Question

A $4\mu F$ capacitor is charged by a $6$ volts supply. It is then disconnected from the supply and connected to another uncharged $2\mu F$ capacitor. Now the energy stored in both the capacitors is

• A. $48\times {10}^{-6}J$
• B. $32\times {10}^{-6}J$
• C. $16\times {10}^{-6}J$
• D. $56\times {10}^{-6}J$

Answer 1

The correct option is A $48\times {10}^{-6}J$

Net charge is $Q=4\times 6+0=24\mu C=24\times {10}^{-6}C$

Charge will remain conserved as the battery is disconnected.

So Energy $=\frac{Q^2}{2C}$

Where $C$ is net capacity.

We know when two capacitor are connected without any battery then they are in parallel and in parallel combination the net resistance is the sum of the two. So $C=4+2=6\mu F=6\times {10}^{-6}F$

So energy $E=48\times {10}^{-6}J$