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Two metal wires of identical dimension are connected in series. If \sigma _{1} and \sigma _{2} are the conductivities of the metal wires respectively, the effective conductivity of the combination is:

  • Option 1)

    \frac{\sigma _{1}+\sigma _{2}}{2\sigma _{1}\sigma _{2}}

  • Option 2)

    \frac{\sigma _{1}+\sigma _{2}}{\sigma _{1}\sigma _{2}}

  • Option 3)

    \frac{\sigma _{1}\sigma _{2}}{\sigma _{1}+\sigma _{2}}

  • Option 4)

    \frac{2\sigma _{1}\sigma _{2}}{\sigma _{1}+\sigma _{2}}

 

Answers (1)

 

In series Grouping -

R_{eq}=R_{1}+R_{2}+R_{3}+cdot cdot cdot +R_{n}

- wherein

R_{eq}- Equivalent Resistance

 

 and,

 

Resistance formula -

R=
ho frac{l}{A}=frac{m}{ne^{2}	au}cdot frac{l}{A}

- wherein


ho = resistivity of material

n= Number of free electrons per unit volume.

 

 

R=R_{1}+R_{2}

R=\frac{L}{\sigma A}

\Rightarrow \frac{1}{\sigma_{eq}}\:.\:\frac{2L}{A}=\frac{.L}{\sigma_{1}A}+\frac{L}{\sigma_{2}A}

\Rightarrow \frac{2}{\sigma_{eq}}=\frac{1}{\sigma_{1}}+\frac{1}{\sigma_{2}}

\Rightarrow \sigma_{eq}=\frac{2\sigma_{1}\sigma_{2}}{\sigma_{1}+\sigma_{2}}


Option 1)

\frac{\sigma _{1}+\sigma _{2}}{2\sigma _{1}\sigma _{2}}

This option is incorrect.

Option 2)

\frac{\sigma _{1}+\sigma _{2}}{\sigma _{1}\sigma _{2}}

This option is incorrect.

Option 3)

\frac{\sigma _{1}\sigma _{2}}{\sigma _{1}+\sigma _{2}}

This option is incorrect.

Option 4)

\frac{2\sigma _{1}\sigma _{2}}{\sigma _{1}+\sigma _{2}}

This option is correct.

Posted by

subam

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