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# A cell can be balanced against 110 cm and 100 cm of potentiometer wire, respectively with and without being short circuited through a resistance of $10\,\Omega$. Its internal resistance isA. $1\,\Omega$B. $0.5\,\Omega$C. $2\,\Omega$D. None

Last updated date: 13th Sep 2024
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Hint: Express the current in the potentiometer wire in terms of emf of the cell and internal resistance. Recall the balance condition for the potentiometer. Substitute the emf of the cell and potential across the wire of the potentiometer in the balance condition to get the value of internal resistance.

Formula used:
Balance condition, $\dfrac{E}{V} = \dfrac{{{l_1}}}{{{l_2}}}$
Here, E is the emf of the cell, V is the potential across the wire of the potentiometer, ${l_1}$ and ${l_2}$ are the points on the wire where the potentiometer is balanced.

Let a cell of emf E is connected across the length of the potentiometer wire and the internal resistance of this potentiometer is r. when we switch on the circuit, the current I flows through the wire of the potentiometer. We can express the current in the potentiometer wire as follows,
$I = \dfrac{E}{{R + r}}$
$\Rightarrow E = I\left( {R + r} \right)$ …… (1)
This is the emf of the cell.

Now, the potential difference across the length of the potentiometer wire is can be expressed using Ohm’s law as,
$V = IR$ …… (2)
Here, R is the resistance.
We know the balance condition for potentiometer,
$\dfrac{E}{V} = \dfrac{{{l_1}}}{{{l_2}}}$
Here, ${l_1}$ and ${l_2}$ are the points on the wire where the potentiometer is balanced.

Using equation (1) and (2) in the above equation, we get,
$\dfrac{{I\left( {R + r} \right)}}{{IR}} = \dfrac{{{l_1}}}{{{l_2}}}$
$\Rightarrow \dfrac{{R + r}}{R} = \dfrac{{{l_1}}}{{{l_2}}}$
$\Rightarrow 1 + \dfrac{r}{R} = \dfrac{{{l_1}}}{{{l_2}}}$
Substituting 110 cm for ${l_1}$, 100 cm for ${l_2}$ and $10\,\Omega$ for R in the above equation, we get,
$1 + \dfrac{r}{{10}} = \dfrac{{110}}{{100}}$
$\Rightarrow \dfrac{r}{{10}} = 1.1 - 1$
$\Rightarrow \dfrac{r}{{10}} = 0.1$
$\therefore r = 1\,\Omega$
Therefore, the internal resistance of the cell is $1\,\Omega$.

So, the correct answer is option A.

Note:The internal resistance of the cell is negligible and in often cases we neglect the internal resistance of the cell if the cell provides a high supply. The internal resistance is the resistance provided by the electrolytic solution in the cell. In the closed circuit while determining the emf of the cell, the Ohm’s law should be expressed such that it has resistance which is the sum of resistance of the circuit and internal resistance.