$A$ circuit has self-inductance $L$ and carries a current $I$. To prevent sparking when the circuit is switched off,a capacitor which can withstand potential difference $V$ is used. The least capacitance is
- A$\frac{IV}{L}$
- B$L\left(\frac{V}{I}\right)^2$
- C$L\left(\frac{I}{V}\right)^2$
- D$\frac{LI}{V}$
Explore More
Similar Questions
If $N$ is the number of turns in a coil,the value of self-inductance varies as
MediumAIPMT 1993
View SolutionWhen the current through an inductor is changed from $2 \,A$ to $6 \,A$ in time $2 \,s$, the emf induced in it is $3 \,V$. Then the inductance of the inductor is (in $\,H$)
The back $e.m.f.$ induced in a coil,when current changes from $1 \ A$ to $0 \ A$ in $1 \ ms$,is $4 \ V$. The self-inductance of the coil is:
Medium
View SolutionIf $R$ and $L$ represent respectively resistance and self-inductance,which of the following combinations has the dimensions of frequency?
Medium
View SolutionAn $e.m.f.$ of $5 \,V$ is produced by a self-inductance when the current changes at a steady rate from $3 \,A$ to $2 \,A$ in $1 \,ms$. The value of self-inductance is:
Vedclass 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