When a parallel plate capacitor is fully char | vedclass

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

Question

(A) Let the capacitance of the capacitor be $C$ and the $EMF$ of the battery be $V$.When the capacitor is fully charged,the charge stored on the plate

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

(A) Let the capacitance of the capacitor be $C$ and the $EMF$ of the battery be $V$.When the capacitor is fully charged,the charge stored on the plates is $Q = CV$.The energy stored in the capacitor is given by $U = \frac{1}{2} CV^2$.The work done by the battery in charging the capacitor is $W = Q 	imes V = (CV) 	imes V = CV^2$.The ratio of the energy stored in the capacitor to the work done by the battery is $\frac{U}{W} = \frac{\frac{1}{2} CV^2}{CV^2} = \frac{1}{2} = 0.5$.

When a parallel plate capacitor is fully charged by a battery,the potential difference between the two plates becomes equal to the electromotive force $(EMF)$ of the battery. What is the ratio of the energy stored in the capacitor to the work done by the battery?

Similar Questions

$A$ parallel plate capacitor has a uniform electric field $E$ in the space between the plates. If the distance between the plates is $d$ and the area of each plate is $A,$ the energy stored in the capacitor is

How does a capacitor store energy? Obtain the formula for the energy stored in a capacitor.

The amount of work done in increasing the voltage across the plates of a capacitor from $5 V$ to $10 V$ is $W$. The work done in increasing it from $10 V$ to $15 V$ will be

The energy required to charge a parallel plate capacitor with plate separation $d$ and plate area $A$ such that the uniform electric field between the plates is $E$,is

When an uncharged capacitor is connected to a battery to charge it fully,

Vedclass Test Series

Exam Paper Generator

Online Exam Module