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

Hint: 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.