A capacitor of capacitance C1=1 μF charged upto a voltage V = 110 V is connected in parallel to the terminals of a circuit consisting of two uncharged capacitors connected in series and possessing capacitances C2=2 μF and C3=3 μF. Then, the amount of charge that will flow through the connecting wires is
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a
40 μC
b
50 μC
c
60 μC
d
110 μC
answer is C.
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Detailed Solution
Initial charge on C1 is Q1=C1V=110 μCLet x charge flow through wireswhere Ceq=C2×C3C2+C3Solve to get x=60 μC