Question

In a potentiometer experiment, it has been mentioned that no current will be passing through the galvanometer when the terminals of the cell are connected across $52cm$ of the potentiometer wire. When the cell has been shunted by a resistance of $5\Omega $, a balance will be found if the cell is connected across $40cm$ of the wire. What will be the internal resistance of this battery?
$\begin{align}
  & A.0.5\Omega \\
 & B.0.6\Omega \\
 & C.1.2\Omega \\
 & D.1.5\Omega \\
\end{align}$

Answer 1

Hint: First of all, the potential of the cell where the balancing point of the cell is equivalent to the product of the constant and the balancing length. Find the equation for the potential when the shunt is connected. Compare both the equations. This will help you in answering this question.

Complete answer:
First of all, the potential of the cell where the balancing point of the cell is equivalent to the product of the constant and the balancing length. This can be written as,
$E=k{{l}_{1}}$
The balancing length of the cell has been given as,
${{l}_{1}}=52cm$
Substituting the balancing length in this equation will give,
$E=k\times 52$……… (1)
When a shunt resistance has been connected, we can say that the electric potential of the cell will become,
$E-\dfrac{Er}{R+r}=k{{l}_{2}}$
Where $r$ be the internal resistance of the cell, $R$ be the shunt resistance connected and ${{l}_{2}}$ be the balancing point at this condition which has been given as,
${{l}_{2}}=40cm$
Substituting this in the equation given as,
$E-\dfrac{Er}{R+r}=k\times 40$
The equation can be written like this,
$\begin{align}
  & E\left( 1-\dfrac{r}{R+r} \right)=k\times 40 \\
 & \Rightarrow E\left( \dfrac{R+r-r}{R+r} \right)=k\times 40 \\
\end{align}$
This equation can be rearranged like this,
$E=40\times K\left( 1+\dfrac{r}{R} \right)$…….. (2)
Now we can compare the equation (1) and (2),
$52=40\times \left( 1+\dfrac{r}{R} \right)$
The values of shunt resistance has been already mentioned as,
$R=5\Omega $
Substituting this value in the equation will give,
$\begin{align}
  & \dfrac{52}{40}=1+\dfrac{r}{5} \\
 & \Rightarrow \dfrac{r}{5}=0.3 \\
 & \therefore r=1.5\Omega \\
\end{align}$
Therefore the internal resistance of the cell has been obtained.

The correct answer has been mentioned as option D.

Note:
Internal resistance of a battery can be referred to as the opposition offered to the flow of current by the cells and the batteries themselves which will cause the production of heat. Internal resistance has been measured in the units of Ohms.