Three parallel plate capacitors `C_(1) = 4 muF,C_(2) =2mu F,C_(3) = 6muF` with respective charges `q_(1) = 20 muC,q_(2) = 10 mu C,q_(3) =5 muC` are connected in series with a battery of `emf 10 V` through an open switch as shown. Now the switch is closed and steady state is reached.
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Options

A. The final charge on `C_(1) = (210)/(11) mu C`
B. The final charge on `C_(2) = (120)/(11) muC`
C. The final charge on `C_(2) = zero`
D. The final charge on `C_(3) = (45)/(11) muC`

Answer

A, B, D

Explanation

Let the final charges on the capacitors be `q_(1),q_(2) &q_(3)`
`(q_(1))/(4)+(q_(2))/(2)+(q_(3))/(6)=0`....(1)
`q_(1)+q_(2)=-30` ....(2)
`q_(1)-q_(3)=15` .....(3)
solve these equations and get the answers.