(C) Let $k$ be the potential gradient of the potentiometer wire.When the cells are in series and in conjunction,the total $e.m.f.$ is $(E_1 + E_2)$. T

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(C) Let $k$ be the potential gradient of the potentiometer wire.When the cells are in series and in conjunction,the total $e.m.f.$ is $(E_1 + E_2)$. The balancing length is $l_1 = 58 \ cm$.So,$(E_1 + E_2) = k \cdot l_1 = 58k$ --- $(1)$When the polarity of $E_2$ is reversed,the total $e.m.f.$ is $(E_1 - E_2)$. The balancing length is $l_2 = 29 \ cm$.So,$(E_1 - E_2) = k \cdot l_2 = 29k$ --- $(2)$Dividing equation $(1)$ by equation $(2)$:$\frac{E_1 + E_2}{E_1 - E_2} = \frac{58k}{29k} = 2$$E_1 + E_2 = 2(E_1 - E_2)$$E_1 + E_2 = 2E_1 - 2E_2$$3E_2 = E_1$Therefore,$\frac{E_1}{E_2} = \frac{3}{1}$.

Class 12 Physics Current Electricity Potentiometer

Medium

In a potentiometer experiment,two cells of $e.m.f.$ $E_1$ and $E_2$ are used in series and in conjunction,and the balancing length is found to be $58 \ cm$ of the wire. If the polarity of $E_2$ is reversed,the balancing length becomes $29 \ cm$. The ratio $\frac{E_1}{E_2}$ of the $e.m.f.$ of the two cells is: (in $:1$)

A
$1$
B
$2$
C
$3$
D
$4$

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Class 12 Questions

Class 12

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Current Electricity

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Potentiometer

The sliding contact of a potentiometer is in the middle of the potentiometer wire having a total resistance $R_p = 1 \Omega$,as shown in the figure. An external resistance of $R_e = 2 \Omega$ is connected via the sliding contact. Find the total current drawn from the $0.9 \text{ V}$ battery. (in $\text{ A}$)

DifficultJEE MAIN 2025

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To measure the internal resistance of a battery,a potentiometer is used. For $R = 10 \ \Omega$,the balance point is observed at $\ell = 500 \ cm$ and for $R = 1 \ \Omega$,the balance point is observed at $\ell = 400 \ cm$. The internal resistance of the battery is approximately: (in $\Omega$)

DifficultJEE MAIN 2024

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The balancing length for a cell is $560 \; cm$ in a potentiometer experiment. When an external resistance of $10 \; \Omega$ is connected in parallel to the cell,the balancing length changes by $60 \; cm$. If the internal resistance of the cell is $\frac{N}{10} \; \Omega$,where $N$ is an integer,then the value of $N$ is:

MediumJEE MAIN 2020

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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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To compare the $EMF$ of two cells using a potentiometer, the balancing lengths obtained are $200 \ cm$ and $150 \ cm$. The least count of the scale is $1 \ cm$. The percentage error in the ratio of the EMFs is . . . . . . .

MediumJEE MAIN 2026

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To measure the internal resistance of a battery,a potentiometer is used. For $R = 10 \ \Omega$,the balance point is observed at $\ell = 500 \ cm$ and for $R = 1 \ \Omega$,the balance point is observed at $\ell = 400 \ cm$. The internal resistance of the battery is approximately: (in $\Omega$)

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 $cm$)

To compare the $EMF$ of two cells using a potentiometer, the balancing lengths obtained are $200 \ cm$ and $150 \ cm$. The least count of the scale is $1 \ cm$. The percentage error in the ratio of the EMFs is . . . . . . .

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