Question

In figure, the battery has a potential difference of 20 V. Find the charge on capacitor 1 (in $\mu C$) :

Answer 1

The corresponding equivalent circuits are shown in figure. The net capacitance for top part is $C_t=2+\frac{4\times 4}{4+4}+2=6\mu F$

The net capacitance for bottom part is $C_b=3+3=6\mu F$

Now $C_t,C_b$ are in series and they have equal capacitance so potential across them will be same i.e $V_t=V_b$

Thus, $V_t+V_b=20$ or $V_t=V_b=10V$

As the capacitors in the bottom part are in parallel so they have same potential i.e 10 V. Thus, potential across $C_1$ will be $V_b=10V$

So, the charge on $C_1$ is $Q_1=C_1{V}_b=3\times 10=30\mu C$