{\mathrm{Al}}\mathrm{L}\left({\mathrm{Ï€²¹}}^2\right)}{\left({\mathrm{Ï€²¹}}^2\right)\left(\mathrm{Ï€}\left({\mathrm{b}}^2-{\mathrm{a}}^2\right)\right)}}\right)}\\ {\mathrm{I}}_{\mathrm{C}}& =& \frac{{\mathrm{iÒÏ}}_{\mathrm{Al}}\mathrm{L}\left({\mathrm{a}}^2\right)}{{\mathrm{ÒÏ}}_{\mathrm{C}}\mathrm{L}\left({\mathrm{b}}^2-{\mathrm{a}}^2\right)+{\mathrm{ÒÏ}}_{\mathrm{Al}}\mathrm{L}\left({\mathrm{a}}^2\right)}\\ {\mathrm{I}}_{\mathrm{C}}& =& \frac{{\mathrm{iÒÏ}}_{\mathrm{Al}}\left({\mathrm{a}}^2\right)}{{\mathrm{ÒÏ}}_{\mathrm{C}}\left({\mathrm{b}}^2-{\mathrm{a}}^2\right)+{\mathrm{ÒÏ}}_{\mathrm{Al}}\left({\mathrm{a}}^2\right)}\end{array}$

The value of the current across the copper wire can be given using the given data in the above formula:

$\begin{array}{rcl}{\mathrm{I}}_{\mathrm{C}}& =& \frac{\left(2\mathrm{A}\right)\left(2.75×{10}^{-8}\mathrm{Î\copyright }.\mathrm{m}\right){\left(0.250×{10}^{-3}\mathrm{m}\right)}^2}{\left(1.69×{10}^{-8}\mathrm{Î\copyright }.\mathrm{m}\right)\left[{\left(0.380×{10}^{-3}\mathrm{m}\right)}^2-\left(0.250×{10}^{-3}{\mathrm{m}}^2\right)\right]+\left(2.75×{10}^{-8}\mathrm{Î\copyright }.\mathrm{m}\right){\left(0.250×{10}^{-3}\mathrm{m}\right)}^2}\\ & =& \frac{0.344×{10}^{-14}\mathrm{A}.\mathrm{Î\copyright }.{\mathrm{m}}^3}{3.10×{10}^{-15}\mathrm{Î\copyright }.{\mathrm{m}}^3}\\ & =& 1.11\mathrm{A}\end{array}$

Hence, the value of the current is 1.11 A.

Now, the current through the aluminum sheath can be given using the above current value of copper in equation (5) as follows:

$\begin{array}{rcl}{\mathrm{I}}_{\mathrm{Al}}& =& \mathrm{I}-{\mathrm{I}}_{\mathrm{C}}\\ & =& 2.00\mathrm{A}-1.11\mathrm{A}\\ & =& 0.89\mathrm{A}\end{array}$

Hence, the current value is 0.89 A.

The length of the composite wire can be calculated using equation (4) in equation (1) as follows:

$\begin{array}{rcl}\mathrm{V}& =& {\mathrm{I}}_{\mathrm{C}}\left(\frac{{\mathrm{ÒÏ}}_{\mathrm{C}}\mathrm{L}}{{\mathrm{Ï€²¹}}^2}\right)\\ 12.0\mathrm{V}& =& \left(1.11\mathrm{A}\right)\left(\frac{\left(1.69×{10}^{-8}\mathrm{Î\copyright }.\mathrm{m}\right)\mathrm{L}}{\mathrm{Ï€}{\left(0.250×{10}^{-3}\mathrm{m}\right)}^2}\right)\\ 12.0\mathrm{V}& =& \left(9.56×{10}^{-2}\mathrm{A}\right)\mathrm{L}\\ \mathrm{L}& =& \frac{12.0}{\left(3.90×{10}^{-2}\right)}\mathrm{m}\\ \mathrm{L}& =& 126\mathrm{m}\end{array}$

Hence, the value of the length is 126 m.