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Two wires of temperature coefficient of resistances $α_{1} a n d α_{2}$ and resistances $R_{1}, R_{2}$ are connected in parallel, effective temperature coefficient of resistance is

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$α_{e f f} = \frac{α_{1} R_{1} + α_{2} R_{2}}{R_{1} + R_{2}}$

$α_{e f f} = \frac{α_{1} R_{2} + α_{2} R_{1}}{R_{1} + R_{2}}$

$α_{e f f} = \frac{α_{1} + α_{2}}{2}$

$α_{e f f} = \frac{2 α_{1} α_{2}}{α_{1} + α_{2}}$

answer is B.

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Detailed Solution

$\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$

$\frac{Δ R}{R^{2}} = \frac{Δ R_{1}}{R_{1}^{2}} + \frac{Δ R_{2}}{R_{2}^{2}}$

$\frac{1}{R} \frac{Δ R}{R} = \frac{1}{R_{1}} \frac{Δ R_{1}}{R_{1}} + \frac{1}{R_{2}} \frac{Δ R_{2}}{R_{2}}$

$\frac{1}{R} α_{e f f} Δ t = \frac{1}{R_{1}} α_{1} Δ t + \frac{1}{R_{2}} α_{2} Δ t$

$\frac{α_{e f f} (R_{1} + R_{2})}{R_{1} R_{2}} = \frac{x_{1}}{R_{1}} + \frac{x_{2}}{R_{2}}$

$α_{e f f} = \frac{α_{1} R_{2} + α_{2} R_{1}}{R_{1} + R_{2}}$

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