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A capacitor of capacitance C₁ = 1 $μ$ F 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 $μ$ F and C₃ = 3 $μ$ F. Then, the amount of charge that will flow through the connecting wires is

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60 $μ$ C

40 $μ$ C

30 $μ$ C

50 $μ$ C

answer is C.

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Detailed Solution

$Q_{1} = C_{1} V_{1} = 110 μC$

$Q_{1} = 10^{- 6} (1 + \frac{6}{5}) V = 2.2 V x 10^{- 6}$

110 $μ$ C = V x 2.2 x10^-6

$∴$ V = 50 V

Q = C₁V = 50 $μ$ C

charge flow in the wire

= Q₁ - Q = (110 - 50) $μ$ C = 60 $μ$ C

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