Correct Answer - A
If `2 muF` where not presen between `B` and `D`, then potential drop across upper `4muF` will be less than potential drop across lower ` 2muF`,i.e.,
.
Now `V_(A)-V_(B)ltV_(A)-V_(D)rArrV_(B)gtV_(D)`
Now if `2muF` is connected between `B` and `D`, charge will flow from `B` and `D` and Finally `V_(B)gtV_(D)`.
`V_(A)=20 V`. At junction `B`,
`q_(2)=q_(1)+q_(3)`
or `(V_(A)-V_(B))=(V_(B)-V_(D))2+(V_(B)-V_(C))2`
or `4(V_(A)-V_(B))+2(V_(D)-V_(B))=2V_B`
At junction D:
`q_(2)=q_(1)+q_(3)`
or `4(V_(D)-V_(C))=2(V_(A)-V_(D))+2(V_(B)-V_(D))`
or `2(V_(A)-V_(D))+2(V_(B)-V_(D))=4V_(D)`.