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A capacitor of 2 muF is charged to a pot...

A capacitor of `2 muF` is charged to a potential of 4V using a battery, and then the battery is disconnected and the changed capacior is connected to an uncjharged caspacitor of `4 muF` capacitance. When the equilibrium is established the total energy stored in the capacitors is

A

`16 muJ`

B

`(16)/(3)muJ`

C

`(32)/(3)muJ`

D

`(32)/(9)muJ`

Text Solution

Verified by Experts

The correct Answer is:
B
`V=(C_(1)V_(1)+C_(2)V_(2))/(C_(1)+C_(2))`
`U_(F)=1/2(C_(1)+C_(2))V^(2)`
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