
The resistance of a wire at temperatures \[{t^ \circ }C\] and \[{0^ \circ }C\] are related by
A. ${R_t} = {R_0}\left( {1 + \alpha t} \right)$
B. ${R_t} = {R_0}\left( {1 - \alpha t} \right)$
C. ${R_t} = {R_0}^2\left( {1 + \alpha t} \right)$
D. ${R_t} = {R_0}^2\left( {1 - \alpha t} \right)$
Answer
161.1k+ views
Hint: To get an expression relating the resistance at two different temperatures we use the relationship between resistance and temperature. The change in resistance with temperature is directly proportional to the resistance at that temperature and the constant of proportionality obtained is known as temperature coefficient of resistance.
Complete step by step solution:
The resistance is related to temperature as the rate of change of resistance with change in temperature is directly proportional to the resistance at that temperature and the proportionality constant is known as the temperature coefficient of resistance denoted by the symbol $\alpha $. That is,
$\dfrac{{dR}}{{dT}} = \alpha R$
If the change in resistance with temperature is small then this relation can be written as,
$\dfrac{{\vartriangle R}}{{\vartriangle T}} = \alpha R$
Given two temperatures are ${0^ \circ }C\,\& \,{t^ \circ }C$. Here, the change in temperature, that is, $\vartriangle T$ is ${\left( {t - 0} \right)^ \circ }C$. So the equation becomes,
$\vartriangle R = \alpha {R_0}\left( {t - 0} \right) \\ $
That is,
$\vartriangle R = \alpha {R_0}t \\ $ ...(1)
Also,
$\vartriangle R = {R_t} - {R_0} \\ $
This implies that,
${R_t} = {R_0} + \vartriangle R \\ $ ...(2)
Substituting equation (1) in equation (2), we get,
\[{R_t} = {R_0} + \alpha {R_0}t \\ \]
Taking ${R_0}$ common on right hand side of the equation, we get,
\[{R_t} = {R_0}\left( {1 + \alpha t} \right)\]
Hence, option A is the correct answer.
Note: Consider a basic model of resistance to better understand the temperature dependency. Atoms and molecules in a conductor obstruct the flow of electrons. It becomes more difficult for the electrons to pass by as these atoms and molecules bounce about more. As a result, resistance typically rises with temperature.
Complete step by step solution:
The resistance is related to temperature as the rate of change of resistance with change in temperature is directly proportional to the resistance at that temperature and the proportionality constant is known as the temperature coefficient of resistance denoted by the symbol $\alpha $. That is,
$\dfrac{{dR}}{{dT}} = \alpha R$
If the change in resistance with temperature is small then this relation can be written as,
$\dfrac{{\vartriangle R}}{{\vartriangle T}} = \alpha R$
Given two temperatures are ${0^ \circ }C\,\& \,{t^ \circ }C$. Here, the change in temperature, that is, $\vartriangle T$ is ${\left( {t - 0} \right)^ \circ }C$. So the equation becomes,
$\vartriangle R = \alpha {R_0}\left( {t - 0} \right) \\ $
That is,
$\vartriangle R = \alpha {R_0}t \\ $ ...(1)
Also,
$\vartriangle R = {R_t} - {R_0} \\ $
This implies that,
${R_t} = {R_0} + \vartriangle R \\ $ ...(2)
Substituting equation (1) in equation (2), we get,
\[{R_t} = {R_0} + \alpha {R_0}t \\ \]
Taking ${R_0}$ common on right hand side of the equation, we get,
\[{R_t} = {R_0}\left( {1 + \alpha t} \right)\]
Hence, option A is the correct answer.
Note: Consider a basic model of resistance to better understand the temperature dependency. Atoms and molecules in a conductor obstruct the flow of electrons. It becomes more difficult for the electrons to pass by as these atoms and molecules bounce about more. As a result, resistance typically rises with temperature.
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