Two wires of resistance R1 and R2 have temperature coefficient of resistance α1 and α2, respectively. These are joined in series. The effective temperature coefficient of resistance is

Rt1=R11+α1t and Rt2=R21+α2t Also Req.=Rt1+Rt2⇒Req=R1+R2+R1α1+R2α2t⇒Req=R1+R21+R1α1+R2α2R1+R2.t So, αeff =R1α1+R2α2R1+R2

Two wires of resistance R1 and R2 have temperature coefficient of resistance α1 and α2, respectively. These are joined in series. The effective temperature coefficient of resistance is

Rt1=R11+α1t and Rt2=R21+α2t Also Req.=Rt1+Rt2⇒Req=R1+R2+R1α1+R2α2t⇒Req=R1+R21+R1α1+R2α2R1+R2.t So, αeff =R1α1+R2α2R1+R2

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Two wires of resistance R₁ and R₂ have temperature coefficient of resistance α₁ and α₂, respectively. These are joined in series. The effective temperature coefficient of resistance is

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$\frac{α_{1} + α_{2}}{2}$

$\sqrt{α_{1} α_{2}}$

$\frac{α_{1} R_{1} + α_{2} R_{2}}{R_{1} + R_{2}}$

$\frac{\sqrt{R_{1} R_{2} α_{1} α_{2}}}{\sqrt{R_{1}^{2} + R_{2}^{2}}}$

answer is C.

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Detailed Solution

$\begin{matrix} R_{t_{1}} = R_{1} (1 + α_{1} t) and R_{t_{2}} = R_{2} (1 + α_{2} t) \\ Also R_{e q .} = R_{t_{1}} + R_{t_{2}} \Rightarrow R_{e q} = R_{1} + R_{2} + (R_{1} α_{1} + R_{2} α_{2}) t \\ \Rightarrow R_{e q} = (R_{1} + R_{2}) \{1 + (\frac{R_{1} α_{1} + R_{2} α_{2}}{R_{1} + R_{2}}) . t\} \\ So, α_{eff} = \frac{R_{1} α_{1} + R_{2} α_{2}}{R_{1} + R_{2}} \end{matrix}$