(C) The total resistance in the primary circuit is $R_{total} = R_{wire} + R_{series} = 8 \ \Omega + 242 \ \Omega = 250 \ \Omega$. The current flowing

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(C) The total resistance in the primary circuit is $R_{total} = R_{wire} + R_{series} = 8 \ \Omega + 242 \ \Omega = 250 \ \Omega$. The current flowing through the potentiometer wire is $I = \frac{V}{R_{total}} = \frac{2.5 \ V}{250 \ \Omega} = 0.01 \ A$. The potential drop across the entire wire is $V_{wire} = I \times R_{wire} = 0.01 \ A \times 8 \ \Omega = 0.08 \ V$. The potential gradient $k$ is the potential drop per unit length: $k = \frac{V_{wire}}{L} = \frac{0.08 \ V}{2.5 \ m} = 0.032 \ V/m$. The potential difference across a length $l = 20 \ cm = 0.2 \ m$ is $V' = k \times l = 0.032 \ V/m \times 0.2 \ m = 0.0064 \ V = 6.4 \ mV$.

Class 12 Physics Current Electricity Potentiometer

Medium

The length of a potentiometer wire is $2.5 \ m$ and its resistance is $8 \ \Omega$. $A$ cell of negligible internal resistance and emf of $2.5 \ V$ is connected in series with a resistance of $242 \ \Omega$ in the primary circuit. The potential difference between two points separated by a distance of $20 \ cm$ on the potentiometer wire is (in $mV$)

TS EAMCET 2025 All TS EAMCET

A
$1.6$
B
$4.8$
C
$6.4$
D
$3.2$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Physics Questions

Chapter

Current Electricity

Subtopic

Potentiometer

Previous Year

TS EAMCET 2025 Paper All TS EAMCET years →

$A$ voltmeter reads the potential difference across the terminals of an old battery as $1.2 \, V$,while a potentiometer reads $1.4 \, V$. The internal resistance of the battery is $40 \, \Omega$. What is the resistance of the voltmeter in $\Omega$?

Medium

View Solution

In a potentiometer experiment,when two cells of emfs $E_1$ and $E_2$ $(E_2 > E_1)$ are connected in series,the balancing length is $160 \ cm$. If one of the cells is reversed,the balancing length decreases by $75 \%$. If $E_1 = 1.2 \ V$,then $E_2 =$ (in $V$)

MediumAP EAMCET 2025

View Solution

In a potentiometer experiment,the balancing length is $8 \ m$ when two cells $E_1$ and $E_2$ are joined in series. When two cells are connected in opposition,the balancing length is $4 \ m$. The ratio of the e.m.f. of the two cells $\left(\frac{E_1}{E_2}\right)$ is

EasyMHT CET 2023

View Solution

In a potentiometer experiment,it is found that no current passes through the galvanometer when the terminals of the cell are connected across $52 \ cm$ of the potentiometer wire. If the cell is shunted by a resistance of $5 \ \Omega$,a balance is found when the cell is connected across $40 \ cm$ of the wire. Find the internal resistance of the cell ........... $\Omega$.

MediumJEE MAIN 2018

View Solution

$A$ cell of internal resistance $1.5\,\Omega$ and of $e.m.f.$ $1.5\,V$ balances at $500\,cm$ on a potentiometer wire. If a wire of $15\,\Omega$ is connected between the balance point and the cell,then the balance point will shift:

Easy

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Similar Questions

$A$ voltmeter reads the potential difference across the terminals of an old battery as $1.2 \, V$,while a potentiometer reads $1.4 \, V$. The internal resistance of the battery is $40 \, \Omega$. What is the resistance of the voltmeter in $\Omega$?

In a potentiometer experiment,when two cells of emfs $E_1$ and $E_2$ $(E_2 > E_1)$ are connected in series,the balancing length is $160 \ cm$. If one of the cells is reversed,the balancing length decreases by $75 \%$. If $E_1 = 1.2 \ V$,then $E_2 =$ (in $V$)

In a potentiometer experiment,the balancing length is $8 \ m$ when two cells $E_1$ and $E_2$ are joined in series. When two cells are connected in opposition,the balancing length is $4 \ m$. The ratio of the e.m.f. of the two cells $\left(\frac{E_1}{E_2}\right)$ is

$A$ cell of internal resistance $1.5\,\Omega$ and of $e.m.f.$ $1.5\,V$ balances at $500\,cm$ on a potentiometer wire. If a wire of $15\,\Omega$ is connected between the balance point and the cell,then the balance point will shift:

Vedclass Test Series

Exam Paper Generator

Online Exam Module