91Ó°ÊÓ

In an electrical circuit, the flow of current varies with the variation in resistance

The resistance of a conductor can be given as:

\(R = {R_0}\left( {1 + \alpha \left( {T - {T_0}} \right)} \right)\)

The value of \({R_0}\) is the resistance at some reference temperature, and the value of \({T_0}\) and the value of \(\alpha \) is the temperature coefficient of resistivity.

The initial temperature of the wire is:\({T_0} = 20^\circ \;{\rm{C}}\).

The resistance of the wire is forty percent greater at \(T = {100^\circ }\;{\rm{C}}\) than at \({T_0}\).

The resistance of the wire is expressed as a function of temperature change by the equation as:

\(R = {R_0}\left( {1 + \alpha \left( {T - {T_0}} \right)} \right)\) (1)

As the resistance of the wire is forty percent at temperature\(T\)than at temperature\({T_0}\), the resistance of the wire will be:

\(\begin{align}R{\rm{ }} &= {\rm{ }}{R_0} + 40\;\% \left( {{R_0}{\rm{ }}} \right)\\ &= {\rm{ }}1.40{R_0}\end{align}\)

Substitute the values in equation 1, and we get:

\(\begin{align}1.40{R_0}{\rm{ }} &= {\rm{ }}{R_0}\left( {1 + \alpha \left( {T - {T_0}} \right)} \right)\\1.40{\rm{ }} &= {\rm{ }}1 + \alpha \left( {T - {T_0}} \right)\end{align}\)

Rearranging and solving for\(\alpha \):

\(\begin{align}1.40 - 1{\rm{ }} &= {\rm{ }}\alpha \left( {T - {T_0}} \right)\\0.40{\rm{ }} &= {\rm{ }}\alpha \left( {T - {T_0}} \right)\\\alpha {\rm{ }} &= {\rm{ }}\frac{{0.40}}{{T - {T_0}}}\end{align}\)

Entering the values for\(T\)and\({T_0}\), we obtain:

\(\begin{align}\alpha {\rm{ }} &= {\rm{ }}\frac{{0.40}}{{{{100}^\circ }\;{\rm{C }} - {{20}^\circ }\;{\rm{C}}}}\\ &= {\rm{ }}5.00 \times {10^{ - 3}}^\circ \;{{\rm{C}}^{ - 1}}\end{align}\)

We do observe that this coefficient of resistivity is related to iron.

Therefore, the wire is made of Iron.