
In an experiment of potentiometer for measuring the internal resistance of primary cell a balancing length l is obtained on the potentiometer wire when the cell is open circuit. Now the cell is short circuited by a resistance R. If R is to be equal to the internal resistance of the cell the balancing length on the potentiometer wire will be
A. l
B. 2l
C. l/2
D. l/4
Answer
162.3k+ views
Hint:In order to solve this problem we have to apply the concept of potentiometer in which potential difference applied across it uniformly distributed across its length. We have to use the expression of potential gradient to answer this question.
Formula used:
The emf of the cell with balancing length $l$ is,
$E = Kl$
Where K is the potential gradient.
The expression of internal resistance is given as,
$r = \left( {\dfrac{{E - V}}{V}} \right)R$
Where, $E$ is the emf of a cell, $V$ is the potential difference applied across it and $R$ is the external resistance.
Complete step by step solution:
The potentiometer is a device used to measure an unknown voltage by comparing it to a known voltage. It can be used to compare the emf of several cells and to determine the emf and internal resistance of the specified cell.
The emf of the cell with balancing length l, $E = Kl$. Now the cell is short circuited by a resistance R. And R is equal to the internal resistance of the cell.
$V = Kl'$
$\Rightarrow r = \left( {\dfrac{{E - V}}{V}} \right)R$
$\Rightarrow Vr = \left( {E - V} \right)R$
Where, internal resistance r = Resistance R.
$V = E - V$
$\Rightarrow 2V = E$
$\Rightarrow 2Kl' = Kl$
$\therefore l' = \dfrac{l}{2}$
Hence, the correct option is option C .
Additional Information: The potentiometer's basic principle states that, if the wire has a uniform cross-sectional area and a uniform current flows through it, the potential drop across any part of the wire will be precisely proportional to the length of the wire. The value of internal resistance that can be measured by potentiometer is:
\[r = \dfrac{{{l_1} - {l_2}}}{{{l_2}}}R\]
Where \[{l_1}\]= length of the potentiometer wire, \[{l_2}\]= length of the given wire, r = internal resistance of the cell and R = resistance of the resistor in the circuit.
Note: The smallest change in the potential difference that a potentiometer can detect is its sensitivity. By lowering the potential gradient, the potentiometer's sensitivity can be improved. That is done by extending the potentiometer wire's length.
Formula used:
The emf of the cell with balancing length $l$ is,
$E = Kl$
Where K is the potential gradient.
The expression of internal resistance is given as,
$r = \left( {\dfrac{{E - V}}{V}} \right)R$
Where, $E$ is the emf of a cell, $V$ is the potential difference applied across it and $R$ is the external resistance.
Complete step by step solution:
The potentiometer is a device used to measure an unknown voltage by comparing it to a known voltage. It can be used to compare the emf of several cells and to determine the emf and internal resistance of the specified cell.
The emf of the cell with balancing length l, $E = Kl$. Now the cell is short circuited by a resistance R. And R is equal to the internal resistance of the cell.
$V = Kl'$
$\Rightarrow r = \left( {\dfrac{{E - V}}{V}} \right)R$
$\Rightarrow Vr = \left( {E - V} \right)R$
Where, internal resistance r = Resistance R.
$V = E - V$
$\Rightarrow 2V = E$
$\Rightarrow 2Kl' = Kl$
$\therefore l' = \dfrac{l}{2}$
Hence, the correct option is option C .
Additional Information: The potentiometer's basic principle states that, if the wire has a uniform cross-sectional area and a uniform current flows through it, the potential drop across any part of the wire will be precisely proportional to the length of the wire. The value of internal resistance that can be measured by potentiometer is:
\[r = \dfrac{{{l_1} - {l_2}}}{{{l_2}}}R\]
Where \[{l_1}\]= length of the potentiometer wire, \[{l_2}\]= length of the given wire, r = internal resistance of the cell and R = resistance of the resistor in the circuit.
Note: The smallest change in the potential difference that a potentiometer can detect is its sensitivity. By lowering the potential gradient, the potentiometer's sensitivity can be improved. That is done by extending the potentiometer wire's length.
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