Question

The two cells are connected in series, in a potentiometer experiment, in such a way so as to support each other and to oppose each other. The balancing lengths in two conditions are obtained as 150 cm and 50 cm respectively. The ratio of EMFs of two cells will be
A) 1:2
B) 2:1
C) 1:4
D) 4:1

Answer 1

Hint
The potential drop across a potentiometer depends on the potential of the cell that is to be measured. The balancing length for a given cell is proportional to the potential of the cell. We will use this relation to find the two balancing lengths and hence the ratio of the EMFs.

Complete step by step answer
We know that for a potentiometer, the EMF of the cell is proportional to the balancing length of the potential meter which can be written as:
 $ \Rightarrow V \propto l $ where $ V $ is the potential and $ l $ is the balancing length of the potentiometer. Let $ {V_1} $ be the EMF of the first cell and $ {V_2} $ be the EMF of the second cell and $ {V_2} > {V_1} $.
Now we’ve been told that for two cells when they are connected to support each other, the balancing length is 150 cm. The potential, in this case, will be equal to $ {V_2} + {V_1} $.
When the two cells are connected to oppose each other, the balancing length is 50 cm and the net potential, in this case, will be equal to $ {V_2} - {V_1} $.
Since $ V \propto l $, taking the ratio of the supporting and the opposing cases, we get
 $ \Rightarrow \dfrac{{{V_2} + {V_1}}}{{{V_2} - {V_1}}} = \dfrac{{150}}{{50}} $
 $ \Rightarrow \dfrac{{{V_2} + {V_1}}}{{{V_2} - {V_1}}} = 3 $
We want to find the ratio of the two EMFs i.e. $ {V_2}/{V_1} $ we can write the expression as:
 $ \Rightarrow \dfrac{{{V_2}}}{{{V_1}}} = \dfrac{{\left( {{V_2} + {V_1}} \right) + \left( {{V_2} - {V_1}} \right)}}{{\left( {{V_2} + {V_1}} \right) - \left( {{V_2} - {V_1}} \right)}} $
On substituting the values we get,
 $ \Rightarrow \dfrac{{{V_2}}}{{{V_1}}} = \dfrac{{3 + 1}}{{3 - 1}} = 2 $
Hence the potential of the two EMFs will have the ratio $ \dfrac{{{V_2}}}{{{V_1}}} = \dfrac{2}{1} = 2:1 $ which corresponds to option (B).

Note
In this question, we only need to use the proportionality relation of the potentiometer balancing length, the voltage of the EMF being measured, which is linear in nature. The ratio of the EMFs is by convention calculated as the ratio of the greater voltage to the lower voltage which is why if we assume $ {V_2} > {V_1} $, we should calculate the ratio of $ {V_2}/{V_1} $ and vice-versa.