
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
233.1k+ 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.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

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

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2025-26

Understanding the Electric Field of a Uniformly Charged Ring

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

Derivation of Equation of Trajectory Explained for Students

Understanding Electromagnetic Waves and Their Importance

