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  • How to solve this problem- - A potentiometer wire AB having length Land resistance 12r is joined to a cell Dof emf and internal - Current Electricity - JEE Main

A potentiometer wire AB having length Land resistance 12r is joined to a cell Dof emf \varepsilon and internal resistance r. A cell Chaving emf \frac{\varepsilon }{2} and internal resistance 3r is connected. The length AJ is connected with the galvanometer as shown in the fig. shows no deflection is:

  • Option 1)

    \frac{11}{12}L

  • Option 2)

    \frac{11}{24 }L

  • Option 3)

    \frac{13}{24}L

  • Option 4)

    \frac{5}{12}L

Answers (1)

best_answer

 

Potential gradient -

V=iR= \left (\frac{e}{R+R_{n}+r} \right )R

x=\frac{V}{L}=\frac{e}{(R+R_{n}+r)}\frac{R}{L}

- wherein

r- internal resistance

AB=L  , AJ=x

So resistance of AB=12r

R resistance of AJ  

 

AJ = \frac{x}{L}12r = R_{AJ}\\\\If \: \: no\: \: deflection \\\\then \: \: i =\: \frac{\epsilon }{r+12r}\: = \frac{\epsilon }{13r}\\\\and \: \: \Delta V_{AJ}=\frac{\epsilon }{2}\\\\\Delta T_{AJ}=\frac{\epsilon }{r}=i\times R_{AJ}\\\\\Delta V_{AJ}=\frac{\epsilon }{r}\: =\: \left ( \frac{\epsilon }{r+12r}\right )\frac{x}{L}\: 12r\\\\\Rightarrow x=\frac{13L}{24}

 


Option 1)

\frac{11}{12}L

Option 2)

\frac{11}{24 }L

Option 3)

\frac{13}{24}L

Option 4)

\frac{5}{12}L

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