Question

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 open. The key `S_(1)` is closed and teh capacitors are charged. If now key `S_(2)` be closed, the charge that will flow across this key is.
A. `2.4 muC`
B. `0.4 muC`
C. `0.2 muC`
D. `1.2 muC`

Answer 1

Correct Answer - A
Before `S_(2)` is closed, `C_(1)` and `C_(3)` are in series and hence the total charge on their common plate is `(+Q-Q)=0`. Similarly for the plates joining `C_(2)` and `C_(4)`. However when `C_(2)` and `C_(4)` is closed, the charges read just to the values calculated in problem `86`. Hence total charge on the linking plates of `C_(1)` and `C_(3)` is `(-8.4 muC + 10.8 muC)=+ 2.4 muC` while the total charge on the linking plates of `C_(2)` and `C_(4)` is `(-16.8 muC + 14.4 muC = -2.4 muC` which shows that a charge of `2.4 muC` has crossed the key `S_(2)` from `(C_(2),C_(4))` to `(C_(1),C_(3))`.