When a cell of $E.M.F.$ $E_1$ is connected to a potentiometer wire,the balancing length is $l_1$. Another cell of $E.M.F.$ $E_2$ $(E_1 > E_2)$ is connected such that the two cells oppose each other,then the balancing length is $l_2$. The ratio $E_1 : E_2$ is
- A$\frac{l_1}{l_1+l_2}$
- B$\frac{l_1+l_1-l_2}{l_1-l_2}$
- C$\frac{l_1+l_2}{l_1}$
- D$\frac{l_1+l_2}{l_1-l_2}$
Explore More
Similar Questions
$A$ $10 \ m$ long wire of resistance $20 \ \Omega$ is connected in series with a battery of e.m.f. $3 \ V$ and a resistance of $10 \ \Omega$. The potential gradient along the wire in $V/m$ is
$A$ potentiometer has a uniform potential gradient. The specific resistance (resistivity) of the material of the potentiometer wire is $10^{-7} \, \Omega \cdot m$, the current passing through it is $0.1 \, A$, and the cross-sectional area of the wire is $10^{-6} \, m^2$. The potential gradient along the potentiometer wire is:
MediumAIPMT 2001
View SolutionIf the length of a potentiometer wire is increased,then the length of the previously obtained balance point will
Easy
View Solution$A$ potentiometer consists of a wire of length $4\, m$ and resistance $10\,\Omega$. It is connected to a cell of $e.m.f.$ $2\, V$. The potential difference per unit length of the wire will be ............. $V/m$.
In the primary circuit of a potentiometer,the current is $0.2 \ A$. The resistivity and cross-sectional area of the potentiometer wire are $4 \times 10^{-7} \ \Omega \cdot m$ and $8 \times 10^{-7} \ m^2$ respectively. The potential gradient will be ......... $V/m$.
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo