A capacitor of capacitance C1 = 1μF charged up to 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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40 μC

50 μC

60 μC

30 μC

answer is C.

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Detailed Solution

Q1=C1V1 =110 μCQ1=10-61+65V=2.2V x 10-6110 μC = V x 2.2 x10-6 ∴V = 50 VQ = C1V = 50 μCcharge flow in the wire= Q1 - Q = (110 - 50) μC = 60 μC