Write the advantages of a potentiometer.
(N/A) potentiometer is a versatile instrument used for measuring potential differences and internal resistance. Its main advantages are:
$1$. It works on the principle of null deflection,meaning no current is drawn from the source of $EMF$ being measured. Therefore,it measures the true $EMF$ of the cell.
$2$. It does not affect the circuit conditions because it does not draw any current from the circuit under test.
$3$. It is highly sensitive and can be used to measure very small potential differences.
$4$. It can be used to compare the EMFs of two cells and to determine the internal resistance of a cell.
$1$. It works on the principle of null deflection,meaning no current is drawn from the source of $EMF$ being measured. Therefore,it measures the true $EMF$ of the cell.
$2$. It does not affect the circuit conditions because it does not draw any current from the circuit under test.
$3$. It is highly sensitive and can be used to measure very small potential differences.
$4$. It can be used to compare the EMFs of two cells and to determine the internal resistance of a cell.
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Resistance of a potentiometer wire is $2 \ \Omega/m$. $A$ cell of e.m.f. $1.5 \ V$ balances at $300 \ cm$. The current through the wire is (in $mA$)
$A$ potentiometer wire of length $100\, cm$ has a resistance of $10\, \Omega$. It is connected in series with a resistance $R$ and an accumulator of emf $2\, V$ and of negligible internal resistance. $A$ source of emf $10\, mV$ is balanced against a length of $40\, cm$ of the potentiometer wire. What is the value of external resistance $R$?
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View Solution$A$ resistance of $4\,\Omega$ and a wire of length $5\,m$ and resistance $5\,\Omega$ are joined in series and connected to a cell of $e.m.f.$ $10\,V$ and internal resistance $1\,\Omega$. $A$ parallel combination of two identical cells is balanced across $300\,cm$ of the wire. The $e.m.f.$ $E$ of each cell is ........... $V$.
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View SolutionIn a potentiometer experiment,the balancing length with a cell is $240 \ cm$. On shunting the cell with a resistance of $2 \ \Omega$,the balancing length becomes $120 \ cm$. The internal resistance of the cell is ................. $\Omega$.
MediumAIEEE 2005
View SolutionIn the determination of the internal resistance of a cell with a potentiometer,the error in the measurement of the balancing length is $\pm 1 \text{ mm}$. When the cell alone is connected in the circuit,the balancing length is obtained at $60 \text{ cm}$ and when the cell is shunted with a resistance of $10 \Omega \pm 2 \%$,the balancing length is obtained at $50 \text{ cm}$. The error in the determination of the internal resistance of the cell is (in $\%$)
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