(B) In a potentiometer,the $E.M.F.$ of a cell is proportional to the balancing length,i.e.,$E \propto l$ or $E = kl$,where $k$ is the potential gradie

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(B) In a potentiometer,the $E.M.F.$ of a cell is proportional to the balancing length,i.e.,$E \propto l$ or $E = kl$,where $k$ is the potential gradient.For the first cell,$E_1 = kl_1$.When the two cells are connected in opposition,the effective $E.M.F.$ is $(E_1 - E_2)$. The new balancing length is $l_2$,so $(E_1 - E_2) = kl_2$.Dividing the two equations: $\frac{E_1}{E_1 - E_2} = \frac{l_1}{l_2}$.Cross-multiplying gives $E_1 l_2 = l_1 E_1 - l_1 E_2$.Rearranging terms: $l_1 E_2 = E_1 (l_1 - l_2)$.Therefore,$\frac{E_1}{E_2} = \frac{l_1}{l_1 - l_2}$.

Class 12 Physics Current Electricity Potentiometer

Difficult

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 such that the two cells oppose each other,then the balancing length is $l_2$. The ratio $E_1 : E_2$ is

MHT CET 2024 All MHT CET

A
$\frac{l_1}{l_1+l_2}$
B
$\frac{l_1+l_1-l_2}{l_1-l_2}$
C
$\frac{l_1+l_2}{l_1}$
D
$\frac{l_1+l_2}{l_1-l_2}$

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$A$ potentiometer wire of length $300\,cm$ is connected in series with a resistance $780\,\Omega$ and a standard cell of emf $4\,V$. $A$ constant current flows through the potentiometer wire. The length of the null point for a cell of emf $20\,mV$ is found to be $60\,cm$. The resistance of the potentiometer wire is ... $\Omega$.

DifficultJEE MAIN 2022

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

MediumMHT CET 2016

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$A$ potentiometer wire of length $1 \, m$ is connected in series with a $490 \, \Omega$ resistance and a $2 \, V$ battery. If the potential gradient is $0.2 \, mV/cm$,then the resistance of the potentiometer wire is ................ $\Omega$. (in $.9$)

Medium

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$A$ Daniel cell is balanced on $125\,cm$ length of a potentiometer wire. Now the cell is short-circuited by a resistance $2\,\Omega$ and the balance is obtained at $100\,cm$. The internal resistance of the Daniel cell is .............. $\Omega$.

Medium

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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 $cm$)

MediumTS EAMCET 2023

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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 such that the two cells oppose each other,then the balancing length is $l_2$. The ratio $E_1 : E_2$ is

Similar Questions

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

$A$ potentiometer wire of length $1 \, m$ is connected in series with a $490 \, \Omega$ resistance and a $2 \, V$ battery. If the potential gradient is $0.2 \, mV/cm$,then the resistance of the potentiometer wire is ................ $\Omega$. (in $.9$)

$A$ Daniel cell is balanced on $125\,cm$ length of a potentiometer wire. Now the cell is short-circuited by a resistance $2\,\Omega$ and the balance is obtained at $100\,cm$. The internal resistance of the Daniel cell is .............. $\Omega$.

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 $cm$)

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