Q&A - Ask Doubts and Get Answers
Q

Solve it, - Current Electricity - JEE Main

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)
178 Views

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.

Exams
Articles
Questions