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(A) Let $R_1 = 100 \, \Omega$ be the resistance of the tungsten wire and $\alpha_1 = 4.5 	imes 10^{-3} \, ^{\circ}C^{-1}$ be its temperature coeffici

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) Let $R_1 = 100 \, \Omega$ be the resistance of the tungsten wire and $\alpha_1 = 4.5 	imes 10^{-3} \, ^{\circ}C^{-1}$ be its temperature coefficient.Let $R_2 = R$ be the resistance of the germanium wire and $\alpha_2 = -5 	imes 10^{-2} \, ^{\circ}C^{-1}$ be its temperature coefficient.The total resistance of the series combination is $R_{eq} = R_1 + R_2$.For the total resistance to be independent of temperature,the change in total resistance with respect to temperature must be zero,i.e.,$\frac{dR_{eq}}{dT} = 0$.Since $R_{eq} = R_1(1 + \alpha_1 \Delta T) + R_2(1 + \alpha_2 \Delta T)$,the condition for the resistance to remain constant is $R_1 \alpha_1 + R_2 \alpha_2 = 0$.Substituting the given values:$(100)(4.5 	imes 10^{-3}) + R(-5 	imes 10^{-2}) = 0$$0.45 = R(5 	imes 10^{-2})$$R = \frac{0.45}{0.05} = 9 \, \Omega$.

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