Question

When a battery of emf E volts is connected across a capacitor of capacitance C, then after some time the potential difference between the plates of the capacitor becomes equal to the battery voltage. The ratio of the work done by the battery and the energy stored in the capacitor when it is fully charged is

Answer 1

Answer:

energy stored in capacitor = 1/2 QV

where's work done by battery to charge the capacitor to V voltage = QV

then ratio = 1/2QV / QV

= 1/2

ratio = 1:2

hope it helps