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

$A$ cell of internal resistance $3 \, \Omega$ and $emf$ $10 \, V$ is connected to a uniform wire of length $500 \, cm$ and resistance $3 \, \Omega$. The potential gradient in the wire is .............. $mV/cm$.

When a potentiometer is connected between the points $A$ and $B$ as shown in the circuit,the balance point is obtained at $64 \ cm$. When it is connected between $A$ and $C$,the balance point is $8 \ cm$. If the potentiometer is connected between $B$ and $C$,the balance point will be: (in $cm$)

To determine the internal resistance of a cell by using a potentiometer,the null point is at $1 \ m$,when shunted by $3 \ \Omega$ resistance and at a length $1.5 \ m$,when cell is shunted by $6 \ \Omega$ resistance. The internal resistance of the cell is (in $\Omega$)

In a potentiometer, the null point is received at the $7^{th}$ wire. If we now want to shift the null point to the $9^{th}$ wire, what should we do?

Vedclass Test Series

Exam Paper Generator

Online Exam Module