Question

For the circuit shown in figure, initially key $K_1$ is closed and key $K_2$ is kept open for a long time. Then $K_1$ was opened and $K_2$ was closed, what will be the charge on capacitors $C_2\&C_3$ finally? $[C=1\mu F]$

• A. $18\mu C,0\mu C$
• B. $0\mu C,18\mu C$
• C. $9\mu C,9\mu C$
• D. $4.5\mu C,4.5\mu C$

Answer 1

The correct option is C $9\mu C,9\mu C$

Let $Q_1$, $Q_2$ and $Q_3$ be the charge on the capacitors, when $K_1$ is closed and $K_2$ is open
Initially: $V\propto \frac{1}{C}$ and $V_1+V_2=E$
$\Rightarrow V_1=3V$ and $V_2=6V$
$\therefore Q_1=C_1{V}_1=6\times 1\times 3=18\mu C$
$Q_2=18\mu C$ and $Q_3=0$
When $K_1$ is open and $K_2$ is closed, let $Q_2^{\prime }$ and $Q_3^{\prime }$ be the final charge on $C_2$ and $C_3$ respectively.
$Q_2=Q_2^{\prime }+Q_3^{\prime }$
$\Rightarrow Q_2=C_{2}V+C_{3}V\Rightarrow V=\frac{Q_2}{C_2+C_3}=\left(\frac{18}{6\times 1}\right)V=3V$
$Q_2^{\prime }=3\times 1\times 3=9\mu C$
$Q_3^{\prime }=3\times 1\times 3=9\mu C$