Question

The figure shows a capacitive circuit, with a switch $S_w$. The charge is $150\times x\mu C$ flows through the switch when it is closed. The value of x is _____

Answer 1

When switch is open, capacitors (6,3) on left are in series and capacitors (3,6) on right series.

Thus, potential across each $3\mu F$ is $V_3=\frac{6}{3+6}\left(200\right)=\frac{400}{3}V$

and potential across each $6\mu F$ is $V_6=\frac{3}{3+6}\left(200\right)=\frac{200}{3}V$

Charge on each $3\mu F$ is $q_3=3\times \frac{400}{3}=400\mu C$ and charge on each $6\mu F$ is $q_6=2\times \frac{200}{3}=400\mu C$

When switch is closed, top capacitors (6,3) are in parallel and bottom (3,6) capacitors are in parallel and these group are in series. i.e, $9\mu F$ and $9\mu F$ are in series.

Voltage across $9\mu F$ is $V_9=\frac{9}{9+9}\times 200=100V$

As top capacitors (6,3) are in parallel so $V_9$ is also potential across each capacitors.

now, $q_3^{\prime }=3\times 100=300\mu C$ and $q_6^{\prime }=6\times 100=600\mu C$

thus, $\Delta q_3=400-300=100\mu C$ and $\Delta q_6=600-400=200\mu C$

Net charge flow through switch is $\Delta q=\Delta q_3+\Delta q_6=100+200=300\mu C$