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the energy stored in the capacitor is...


the energy stored in the capacitor is

A

`12muJ`

B

`24muJ`

C

`36muJ`

D

`48muJ`

Text Solution

Verified by Experts

The correct Answer is:
B

We have
`V_(A)-V_(B)=ixx(4+1)`
`8-3=ixx5`
`5=ixx5`
`i=1A`
`V_(A)-V_(P)=4xx1`
`8-V_(P)=4,V_(P)=4` volt
Now `V_(C)=0`. So Energy stored in the capacitor
is `xi=(1)/(2)xx3xx16=24muJ`
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