A brass disc and a carbon disc of same radius | vedclass

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(C) The resistance of a conductor is given by $R = \frac{\rho l}{A}$.Since the discs are connected in series,the total resistance is $R = R_B + R_C$.F

Vedclass · Accepted Answer

$\left| \frac{\alpha_B \rho_C}{\alpha_C \rho_B} \right|$

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(C) The resistance of a conductor is given by $R = \frac{\rho l}{A}$.Since the discs are connected in series,the total resistance is $R = R_B + R_C$.For the resistance to be independent of temperature,the change in resistance with respect to temperature must be zero: $\frac{dR}{dT} = 0$.$\frac{dR_B}{dT} + \frac{dR_C}{dT} = 0$.Since $R = \frac{\rho l}{A}$,the change in resistance is $\Delta R = R \alpha \Delta T = \frac{\rho l}{A} \alpha \Delta T$.Thus,$\frac{\rho_B l_B}{A} \alpha_B \Delta T + \frac{\rho_C l_C}{A} \alpha_C \Delta T = 0$.Since $\alpha$ for carbon is negative and for brass is positive,we consider the magnitudes: $\frac{\rho_B l_B}{A} \alpha_B = \frac{\rho_C l_C}{A} |\alpha_C|$.Therefore,the ratio of thicknesses $\frac{l_B}{l_C} = \frac{\rho_C |\alpha_C|}{\rho_B \alpha_B}$.

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