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
Verified
120.6k+ 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
Structure of Atom: Key Models, Subatomic Particles, and Quantum Numbers
Young's Double Slit Experiment Step by Step Derivation
Difference Between Circuit Switching and Packet Switching
Difference Between Mass and Weight
JEE Main Participating Colleges 2024 - A Complete List of Top Colleges
JEE Main Maths Paper Pattern 2025 – Marking, Sections & Tips
Trending doubts
JEE Mains 2025: Check Important Dates, Syllabus, Exam Pattern, Fee and Updates
JEE Main Login 2045: Step-by-Step Instructions and Details
JEE Main Chemistry Question Paper with Answer Keys and Solutions
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking
JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions
JEE Main Chemistry Exam Pattern 2025
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
Learn About Angle Of Deviation In Prism: JEE Main Physics 2025
JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics
Dual Nature of Radiation and Matter Class 12 Notes CBSE Physics Chapter 11 (Free PDF Download)
Electric field due to uniformly charged sphere class 12 physics JEE_Main