InterviewSolution
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InterviewSolution
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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 key `S_(2)` is closed. Then the key `S_(1)` is now closed. Then the charges on the capacitors are.A. `q_(1)=q_(2)=24 muC,q_(3)= q_(4) = 12 mu C`B. `q_(1)=q_(2) =12 muC,q_(3)=q_(4) =24 muC`C. `q_(1)=10.8 muC,q_(2) = 14.4 muC, 3q_(3) =2q_(4),2q_(4) =25.2 muC`D. `2q_(1)=q_(2)=16.8 muC,4q_(3) =3q_(4) = 43.3 mu C` |
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Answer» Correct Answer - D If initially `S_(2)` is closed, `C_(1)` and `C_(2)` are in parallel, as also `C_(3)` and `C_(4)`. Effectively therefore `(C_(1)+C_(2))` and `(C_(2)+C_(4))` are in series when key `S_(1)` is closed. If `V` be the potential of the four common plates joined by `S_(2)`, when `S_(1)` is closed. Hence `q_(1)=(12-3.6)1=8.4 muC`, `q_(2)=(12-3.6)2 =16.8 muC` `q_(3)=(3.6)3=10.8 muC,q_(4)=(3.6)4=14.4 muC`. |
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