Question

In the potentiometer circuit suppose, $ \in = 12V$, $l = 1{\text{m}}$. If the galvanometer current is zero when $x = 0.75{\text{m}}$
A) Find the potential difference $V$
B) If $V = emf$of the battery, find its internal resistance.

Answer 1

Hint:In this question, draw the circuit diagram and then use the definition and expression of potential gradient and internal resistance and then find the potential difference and the internal resistance of the circuit.

Complete step by step solution:
Let us consider the diagram of the potentiometer as,

In this question, the null point is given as $X = 0.75{\text{m}}$

A) We know that potential gradient of a cell is the potential drop per unit length. It is expressed as $ \Rightarrow K = \dfrac{V}{l}$
where $K$ is potential gradient, $V$ is voltage and $l$ is the length
Potential gradient,
Now, we substitute the values in the above equation as,
$ \Rightarrow K = \dfrac{{12}}{1} = 12$
Now we will calculate the potential difference as,
$ \Rightarrow V = KL$
Where, $L$ is the null point distance
Now, we will substitute the values as,
$ \Rightarrow V = 12 \times 0.75$
After calculation, we will get
$\therefore V = 9\;{\text{V}}$

Hence, the potential difference is $9V.$

B) As we know that the internal resistance of a cell is opposition to the current flow provided by the cells and batteries which results in the generation of heat in the circuit.
The expression of internal resistance in an electrical circuit is $E = I\left( {r + R} \right)$
$ \Rightarrow E = Ir + IR$
$ \Rightarrow E = Ir + V$
Where $r$ - the internal resistance of an electrical circuit, $E$ is the electromotive force, so in this case $V$ is the E. M.F of the cell.
Hence,
$E = V + Ir$
$ \Rightarrow E = E + Ir$
After calculation we get
$\therefore r = 0$

Thus, the internal resistance of the circuit is Zero.

Note:The potential gradient mainly depends on various properties such as-i) Current passing through the potentiometer wire ii) Specific resistance of the material of the potentiometer iii) Area of cross section of the potentiometer wire. Internal resistance is used to limit the current flow in a circuit to prevent the short circuit problem in an electrical circuit.