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Q31PE
Expert-verifiedOf what material is a resistor made if its resistance is \({\bf{40}}{\bf{.0}}\;{\bf{\% }}\)greater at than at \({\bf{20}}{\bf{.0°}}\;{\bf{C}}\)?
The wire is made from the material known as Iron.
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.
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