Q&A - Ask Doubts and Get Answers
Q

I have a doubt, kindly clarify. - Current Electricity - NEET-3

A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery used across the potentiometer wire , has an emf of 2.0V and a negligible internal resistance. The potentiometer wire itself is 4m long, when the resistance R, connected across the given cell, has value of

(i) infinity              (ii) 9.5\Omega

The balancing lengths', on the potentiometer  wire are found to be 3 m and 2.85m, respectivily. The value of internal resistance of the cell is

  • Option 1)

    0.25\Omega

  • Option 2)

    0.95\Omega

  • Option 3)

    0.5\Omega

  • Option 4)

    0.75\Omega

 
Answers (1)
111 Views

 

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 across wire =2V

\therefore potential/length =\frac{2}{4}=0.5\: V/m

(\phi )

{e}'=\phi l_{1} ------(1)

v=\phi\: l_{2}--------(2)

=> {e}'=l(r+R)

r=R(\frac{l_{1}}{l_{2}}-1) = 9.5 (\frac{.3}{2.85} -1)

r=0.5\Omega


Option 1)

0.25\Omega

This is incorrect option 

Option 2)

0.95\Omega

This is incorrect option 

Option 3)

0.5\Omega

This is correct option 

Option 4)

0.75\Omega

This is incorrect option 

Exams
Articles
Questions