Question

A capacitor is charged by connecting a battery across its pates. It stores energy $u$. How the battery is disconnected and another identical capacitor is connected across it, then the energy by both the capacitors of the system will be:-

• A. $u$
• B. $\frac{u}{4}$
• C. $2u$
• D. $\frac{3}{2}u$

Answer 1

The correct option is B $\frac{u}{4}$

$\begin{array}{c}Letthevoltagecrossthecapacitorbevafterching\\ Thus,u=\frac{1}{2}c{v}^{2}wherecisitscapaci\tan ce.\\ so,v=\sqrt{\frac{2u}{c}}\\ whereanotherindenticalcapacitorisconnedctedinparallel\\ Atequilibrium\\ \frac{Q-q}{c}=\frac{q}{c}\\ \Rightarrow \frac{2q}{c}=\frac{Q}{c}\\ \Rightarrow q=\frac{Q}{2}\\ \Rightarrow c{v}^{\prime }=\frac{cv}{2}\\ \Rightarrow v^{\prime }=\frac{v}{2}\\ sofinalvoltagebecomes\left(\frac{v}{2}\right)\\ Energyinboththeeapacitorsnow=2\times \frac{1}{2}c{v}^2\\ =c\times {\left(\frac{v}{2}\right)}^2\\ =\frac{c\times \left(\frac{2u}{c}\right)}{4}=\frac{u}{2}\\ Energyinonecapacitor=\left(\frac{u}{2}\right)\frac{1}{2}=\frac{u}{4}\\ Hence,optionBisthecorrectanswer.\end{array}$