Question

Two cells of e.m.f. ${E_1}$ and ${E_2}$ are joined in series and the balancing length of the potential wire is $625$ cm. If the terminals of ${E_1}$ are reversed, the balancing length obtained is $125$ cm. Given ${E_2} > {E_1}$ , the ratio ${E_1}:{E_2}$ will be
(A) $2:3$
(B) $5:1$
(C) $3:2$
(D) $1:5$

Answer 1

The balancing length of the potentiometer when the two cells are connected in series is given. When the cell ${E_1}$ is connected in reverse manner the balancing length is given. The balancing length in the potentiometer is directly proportional to the sum of the emf of the cells. Using this, we can find the required relation.

Complete step by step solution: The working principle of the potentiometer depends on the potential across any portion of the wire which is directly proportional to the length of the wire. Potentiometer can be used to find the emf of an unknown cell. Potentiometer is also used to determine the internal resistance of the cell.
The balancing length of the potentiometer is proportional to the net emf of the cells.
When the cells are connected in series, the net emf is ${E_2} + {E_1}$ and the balancing length is $652$ cm.
When the cell ${E_1}$ is reversed, the net emf of the cell will be ${E_2} - {E_1}$ and this balancing length is given as $125$ cm. As balancing length is proportional to net emf thus, we have:
${E_2} + {E_1}\,\alpha \,625$ --equation $1$
And ${E_2} - {E_1}\,\alpha \,125$ --equation $2$
Dividing equation $1$ by equation $2$ , we get
$\dfrac{{{E_2} + {E_1}}}{{{E_2} - {E_1}}}\, = \dfrac{{625\,}}{{125}}$
$\Rightarrow \dfrac{{{E_2} + {E_1}}}{{{E_2} - {E_1}}}\, = \dfrac{5}{1}$
We need to find the ratio ${E_1}:{E_2}$ , solving the above equation we get.
$\Rightarrow {E_2} + {E_1}\, = 5\left( {{E_2} - {E_1}} \right)$
$\Rightarrow {E_2} + {E_1}\, = 5{E_2} - 5{E_1}$
$\Rightarrow 6{E_1}\, = 4{E_2}$
$\Rightarrow \dfrac{{{E_1}}}{{{E_2}}} = \dfrac{4}{6} = \dfrac{2}{3}$
$\Rightarrow {E_1}:{E_2} = 2:3$
The required ratio ${E_1}:{E_2}$ is $2:3$

Therefore, $1$ is the correct option.

Note: The net emf of the potentiometer is proportional to the balancing length of the potentiometer. It is given that ${E_2} > {E_1}$ thus, we must take the difference as $\left( {{E_2} - {E_1}} \right)$ . When the cell is reversed the balancing length decreases as the net emf of the cell decreases.