Question

A cell of emf \[{e_1}\] in the secondary circuit is gives null deflection for \[1.5\,{\text{m}}\] length of potentiometer of wire length \[{\text{10 m}}\]. If another cell of emf \[{e_2}\] is connected in series with \[{e_1}\]then null deflection was obtained for \[2.5\,{\text{m}}\] length. Then \[{e_1}:{e_2}\] is:
A. \[3:5\]
B. \[5:3\]
C. \[3:2\]
D. \[2:3\]

Answer 1

Hint:Write down the given values. Recall the concepts of potentiometer. Use the relation used in case of potentiometer, between null point length and emf. It is given that another cell is connected in series, taking out their resultant emf. Use these values to find the ratio of the first and second emf.

Complete step by step answer:
Given, null point length of cell of emf \[{e_1}\] is \[{l_1} = 1.5\,{\text{m}}\]
Null point length when another cell of emf \[{e_2}\] connected in series with first cell is \[{l_2} = 2.5\,{\text{m}}\].

From the theory of potentiometer we have,
\[\dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{L_1}}}{{{L_2}}}\] (i)
where \[{E_1}\] is the emf of the first cell and \[{L_1}\] is its balancing length and \[{E_2}\] is the emf of the second cell and \[{L_2}\] is its balancing length. Here, \[{E_1} = {e_1}\] , \[{L_1} = {l_1}\]

It is said that a cell of emf \[{e_2}\] is connected in series with the first cell so emf of second cell will be,
\[{E_2} = {e_1} + {e_2}\] and \[{L_2} = {l_2}\].
Using these values in equation (i), we have
\[\dfrac{{{e_1}}}{{{e_1} + {e_2}}} = \dfrac{{{l_1}}}{{{l_2}}}\]
Putting the values of \[{l_1}\] and \[{l_2}\] we get
\[\dfrac{{{e_1}}}{{{e_1} + {e_2}}} = \dfrac{{1.5}}{{2.5}} \\
\Rightarrow \dfrac{1}{{1 + \dfrac{{{e_2}}}{{{e_1}}}}} = \dfrac{3}{5} \\
\Rightarrow 1 + \dfrac{{{e_2}}}{{{e_1}}} = \dfrac{5}{3} \\
\Rightarrow \dfrac{{{e_2}}}{{{e_1}}} = \dfrac{2}{3} \\
\therefore\dfrac{{{e_1}}}{{{e_2}}} = \dfrac{3}{2} \]
Therefore, the ratio of the emf is \[{{\text{e}}_1}:{e_2} = 3:2\].

Hence, the correct answer is option C.

Note: Potentiometer is a three terminal device used for measuring the value of unknown voltage with the help of a known voltage by manually varying the resistance. Many times students get confused between a rheostat and a potentiometer. A rheostat is a two terminal variable resistor, when this variable resistor is used as a potential divider by adding one more terminal then it is known as potentiometer.