The coefficient of linear expansion of the ma | vedclass

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Question

(A) The resistance of a conductor is given by $R = \frac{\rho L}{A}$.Given,linear expansion coefficient is $\alpha$. Thus,length $L = L_0(1 + \alpha \

Vedclass · Accepted Answer

$\alpha_R = \alpha_\rho - \alpha$

Vedclass · Answer

(A) The resistance of a conductor is given by $R = \frac{\rho L}{A}$.Given,linear expansion coefficient is $\alpha$. Thus,length $L = L_0(1 + \alpha \Delta T)$ and area $A = A_0(1 + 2\alpha \Delta T)$.Resistivity varies as $\rho = \rho_0(1 + \alpha_\rho \Delta T)$.Substituting these into the resistance formula:$R = \frac{\rho_0(1 + \alpha_\rho \Delta T) L_0(1 + \alpha \Delta T)}{A_0(1 + 2\alpha \Delta T)}$$R = R_0(1 + \alpha_\rho \Delta T)(1 + \alpha \Delta T)(1 + 2\alpha \Delta T)^{-1}$Using binomial approximation $(1+x)^n \approx 1+nx$ for small $\Delta T$:$R \approx R_0(1 + \alpha_\rho \Delta T)(1 + \alpha \Delta T)(1 - 2\alpha \Delta T)$Neglecting higher-order terms of $\Delta T$:$R \approx R_0(1 + \alpha_\rho \Delta T + \alpha \Delta T - 2\alpha \Delta T)$$R \approx R_0(1 + (\alpha_\rho - \alpha) \Delta T)$Comparing this with $R = R_0(1 + \alpha_R \Delta T)$,we get $\alpha_R = \alpha_\rho - \alpha$.

The coefficient of linear expansion of the material of a resistor is $\alpha$. Its temperature coefficient of resistivity and resistance are $\alpha_\rho$ and $\alpha_R$ respectively. Then the correct relation is:

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