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)
D)
\[0.75\,\,\Omega \]
Correct Answer:
C Solution :
\[0.5\,\,\Omega \]
[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, \[r=R\left( \frac{{{l}_{1}}}{{{l}_{2}}}-1 \right)=9.5\left( \frac{3}{2.85}-1 \right)=9.5(1.05-1)\] \[=9.5\times 0.05=0.475\simeq 0.5\Omega \]
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