In the network shown, each capacitor has capacitance is 2μF.Then the charge flown through the battery after the switch S is closed, is
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a
6 μC
b
4 μC
c
8 μC
d
9 μC
answer is B.
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Detailed Solution
Initial charge stored in the system Qi=C2×E=22×12 μC=12 μC.Final charge stored in the system,Qf=C×2C(C+2C)×E=2CE3=2×2×123 μC=16 μC.Charge flown through the system,∆Q=(16-12) μC =4 μC.