(A) The potential difference across the balancing length $l$ of the potentiometer wire is equal to the $emf$ of the thermocouple $(E_{th})$.Given: Bal

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(A) The potential difference across the balancing length $l$ of the potentiometer wire is equal to the $emf$ of the thermocouple $(E_{th})$.Given: Balancing length $l = 5 \, m$, resistance per unit length $\rho = 1.2 \, \Omega/m$, and $E_{th} = 2.4 \, mV = 2.4 \times 10^{-3} \, V$.The potential drop across the balancing length is $V = i \times R_{wire} = i \times (\rho \times l)$.Since the galvanometer deflection is zero, $V = E_{th}$.Therefore, $i \times (1.2 \, \Omega/m \times 5 \, m) = 2.4 \times 10^{-3} \, V$.$i \times 6 \, \Omega = 2.4 \times 10^{-3} \, V$.$i = \frac{2.4 \times 10^{-3}}{6} \, A = 0.4 \times 10^{-3} \, A = 4 \times 10^{-4} \, A$.

Class 12 Physics Current Electricity Potentiometer

Difficult

In the following circuit, a $10 \, m$ long potentiometer wire with resistance $1.2 \, \Omega/m$, a resistance $R_1$, and an accumulator of $emf$ $2 \, V$ are connected in series. When the $emf$ of the thermocouple is $2.4 \, mV$, the deflection in the galvanometer is zero at a balancing length of $5 \, m$. The current supplied by the accumulator is:

A
$4 \times 10^{-4} \, A$
B
$8 \times 10^{-4} \, A$
C
$4 \times 10^{-3} \, A$
D
$8 \times 10^{-3} \, A$

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Current Electricity

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Potentiometer

$A$ circuit for the measurement of resistance by a potentiometer is shown. The galvanometer is first connected at point $A$ and zero deflection is observed at a length $PJ = 10 \ cm$. In the second case,it is connected at point $C$ and zero deflection is observed at a length of $30 \ cm$ from $P$. Then the unknown resistance $X$ is:

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Two cells $A$ and $B$ are connected in the secondary circuit of a potentiometer one at a time,and the balancing lengths are $400 \ cm$ and $440 \ cm$ respectively. The emf of cell $A$ is $1.08 \ V$. The emf of the second cell $B$ in volts is:

DifficultTS EAMCET 2004

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In a potentiometer experiment, the balancing length with a cell is $250 \, cm$. On shunting the cell with a resistance of $2 \, \Omega$, the balancing length becomes $125 \, cm$. The internal resistance of the cell is:

MediumAP EAMCET 2021

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$A$ $6 \, V$ battery of negligible internal resistance is connected across a uniform wire $AB$ of length $1 \, m$. The positive terminal of another battery of emf $4 \, V$ and internal resistance $1 \, \Omega$ is joined to point $A$ as shown in the figure. The ammeter shows zero deflection when the jockey touches the wire at point $C$. The length $AC$ is equal to:

Medium

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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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Two cells $A$ and $B$ are connected in the secondary circuit of a potentiometer one at a time,and the balancing lengths are $400 \ cm$ and $440 \ cm$ respectively. The emf of cell $A$ is $1.08 \ V$. The emf of the second cell $B$ in volts is:

In a potentiometer experiment, the balancing length with a cell is $250 \, cm$. On shunting the cell with a resistance of $2 \, \Omega$, the balancing length becomes $125 \, cm$. The internal resistance of the cell is:

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:

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