When a cell of E.M.F. 'E_1' is connected to a | vedclass

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(C) In a potentiometer, the $E$.$M$.$F$. of a cell is directly proportional to the balancing length, i.e.,$E = k \cdot l$, where $k$ is the potential

Vedclass · Accepted Answer

$(l_1) : (l_1 - l_2)$

Vedclass · Answer

(C) In a potentiometer, the $E$.$M$.$F$. of a cell is directly proportional to the balancing length, i.e.,$E = k \cdot l$, where $k$ is the potential gradient.$1$. When the cell $E_1$ is connected, the balancing length is $l_1$, so $E_1 = k \cdot l_1$.$2$. When the cells $E_1$ and $E_2$ are connected in opposition, the effective $E$.$M$.$F$. is $(E_1 - E_2)$, and the balancing length is $l_2$, so $(E_1 - E_2) = k \cdot l_2$.$3$. Dividing the two equations: $\frac{E_1}{E_1 - E_2} = \frac{k \cdot l_1}{k \cdot l_2} = \frac{l_1}{l_2}$.$4$. Rearranging the equation: $E_1 \cdot l_2 = l_1 \cdot (E_1 - E_2) \implies E_1 \cdot l_2 = E_1 \cdot l_1 - E_2 \cdot l_1$.$5$. Grouping terms: $E_2 \cdot l_1 = E_1 \cdot l_1 - E_1 \cdot l_2 = E_1(l_1 - l_2)$.$6$. Therefore, $\frac{E_1}{E_2} = \frac{l_1}{l_1 - l_2}$.Thus, the ratio $E_1 : E_2$ is $(l_1) : (l_1 - l_2)$.

When a cell of $E$.$M$.$F$. '$E_1$' is connected to a potentiometer wire, the balancing length is '$l_1$'. Another cell of $E$.$M$.$F$. '$E_2$' $(E_1 > E_2)$ is connected along with $E_1$ such that the two cells oppose each other, and the balancing length is '$l_2$'. The ratio $E_1 : E_2$ is

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