(B) Let $k$ be the potential gradient of the potentiometer wire.When the potentiometer is connected between $A$ and $B$,the potential difference measu

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(B) Let $k$ be the potential gradient of the potentiometer wire.When the potentiometer is connected between $A$ and $B$,the potential difference measured is $E_1$. Thus,$E_1 = k \times 3.60$.When the potentiometer is connected between $A$ and $C$,the cells are in series opposition because they are connected with opposite polarities. The net e.m.f. is $E_1 - E_2$. Thus,$E_1 - E_2 = k \times 0.90$.Dividing the two equations: $E_1 / (E_1 - E_2) = 3.60 / 0.90 = 4$.This implies $E_1 = 4(E_1 - E_2) = 4E_1 - 4E_2$.Rearranging gives $3E_1 = 4E_2$,so $E_1 / E_2 = 4/3$.

Class 12 Physics Current Electricity Potentiometer

Medium

Two cells of e.m.f.s $E_1$ and $E_2$ $(E_1 > E_2)$ are connected as shown in the figure. When the potentiometer is connected between $A$ and $B$,the balancing length of the potentiometer wire is $3.60 \ m$. On connecting the potentiometer between $A$ and $C$,the balancing length is $0.90 \ m$. The ratio $E_1 / E_2$ is

MHT CET 2025 All MHT CET

A
$5:4$
B
$4:3$
C
$3:4$
D
$4:5$

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$A$ cell of $emf$ $2 \,V$ and internal resistance $5 \,\Omega$ is connected to a wire of length $100 \,cm$ and resistance $15 \,\Omega$. What is the potential gradient along the wire (in $,V/cm$)?

Medium

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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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$A$ voltmeter reads the potential difference across the terminals of an old battery as $1.2 \, V$,while a potentiometer reads $1.4 \, V$. The internal resistance of the battery is $40 \, \Omega$. What is the resistance of the voltmeter in $\Omega$?

Medium

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In an experiment to find the emf of a cell using a potentiometer, the length of the null point for a cell of emf $1.5 \text{ V}$ is found to be $60 \text{ cm}$. If this cell is replaced by another cell of emf $E$, the length of the null point increases by $40 \text{ cm}$. The value of $E$ is $x/10 \text{ V}$. The value of $x$ is

DifficultMHT CET 2026

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The circuit shown here is used to compare the $e.m.f.$ of two cells ${E_1}$ and ${E_2}$ $(E_1 > E_2)$. The null point is at $C$ when the galvanometer is connected to ${E_1}$. When the galvanometer is connected to ${E_2}$,the null point will be

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$A$ cell of $emf$ $2 \,V$ and internal resistance $5 \,\Omega$ is connected to a wire of length $100 \,cm$ and resistance $15 \,\Omega$. What is the potential gradient along the wire (in $,V/cm$)?

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 $\Omega$)

$A$ voltmeter reads the potential difference across the terminals of an old battery as $1.2 \, V$,while a potentiometer reads $1.4 \, V$. The internal resistance of the battery is $40 \, \Omega$. What is the resistance of the voltmeter in $\Omega$?

The circuit shown here is used to compare the $e.m.f.$ of two cells ${E_1}$ and ${E_2}$ $(E_1 > E_2)$. The null point is at $C$ when the galvanometer is connected to ${E_1}$. When the galvanometer is connected to ${E_2}$,the null point will be

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