question_answer

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 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance, R, connected across the given cell, has values of [NEET 2014]

(i) infinity                       (ii) \[9.5\,\,\Omega \]

the ‘balancing lengths’, on the potentiometer wire are found to be 3 m and 2.85 m, respectively.

The value of internal resistance of the cell is

A)  \[0.25\,\,\Omega \]        

B)       \[0.95\,\,\Omega \]

C)

\[0.5\,\,\Omega \]           D)       \[0.75\,\,\Omega \] Correct Answer: C Solution :

[c] Given, \[e=2V\] and \[l=4m\]

Potential drop per unit length

\[\phi =\frac{e}{l}=\frac{2}{4}=0.5V/m\]

For the first case, \[\Rightarrow \]\[e'=\phi {{l}_{1}}\]                     (i)

(\[e'\to \]emf of the given cell)

For the second case,

\[V=\phi {{l}_{2}}\]                                          (ii)

From Eqs. (i) and (ii),

\[e'/V={{l}_{1}}/{{l}_{2}}\]

\[e'=l(r+R)\] and \[V=lR\] for the second case

So,  \[r=R\left( \frac{{{l}_{1}}}{{{l}_{2}}}-1 \right)=9.5\left( \frac{3}{2.85}-1 \right)=9.5(1.05-1)\]

\[=9.5\times 0.05=0.475\simeq 0.5\Omega \]