Question

In figure, the battery has a potential difference of 20 V. The charge on capacitor 3 (in $\mu C$) is $10X$. The value of $X$ is:

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 top part are in parallel so they have same potential i.e 10 V. Now $C_3,4\mu F$ capacitors are in series and they have same capacitance so $V_{C_3}=V_4$. Thus, $V_{C_3}+V_4=V_t=10V$

or $V_{C_3}=V_4=5V$

So, the charge on $C_3$ is $Q_3=C_1{V}_{C_3}=4\times 5=20\mu C$