(C) The total resistance in the primary circuit is $R_{total} = R_{wire} + R_{series} = 8 \ \Omega + 242 \ \Omega = 250 \ \Omega$. The current flowing

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(C) The total resistance in the primary circuit is $R_{total} = R_{wire} + R_{series} = 8 \ \Omega + 242 \ \Omega = 250 \ \Omega$. The current flowing through the potentiometer wire is $I = \frac{V}{R_{total}} = \frac{2.5 \ V}{250 \ \Omega} = 0.01 \ A$. The potential drop across the entire wire is $V_{wire} = I \times R_{wire} = 0.01 \ A \times 8 \ \Omega = 0.08 \ V$. The potential gradient $k$ is the potential drop per unit length: $k = \frac{V_{wire}}{L} = \frac{0.08 \ V}{2.5 \ m} = 0.032 \ V/m$. The potential difference across a length $l = 20 \ cm = 0.2 \ m$ is $V' = k \times l = 0.032 \ V/m \times 0.2 \ m = 0.0064 \ V = 6.4 \ mV$.

Class 12 Physics Current Electricity Potentiometer

Medium

The length of a potentiometer wire is $2.5 \ m$ and its resistance is $8 \ \Omega$. $A$ cell of negligible internal resistance and emf of $2.5 \ V$ is connected in series with a resistance of $242 \ \Omega$ in the primary circuit. The potential difference between two points separated by a distance of $20 \ cm$ on the potentiometer wire is (in $mV$)

TS EAMCET 2025 All TS EAMCET

A
$1.6$
B
$4.8$
C
$6.4$
D
$3.2$

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Current Electricity

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Potentiometer

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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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In the figure,the potentiometer wire $AB$ of length $L$ and resistance $9r$ is joined to the cell $D$ of $emf$ $\varepsilon$ and internal resistance $r$. The cell $C$'s $emf$ is $\frac{\varepsilon}{2}$ and its internal resistance is $2r$. The galvanometer $G$ will show no deflection when the length $AJ$ is

Medium

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In the potentiometer circuit as shown in the figure,the balance length $l = 60 \ cm$ when switch $S$ is open. When switch $S$ is closed and the value of $R$ is $5 \ \Omega$,the balance length $l' = 50 \ cm$. The internal resistance of the cell $C'$ is : .............. $\Omega$

Medium

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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 $\Omega$)

DifficultMHT CET 2021

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$A$ potentiometer wire of length $100 \ cm$ and resistance $3 \ \Omega$ is connected in series with a resistance of $8 \ \Omega$ and an accumulator of $4 \ V$ whose internal resistance is $1 \ \Omega$. $A$ cell of e.m.f. $E$ is balanced by $50 \ cm$ length of the wire. The e.m.f. of the cell is: (in $V$)

MediumMHT CET 2020

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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 . . . . . . .

In the potentiometer circuit as shown in the figure,the balance length $l = 60 \ cm$ when switch $S$ is open. When switch $S$ is closed and the value of $R$ is $5 \ \Omega$,the balance length $l' = 50 \ cm$. The internal resistance of the cell $C'$ is : .............. $\Omega$

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 $\Omega$)

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