In a potentiometer experiment, the balancing with a cell is at length $250 \mathrm{~cm}$. In shunting the…
In a potentiometer experiment, the balancing with a cell is at length $250 \mathrm{~cm}$. In shunting the cell with a resistance of $2 \Omega$, the balancing length becomes $125 \mathrm{~cm}$. The internal resistance of the cell is
$2 \Omega$
$4 \Omega$
$0.5 \Omega$
$1 \Omega$
Solution
Given, balancing length with emf of cell,
$l_1=250 \mathrm{~cm}$
Balancing length with cell and shunt resistance,
$l_2=125 \mathrm{~cm}$
Shunt resistance, $R=2 \Omega$
Let $r$ be the internal resistance of cell. Then, using relation,
$r=R\left(\frac{l_1}{l_2}-1\right)$
By substituting the values, we get
$r=2\left(\frac{250}{125}-1\right)=2 \Omega$