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A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. E_{o} and a resistance r_{1}. An unknown e.m.f.E is balanced at a length l of the potentiometer wire. The e.m.f. E will be given by:

  • Option 1)

    \frac{E_{o}r}{\left ( r+r_{1} \right )}.\frac{l}{L}

  • Option 2)

    \frac{E_{o}l}{L}

  • Option 3)

    \frac{LE_{o}r}{\left ( r+r_{1} \right )l}

  • Option 4)

    \frac{LE_{o}r}{lr_1}

 

Answers (1)

best_answer

 

Determine the internal resistance -

r=(frac{l_{1}-l_{2}}{l_{2}})R

r=(frac{E}{V}-1)R

frac{E}{V}=frac{l_{1}}{l_{2}}

- wherein

 

 potential difference across potentiometer wire is V=I r

\frac{E_{o}}{r+r_{1}}.r

Potential difference across length l is 

E=(\frac{V}{l})l =\frac{E_{o}rl}{(r+r_{1})L}


Option 1)

\frac{E_{o}r}{\left ( r+r_{1} \right )}.\frac{l}{L}

This is correct option 

Option 2)

\frac{E_{o}l}{L}

This is incorrect option 

Option 3)

\frac{LE_{o}r}{\left ( r+r_{1} \right )l}

This is incorrect option 

Option 4)

\frac{LE_{o}r}{lr_1}

Posted by

Aadil

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