A capacitor of capacitance $C_{1} = 1 μ F$ 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 the $C_{2} = 2.0 μ F a n d C_{3} = 3.0 μ F$ capacitance . What charge will flow through the connecting wires in $μ C$ .

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answer is 60.

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Detailed Solution

$C_{23} = \frac{2 \times 3}{2 + 3} = 1.2 μ F$

$q_{total} = C_{1} V = 110 μ C$

Common potential in parallel is given by

$V = \frac{Total charg e}{Total capacity}$

$= \frac{110}{1 + 1.2} = 50 V$

$q_{23} = (C_{23}) V = 60 μ C$

So, this much charge flows through the switch.

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