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) =30 muC`
B. The final charge on `C_(2) = 20 muC`
C. The final charge on `C_(2) = zero`
D. The final charge on `C_(3) = 15 muC`

Answer

A, C, D

Explanation

After the switch is closed, let a charge `Q` be driven by the battery. Now, new charges on capacitors would become,
`Q_(1)=(20+Q)muC ,Q_(2)=(10-Q)muC`
`Q_(3)=(5+Q)muC (20+Q)/(4)-(10-Q)/(2)+(5+Q)/(6) =10`
`60+3Q-60+6Q +10+2Q=120`
`11Q =110 , Q=10 muC`.