Question

A $10\mu F$ capacitor charged to $20V$ is connected across a $15V$ battery through a switch such that positive terminal of battery is connected to negative charged plate of capacitor. When switch is closed, work done by battery of $15V$ in long time will be

• A. $2250\mu J$
• B. $-3000\mu J$
• C. $30000\mu J$
• D. $5250\mu J$

Answer 1

The correct option is D $5250\mu J$


Intial charge $Q$ on capacitor is $Q=CV=10\times 20=200\mu C$

On closing switch final charge on capacitor is $q=C\epsilon =10\times 15\mu C=150\mu C$
So charge flown through battery is $\Delta q=q+Q=350\mu C$
So work of battery is $W=\epsilon \Delta q$
$W=15\times 350\mu J=5250\mu J$