Question

In an experiment with a potentiometer to measure the internal resistance of a cell, when the cell in the secondary circuit is shunted by \[5\,\Omega \] , the null point is \[220\,{\text{cm}}\] . When the cell is shunted by \[20\,\Omega \] resistance the null point is \[300\,{\text{cm}}\] , find the internal resistance of the cell.

Answer 1

Hint:First of all, we will find a formula which gives the internal resistance of the cell, which relates balancing lengths. Using the two conditions given, we will find two equations. We will equate both the equations and find the balance length when the cell is open circuited and thereby find the internal resistance by putting the required found out values.

Complete step by step solution:
In the given problem, we are supplied with the following data:
There is an experimental set up with a potentiometer to measure the internal resistance of a cell.If the cell in the secondary circuit is shunted by \[5\,\Omega \],the null point is \[220\,{\text{cm}}\].If the cell in the secondary circuit is shunted by \[20\,\Omega\] , the null point is \[300\,{\text{cm}}\].We are asked to find the internal resistance of the cell.
To begin with, we know that a potentiometer is a device which is used to measure the internal resistance of a cell. During the null point, there is no current flow because of zero potential difference.We have a formula which gives the internal resistance of a cell, which is shown below:
\[r = R\dfrac{{\left( {E - V} \right)}}{V}\] …… (1)
Where,
\[r\] indicates internal resistance of the cell.
\[R\] indicates an external resistance.
\[V\] indicates potential difference across the connected cell but less than \[E\] .
\[E\] indicates the emf of the cell.
Again, we know:
\[E = k{l_1}\] and
\[V = k{l_2}\]
Where,
\[k\] indicates the potential gradient along the wire.
We can now modify the equation (1) as follows:
$r = R\dfrac{{\left( {E - V} \right)}}{V} \\
\Rightarrow r = R \times \dfrac{{k{l_1} - k{l_2}}}{{k{l_2}}} \\$
\[ \Rightarrow r = R \times \dfrac{{{l_1} - {l_2}}}{{{l_2}}}\] …… (2)
Where,
\[{l_2}\] indicates the balancing point.
We know,
\[100\,{\text{cm}} = 1\,{\text{m}}\]
So, \[220\,{\text{cm}} = 2.2\,{\text{m}}\] and \[300\,{\text{cm}} = 3\,{\text{m}}\]
If the cell in the secondary circuit is shunted by \[5\,\Omega \],the null point is \[220\,{\text{cm}}\].So, we have, from equation (1):
\[ \Rightarrow r = 5 \times \dfrac{{{l_1} - 2.2}}{{2.2}}\] …… (3)
If the cell in the secondary circuit is shunted by \[20\,\Omega \],the null point is \[300\,{\text{cm}}\].So, we have from equation (1):
\[ \Rightarrow r = 20 \times \dfrac{{{l_1} - 3}}{3}\] …… (4)
Now, we equate equations (3) and (4), and we get:
$5 \times \dfrac{{{l_1} - 2.2}}{{2.2}} = 20 \times \dfrac{{{l_1} - 3}}{3} \\
\Rightarrow \dfrac{{{l_1} - 2.2}}{{{l_1} - 3}} = \dfrac{{20}}{3} \times \dfrac{{2.2}}{5} \\
\Rightarrow \dfrac{{{l_1} - 2.2}}{{{l_1} - 3}} = 2.93 \\
\Rightarrow {l_1} = 3.41\,{\text{m}}$
So, the balancing length when the cell is open circuited with the resistance box is found to be \[3.41\,{\text{m}}\] .
Now, we again use the equation (3) to find the internal resistance of the cell:
$r = 5 \times \dfrac{{{l_1} - 2.2}}{{2.2}} \\
\Rightarrow r = 5 \times \dfrac{{3.41 - 2.2}}{{2.2}} \\
\therefore r = 2.75\,\Omega$

Hence the internal resistance of the cell is found out to be \[2.75\,\Omega \].

Note: While solving this problem, remember that in case of null point there is no current flow due to zero potential difference. Potentiometer also acts as a variable resistor or rheostat. The wire that is to be used should have a uniform area of cross section.