$A$ capacitor of capacitance $C_0$ is charged to a potential $V_0$ and is connected with another capacitor of capacitance $C$ as shown. After closing the switch $S$,the common potential across the two capacitors becomes $V$. The capacitance $C$ is given by
- A$\frac{C_0(V_0 - V)}{V_0}$
- B$\frac{C_0(V - V_0)}{V_0}$
- C$\frac{C_0(V + V_0)}{V}$
- D$\frac{C_0(V_0 - V)}{V}$
Explore More
Similar Questions
The events $A$ and $B$ have probabilities $0.25$ and $0.50$,respectively. The probability that both $A$ and $B$ occur simultaneously is $0.14$. Then,the probability that neither $A$ nor $B$ occurs is:
$A$ particle of mass $m$ and charge $q$ is placed at rest in a uniform electric field $E$ and then released. The $KE$ attained by the particle after moving a distance $y$ is
The binding energy per nucleon versus mass number curve for nuclei is shown in the figure. $W, X, Y$ and $Z$ are four nuclei indicated on the curve. The process that would release energy is
$A$ black body radiates energy at the rate of $E \text{ W/m}^2$ at a high temperature $T \text{ K}$. When the temperature is reduced to $\left(\frac{T}{2}\right) \text{ K}$,the radiant energy is
When $KMnO_4$ solution is added to hot oxalic acid solution,the decolourisation is slow in the beginning but becomes instantaneous after some time. This is because
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