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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

Explain the comparison of the electromotive force (emf) of two cells using a potentiometer with a necessary diagram.

In a potentiometer experiment, a null point is obtained at a particular point for a cell on a potentiometer wire of length '$x$' cm. If the length of the potentiometer wire is increased by a few cm without changing the cell or the driving source, the balancing length will:

$A$ potentiometer consists of a wire of length $4\, m$ and resistance $10\,\Omega$. It is connected to a cell of $e.m.f.$ $2\, V$. The potential difference per unit length of the wire will be ............. $V/m$.

With a potentiometer,null points were obtained at $140 \, cm$ and $180 \, cm$ with cells of $emf$ $1.1 \, V$ and one unknown $X \, V$. The unknown $emf$ is .............. $V$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module