Question

A copper wire has a resistance of \(0.500{\rm{ }}\Omega \) at \({20.0^ \circ }{\rm{ }}C\), and an iron wire has a resistance of \(0.525{\rm{ }}\Omega \) at the same temperature. At what temperature are their resistances equal?

Short answer

The resistance is equal at the temperature \(T{\rm{ }} = {\rm{ }} - {17^o}{\rm{ }}C\).

Step-by-step solution

Define Resistance

In an electrical circuit, resistance is a measure of the resistance to current flow. The Greek letter omega\({\rm{(\Omega )}}\)is used to represent resistance in ohms.

Concepts and Principles

The resistance of a conductor varies approximately linearly with temperature according to the expression given as:

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

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 given data

• Resistance of the copper wire at temperature\({T_0}{\rm{ }} = {\rm{ }}20^\circ {\rm{ }}C\)is:

\({R_{Cu,0}}{\rm{ }} = {\rm{ }}0.500{\rm{ }}\Omega \).

• Resistance of the iron wire at temperature\({T_0}{\rm{ }} = {\rm{ }}20^\circ {\rm{ }}C\)is:

\({R_{Fe,0}}{\rm{ }} = {\rm{ }}0.525{\rm{ }}\Omega \).

• The temperature coefficient of resistivity for copper is:\({\alpha _{Cu}}{\rm{ }} = {\rm{ }}3.9{\rm{ }}x{\rm{ }}{10^{ - 3}}{\rm{ }}^\circ {\rm{ }}{C^{ - 1}}\).
• The temperature coefficient of resistivity for iron is: \({\alpha _{Fe}}{\rm{ }} = {\rm{ }}5.0{\rm{ }}x{\rm{ }}{10^{ - 3}}{\rm{ }}^\circ {\rm{ }}{C^{ - 1}}\).

Evaluating the temperature at which the resistance of both wires are equal

The value of\({R_{Cu}}\)be the resistance of the copper wire at temperature\(T\).

The value of\({R_{Fe}}\)be the resistance of the iron wire at the same temperature.

Setting the value\({R_{Cu}}\)equal to the value\({R_{Fe}}\):

\({R_{Cu}}{\rm{ }} = {\rm{ }}{R_{Fe}}\)

Substituting the values\({R_{Cu}}\)and\({R_{Fe}}\)from the above equation to find the resistances of the two wires as a function of temperature change, we get:

\({R_{Cu,0}}{\rm{ }}(1 + {\alpha _{Cu}}(T - {T_0})){\rm{ }} = {\rm{ }}{R_{Fe,0}}{\rm{ }}(1 + {\alpha _{Fe}}(T - {T_0}))\)

Rearranging and solving for the value of\({\rm{T}}\)as:

\(\begin{aligned}{}{R_{Cu,0}} + {R_{Cu,0}}{\alpha _{Cu}}\left( {T - {T_0}} \right){\rm{ }} &= {\rm{ }}{R_{Fe,0}} + {R_{Fe,0}}{\alpha _{Fe}}\left( {T - {T_0}} \right)\\{R_{Cu,0}} + {R_{Cu,0}}{\alpha _{Cu}}T - {R_{Cu,0}}{\alpha _{Cu}}{T_0}{\rm{ }} &= {\rm{ }}{R_{Fe,0}} + {R_{Fe,0}}{\alpha _{Fe}}T - {R_{Fe,0}}{\alpha _{Fe}}{T_0}\\{R_{Cu,0}}{\alpha _{Cu}}T - {R_{Fe,0}}{\alpha _{Fe}}T{\rm{ }} &= {\rm{ }}{R_{Fe,0}} - {R_{Fe,0}}{\alpha _{Fe}}{T_0} - {R_{Cu,0}} + {R_{Cu,0}}{\alpha _{Cu}}{T_0}\\\left( {{R_{Cu,0}}{\alpha _{Cu}} - {R_{Fe,0}}{\alpha _{Fe}}} \right)T{\rm{ }} &= {\rm{ }}\left( {{R_{Cu,0}}{\alpha _{Cu}} - {R_{Fe,0}}{\alpha _{Fe}}} \right){T_0} + {R_{Fe,0}} - {R_{Cu,0}}\\T{\rm{ }} &= {\rm{ }}\frac{{\left( {{R_{Cu,0}}{\alpha _{Cu}} - {R_{Fe,0}}{\alpha _{Fe}}} \right){T_0} + {R_{Fe,0}} - {R_{Cu,0}}}}{{{R_{Cu,0}}{\alpha _{Cu}} - {R_{Fe,0}}{\alpha _{Fe}}}}\\T{\rm{ }} &= {\rm{ }}{T_0} + \frac{{{R_{Fe,0}} - {R_{Cu,0}}}}{{{R_{Cu,0}}{\alpha _{Cu}} - {R_{Fe,0}}{\alpha _{Fe}}}}\end{aligned}\)

Putting the numerical values and we get:

\(\begin{aligned}{}T{\rm{ }} &= {\rm{ }}20^\circ {\rm{ }}C + \frac{{0.525{\rm{ }}\Omega - 0.500{\rm{ }}\Omega }}{{(0.500{\rm{ }}\Omega )(3.9{\rm{ }}x{\rm{ }}{{10}^{ - 3}}{\rm{ }}^\circ {C^{ - 1}}) - (0.525{\rm{ }}\Omega )(5.0{\rm{ }}x{\rm{ }}{{10}^{ - 3}}{\rm{ }}^\circ {\rm{ }}{C^{ - 1}})}}\\ &= {\rm{ }} - {17^o}{\rm{ }}C\end{aligned}\)

Therefore, the resistance is equal at the temperature of \(T{\rm{ }} = {\rm{ }} - {17^o}{\rm{ }}C\).