Question

Each capacitance shown in the figure is in $\mu F$. Find the charge on $6\mu F$ in $\mu C$:

Answer 1

As $3\mu F$ and group $(6,1.5,1.5)\mu F$ are in parallel so voltage across $(6,1.5,1.5)\mu F$ is also $V=2V$.
now equivalent capacitance for $(6,1.5,1.5)\mu F$ is $C_{eq}=\frac{6(1.5+1.5)}{6+(1.5+1.5)}=\frac{18}{9}=2\mu F$
$\therefore Q_{eq}=C_{eq}V=2\times 2=4\mu C$
As $6\mu F$ and $(1.5,1.5)\mu F$ are in series so charge on $6\mu F$ and $(1.5,1.5)\mu F$ is equal to the $Q_{eq}$.
Thus charge on $6\mu F$ is $Q_{eq}=4\mu C$