Two capacitors of 10 pF and 20 pF are connected to 200 V and 100 V sources, respectively. If they are
connected by the wire, then what is the common potential of the capacitors? [KCET 2014]

Given, ${\mathrm{C}}_1=10\mathrm{pF}=10\times {10}^{-12} \mathrm{F}$

           ${\mathrm{C}}_2=20\mathrm{pF}=20\times {10}^{-12} \mathrm{F}$

           V₁ = 200 V and V₂ = 100 V

where, C₁ = capacitance of 1st capacitor,

            C₂ = capacitance of 2nd capacitor,

            V₁ = voltage across 1st capacitor

and      V₂ = voltage across 2nd capacitor.

We know that, ${\mathrm{V}}_1=\frac{{\mathrm{q}}_1}{{\mathrm{C}}_1}\text{ and }{\mathrm{V}}_2=\frac{{\mathrm{q}}_2}{{\mathrm{C}}_2} \Rightarrow {\mathrm{q}}_1={\mathrm{V}}_1{\mathrm{C}}_1$      …(i)

                         q₂ = V₂C₂

So, common potential of capacitors,

                        $\mathrm{V}=\frac{{\mathrm{q}}_1+{\mathrm{q}}_2}{{\mathrm{C}}_1+{\mathrm{C}}_2}=\frac{{\mathrm{V}}_1{\mathrm{C}}_1+{\mathrm{V}}_2{\mathrm{C}}_2}{{\mathrm{C}}_1+{\mathrm{C}}_2}$

                           $=\frac{200\times 10\times {10}^{-12}+100\times 20\times {10}^{-12}}{10\times {10}^{-12}+20\times {10}^{-12}}$

                           $=\frac{200\times 10+100\times 20}{10+20}$

                           $=\frac{2000+2000}{30}=\frac{4000}{30}=133.3 \mathrm{V}$