
In a potentiometer arrangement for determining the emf of a cell, the balance point of the cell in an open circuit is $350\,cm$. When a resistance of $9\,\Omega $ is used in the external circuit of the cell, the balance point shifts to $300\,cm$. Determine the internal resistance of the cell.
Answer
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Hint: Use the relation of the potentiometer given below and substitute the formula of the potential difference in the circuit in it. Substitute the known values of the parameters in the obtained relation to find the value of the internal resistance of the circuit.
Useful formula:
(1) In the potentiometer, the following relation exists.
$ \dfrac{E}{{{V_{\operatorname{int} }}}} = \dfrac{{{l_1}}}{{{l_2}}}$
Where $E$ is the emf, ${V_{\operatorname{int} }}$ is the internal voltage, ${l_1}$ is the length of the initial null point and ${l_2}$ is the final null point in the circuit.
(2) The potential difference is given by
$V = \dfrac{{ER}}{{R + r}}$
Where $V$ is the potential difference, $R$ is the external resistance and $r$ is the internal resistance of the circuit.
Complete step by step solution:
It is given that the
Length of the initial null point, ${l_1} = 350\,cm$
Length of the final null point, ${l_2} = 300\,cm$
The external resistance, $R = 9\,\Omega $
Using the relation (1),
$\dfrac{E}{{{V_{\operatorname{int} }}}} = \dfrac{{{l_1}}}{{{l_2}}}$
Substituting the formula (2) in the above step.
$\dfrac{E}{{\dfrac{{ER}}{{R + r}}}} = \dfrac{{{l_1}}}{{{l_2}}}$
By cancelling the similar terms,
$\dfrac{{R + r}}{R} = \dfrac{{{l_1}}}{{{l_2}}}$
By cross multiplying both the sides of the equation.
$\left( {R + r} \right){l_2} = {R_{}}{l_1}$
By grouping the similar terms in one side of the equation.
$r = \dfrac{{\left( {{l_1} - {l_2}} \right)R}}{{{l_2}}}$
Substituting the known values in the above equation.
$r = \dfrac{{\left( {350 - 300} \right)9}}{{300}}$
By performing the simple arithmetic operation.
$r = 1.5\,\Omega $
Hence the internal resistance of the cell is obtained as $1.5\,\Omega $.
Note: Potentiometer is the device with the three variable pins which may be of resistor and the wiper. It also works as the voltage divider or the rheostat. Remember the formula of the potentiometer relation which has the ratio of the initial length to the final length.
Useful formula:
(1) In the potentiometer, the following relation exists.
$ \dfrac{E}{{{V_{\operatorname{int} }}}} = \dfrac{{{l_1}}}{{{l_2}}}$
Where $E$ is the emf, ${V_{\operatorname{int} }}$ is the internal voltage, ${l_1}$ is the length of the initial null point and ${l_2}$ is the final null point in the circuit.
(2) The potential difference is given by
$V = \dfrac{{ER}}{{R + r}}$
Where $V$ is the potential difference, $R$ is the external resistance and $r$ is the internal resistance of the circuit.
Complete step by step solution:
It is given that the
Length of the initial null point, ${l_1} = 350\,cm$
Length of the final null point, ${l_2} = 300\,cm$
The external resistance, $R = 9\,\Omega $
Using the relation (1),
$\dfrac{E}{{{V_{\operatorname{int} }}}} = \dfrac{{{l_1}}}{{{l_2}}}$
Substituting the formula (2) in the above step.
$\dfrac{E}{{\dfrac{{ER}}{{R + r}}}} = \dfrac{{{l_1}}}{{{l_2}}}$
By cancelling the similar terms,
$\dfrac{{R + r}}{R} = \dfrac{{{l_1}}}{{{l_2}}}$
By cross multiplying both the sides of the equation.
$\left( {R + r} \right){l_2} = {R_{}}{l_1}$
By grouping the similar terms in one side of the equation.
$r = \dfrac{{\left( {{l_1} - {l_2}} \right)R}}{{{l_2}}}$
Substituting the known values in the above equation.
$r = \dfrac{{\left( {350 - 300} \right)9}}{{300}}$
By performing the simple arithmetic operation.
$r = 1.5\,\Omega $
Hence the internal resistance of the cell is obtained as $1.5\,\Omega $.
Note: Potentiometer is the device with the three variable pins which may be of resistor and the wiper. It also works as the voltage divider or the rheostat. Remember the formula of the potentiometer relation which has the ratio of the initial length to the final length.
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