
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
143.1k+ views
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.
Recently Updated Pages
How to find Oxidation Number - Important Concepts for JEE

How Electromagnetic Waves are Formed - Important Concepts for JEE

Electrical Resistance - Important Concepts and Tips for JEE

Average Atomic Mass - Important Concepts and Tips for JEE

Chemical Equation - Important Concepts and Tips for JEE

Concept of CP and CV of Gas - Important Concepts and Tips for JEE

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Degree of Dissociation and Its Formula With Solved Example for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Physics Average Value and RMS Value JEE Main 2025

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main
