Two resistors with temperature coefficients of $α_{1} and α_{2}$ resistance have resistance $R_{1} and R_{2}$ at $0^{0} C$ . Find the temperature coefficient of the compound resistor consisting of 2 resistors connected in parallel.

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

$\frac{α_{1} + α_{2}}{2}$

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

$\frac{2 α_{1} α_{2}}{α_{1} + α_{2}}$

answer is D.

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Detailed Solution

Initially $\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} \Rightarrow R = \frac{R_{1} R_{2}}{R_{1} + R_{2}}$
Differentiating on both sides
$- \frac{dR}{R^{2}} = - \frac{{dR}_{1}}{R_{1}^{2}} - \frac{{dR}_{2}}{R_{2}^{2}}$

$\Rightarrow \frac{dR}{R} = R [\frac{R_{1} α_{1} t}{R_{1}^{2}} + \frac{R_{2} α_{2} t}{R_{2}^{2}}]$

$\Rightarrow \frac{dR}{R} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} [\frac{R_{2} α_{1} + R_{1} α_{2}}{R_{1} R_{2}}] t$

$∴ α = \frac{R_{1} α_{2} + R_{2} α_{1}}{R_{1} + R_{2}}$

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