Question

In a potentiometer experiment, balancing length is found to be $120\,cm$ for a cell ${E_1}$ of e.m.f $2\,V$. What will be the balancing length for another cell ${E_2}$ of e.m.f $1.5\,V$ ? (No other changes are made in the experiment.)

Answer 1

Use the condition of the potentiometer that the potential gradient is always constant in the balance condition of a potentiometer to find the balancing length for the other cell.Potentiometer working can be explained when the potentiometer is understood. It is defined as a three-terminal resistor having either sliding or rotating contact that forms an adjustable voltage divider.

Formula used:
The condition of the potentiometer is given as,
$K = \dfrac{E}{l}$
Here, $l$ is the balancing length of the potentiometer $e$ is the e.m.f of the cell used for balancing the potentiometer.

Complete step by step answer:
Since, nothing is changed except the cells in the experiment including potentiometer so, from this we can write, $\dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{l_1}}}{{{l_2}}}$. Where, ${E_1}$ is the e.m.f of the first cell ${E_2}$ is the e.m.f of second cell ${l_1}$ is the balancing length for first cell and ${l_2}$ is the balancing length for the second cell.

So, here we have the e.m.f of the first cell, ${E_1} = 2\,V$ and the balancing length is ${l_1} = 120\,cm$. Also e.m.f of the second cell, ${E_2} = 1.5\,V$. Hence, putting the values of the e.m.f of the first cell, ${E_1} = 2\,V$ and the balancing length is ${l_1} = 120\,cm$ and e.m.f of the second cell, ${E_2} = 1.5\,V$ in the equation we get,
$\dfrac{2}{{1.5}} = \dfrac{{120}}{{{l_2}}}$
On simplifying we get,
${l_2} = \dfrac{{120 \times 1.5}}{2}$
$\Rightarrow {l_2} = 60 \times 1.5$
$\therefore {l_2} = 90\,cm$

Hence, the balancing length of the cell ${E_2}$ is $90\,cm$.

Note: The balance condition of a potentiometer is acquired when the current through the galvanometer is zero. So, the net current through the galvanometer is zero. For, a cell of e.m.f $E$ with balancing length $l$, having the potentiometer of $L$with a source of ${E_0}$volt, the balance condition is, $E = l\dfrac{{{E_0}}}{L}$. Or, $\dfrac{E}{l} = \dfrac{{{E_0}}}{L} = K$. So, the potential gradient is constant for a particular potentiometer.