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(B) Let the resistance of each conductor at $0^{\circ} C$ be $R$.For series combination:$R_{eq} = R_1 + R_2$$2R(1 + \alpha_{eq} \Delta 	heta) = R(1 +

Vedclass · Accepted Answer

$\frac{\alpha_1+\alpha_2}{2}, \frac{\alpha_1+\alpha_2}{2}$

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(B) Let the resistance of each conductor at $0^{\circ} C$ be $R$.For series combination:$R_{eq} = R_1 + R_2$$2R(1 + \alpha_{eq} \Delta 	heta) = R(1 + \alpha_1 \Delta 	heta) + R(1 + \alpha_2 \Delta 	heta)$$2 + 2\alpha_{eq} \Delta 	heta = 2 + (\alpha_1 + \alpha_2) \Delta 	heta$$\alpha_{eq} = \frac{\alpha_1 + \alpha_2}{2}$For parallel combination:$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$$\frac{1}{\frac{R}{2}(1 + \alpha_{eq} \Delta 	heta)} = \frac{1}{R(1 + \alpha_1 \Delta 	heta)} + \frac{1}{R(1 + \alpha_2 \Delta 	heta)}$$\frac{2}{1 + \alpha_{eq} \Delta 	heta} = \frac{1}{1 + \alpha_1 \Delta 	heta} + \frac{1}{1 + \alpha_2 \Delta 	heta}$Using the binomial approximation $(1+x)^{-1} \approx 1-x$ for small $\Delta 	heta$:$2(1 - \alpha_{eq} \Delta 	heta) = (1 - \alpha_1 \Delta 	heta) + (1 - \alpha_2 \Delta 	heta)$$2 - 2\alpha_{eq} \Delta 	heta = 2 - (\alpha_1 + \alpha_2) \Delta 	heta$$\alpha_{eq} = \frac{\alpha_1 + \alpha_2}{2}$

Two conductors have the same resistances at $0^{\circ} C$ but their temperature coefficients of resistance are $\alpha_1$ and $\alpha_2$. The respective temperature coefficients for their series and parallel combinations are :

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