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Two conductors have the same resistance at 0°C but their temperature coefficients of resistance are $\alpha _{1}\: and \: \alpha _{2}$ . The respective temperature coefficients of their series and parallel combinations are nearly

Option 1)
$\frac{\alpha _{1}+\alpha _{2}}{2},\frac{\alpha _{1}+\alpha _{2}}{2}$

Option 2)
$\frac{\alpha _{1}+\alpha _{2}}{2},\alpha _{1}+\alpha _{2}$

Option 3)
$\alpha _{1}+\alpha _{2},\frac{\alpha _{1}+\alpha _{2}}{2},$

Option 4)
$\alpha _{1}+\alpha _{2},\frac{\alpha _{1}\alpha _{2}}{\alpha _{1}+\alpha _{2}}$

Answers (1)

best_answer

As we learnt in

Temperature co-efficient of Resistivity -

$\alpha=\frac{R_{T}-R_{o}}{R_{o}(T-T_{o})}$

- wherein

Where the value of $\alpha$ is different at different temperatures

Series Grouping -

Potential - Different

Current - Same

- wherein

Parallel Grouping -

Potential - Same

Current - Different

- wherein

$R_{1}= R_{^{\circ}}\left ( 1+\alpha _{1}\Delta t \right )$

$R_{2}= R_{^{\circ}}\left ( 1+\alpha _{2}\Delta t \right )$

in series $R_{1} + R_{2}$

$R=R_{o}[2+(\alpha_{1}+\alpha_{2})\Delta t]$

$2R_{o}[1+(\frac{\alpha_{1}+\alpha_{2}}{2})\Delta t]$

$\therefore \alpha _{eq}= \frac{\alpha_{1}+\alpha_{2}}{2}$

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

$\frac{1}{R}=\frac{1}{R_{o(1+\alpha_{1}\Delta t)}}+ \frac{1}{R_{o}(1+\alpha_{2}\Delta t)}$

$\alpha _{eq}=\frac{\alpha_{1}+\alpha_{2}}{2}$

Option 1)

$\frac{\alpha _{1}+\alpha _{2}}{2},\frac{\alpha _{1}+\alpha _{2}}{2}$

This option is correct.

Option 2)

$\frac{\alpha _{1}+\alpha _{2}}{2},\alpha _{1}+\alpha _{2}$

This option is incorrect.

Option 3)

$\alpha _{1}+\alpha _{2},\frac{\alpha _{1}+\alpha _{2}}{2},$

This option is incorrect.

Option 4)

$\alpha _{1}+\alpha _{2},\frac{\alpha _{1}\alpha _{2}}{\alpha _{1}+\alpha _{2}}$

This option is incorrect.

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