Question

A capacitor of capacitance $C_1=1\mu F$ charged upto a voltage $V=110V$ is connected in parallel to the terminals of a circuit consisting of two uncharged capacitors connected in series and possessing capacitances $C_2=2\mu F$ and $C_3=3\mu F$. Then, the amount of charge that will flow through the connecting wires is :

• A. $40\mu C$
• B. $50\mu C$
• C. $60\mu C$
• D. $110\mu C$

Answer 1

Answer: (C)

$60\mu C$

Initial charge on $C_1$ is $Q_1=C_{1}V=110\mu C$

Let $x$ charge flow through wires.

$\frac{Q_1-x}{C_1}=\frac{x}{C_{eq}}$

$C_{eq}=\frac{C_2\times C_3}{C_2+C_3}=\frac{2\times 3}{2+3}=\frac{6}{5}\mu F$

$\frac{Q_1-x}{C_1}=\frac{x}{C_{eq}}$

$\frac{110-x}{1}=\frac{5x}{6}$

$\Rightarrow x=60\mu C$