A wire of cross-section area A,length L_1,res | vedclass

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(B) Let the initial resistances of the wires be $R_1$ and $R_2$ respectively.$R_1 = \frac{\rho_1 L_1}{A}$ and $R_2 = \frac{\rho_2 L_2}{A}$.For a small

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$\rho_1 L_1 \alpha_1 + \rho_2 L_2 \alpha_2 = 0$

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(B) Let the initial resistances of the wires be $R_1$ and $R_2$ respectively.$R_1 = \frac{ho_1 L_1}{A}$ and $R_2 = \frac{ho_2 L_2}{A}$.For a small temperature change $\Delta T$,the new resistances are $R_1' = R_1(1 + \alpha_1 \Delta T)$ and $R_2' = R_2(1 + \alpha_2 \Delta T)$.The total resistance $R = R_1' + R_2'$ is independent of temperature if the change in total resistance is zero.$\Delta R = (R_1' + R_2') - (R_1 + R_2) = 0$.Substituting the expressions:$R_1(1 + \alpha_1 \Delta T) + R_2(1 + \alpha_2 \Delta T) = R_1 + R_2$.$R_1 \alpha_1 \Delta T + R_2 \alpha_2 \Delta T = 0$.Since $\Delta T 
eq 0$,we have $R_1 \alpha_1 + R_2 \alpha_2 = 0$.Substituting $R_1$ and $R_2$:$\frac{ho_1 L_1}{A} \alpha_1 + \frac{ho_2 L_2}{A} \alpha_2 = 0$.$ho_1 L_1 \alpha_1 + ho_2 L_2 \alpha_2 = 0$.

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