Question

A capacitor C is charged to a potential V by a battery of emf V. It is then disconnected from the battery and again connected with it's polarity reversed to the battery.

• A. The work done by the battery is $2C{V}^2$
• B. The total charge that passes through the battery is 2CV.
• C. The initial energy of the capacitor is greater than the final energy of the capacitor.
• D. The heat generated in the circuit is $2C{V}^2$.

Answer 1

Answer: (A), (B), (D)

The work done by the battery is $2C{V}^2$
The total charge that passes through the battery is 2CV.
The heat generated in the circuit is $2C{V}^2$.

The capacitor is charged twice by $\left(\Delta q\right)=CV$
Work done by the battery $=\left(2\Delta q\right)V=\left(2CV\right)V=2C{V}^2$
Charge that passes through the battery after it is disconnected is $2\Delta q=2CV$
As energy stored in capacitor does not change, So work done by battery $=$ heat generated.