A circuit has a section AB shown in figure. The emf of the cell is 10 V and the capacitors have capacitances ${\mathrm{C}}_1=1\mathrm{\mu F}$ and ${\mathrm{C}}_2=2\mathrm{\mu F}$. The potential difference ${\mathrm{V}}_{\mathrm{AB}}=5\mathrm{V}.$ Find the charges on the capacitors (in $\mathrm{\mu }$C).

Let the charge on the left plate of the capacitor is q, hence the charge on right plate of capacitor C₂ will be -q as charge supplied by negative plate of capacitor will be -q. 

Moving from point A in the section of circuit AB 

${\mathrm{V}}_{\mathrm{A}}-\frac{\mathrm{q}}{1}+10-\frac{\mathrm{q}}{2}={\mathrm{V}}_{\mathrm{B}}$

${\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}=\mathrm{q}+\frac{\mathrm{q}}{2}-10\Rightarrow 5=\frac{3}{2}\mathrm{q}-10$

Hence the charges on the capacitors $\mathrm{q}=10\mathrm{\mu C}$