Four capacitors `C_(1)(=1 muF),C_(2)(=2muF),C_(3)(=3muF)` and `C_(4)(=4 muF)` are connected in a network as shown in the diagram. The emf of the battery is `E = 12 V` and its internal resistance is negligible. The keys `S_(1)` and `S_(2)` can be independetly put on or off. Indicate the charge on the capacitors by `q_(1),q_(2),q_(3)` and `q_(4)` respectively and the potential drops across them by `V_(1),V_(2),V_(3)` and `V_(4)` respectively.

Initially both the keys are open. Then the key `S_(1)` is closed. Then the charges on the capacitors are.

Options

A. `q_(1)=q_(2)=16 mu C,q_(3)=q_(4) = 9 muC`
B. `q_(1) =q_(3) =16 muC,q_(2)=q_(4) = 9 mu C`
C. `q_(1)=q_(4) =9 muC, q_(2)=q_(3) =16 muC`
D. `q_(1)=q_(3)=9 muC,q_(2) =q_(4) =16 mu C`

Answer

D

Explanation

Refer to the circuit network in the problem.
With both keys `S_(1)` and `S_(2)` initially open, it now `S_(1)` is closed, the capacitance `C_(1)` and `C_(3)` are in series as also the capacitors `C_(2)` and `C_(4)`
Hence `q_(1)=q_(3)=(VC_(1)C_(3))/(C_(1)+C_(3))=(12 xx 1xx3)/((1+3))=9 muC`
Also `q_(2)=q_(4)=(VC_(2)C_(4))/(C _(2)+C_(4))=(12 xx 2xx4)/((2+4))=16 muC`.