Explain series combination of capacitors. Derive the formula for equivalent capacitance in each combination.

In series combination second plate of first capacitor is connected to first plate of second capacitor and second plate of second capcitor is connected to first plate of third capacitor and so on first plate of first capacitor and second plate of last capacitor is connected to opposite terminals of battery. Let c₁,c₂,c₃ are three capacitors connected in series. In series combination charge remains constant, potential in series combination.
$\begin{array}{l}\mathrm{V}={\mathrm{V}}_1+{\mathrm{V}}_2+{\mathrm{V}}_3\text{ but }\\ {\mathrm{V}}_1=\frac{\mathrm{Q}}{{\mathrm{c}}_1},{\mathrm{V}}_2=\frac{\mathrm{Q}}{{\mathrm{c}}_2},{\mathrm{V}}_3=\frac{\mathrm{Q}}{{\mathrm{c}}_3}\Rightarrow \mathrm{V}=\frac{\mathrm{Q}}{{\mathrm{c}}_1}+\frac{\mathrm{Q}}{{\mathrm{c}}_2}+\frac{\mathrm{Q}}{{\mathrm{c}}_3}\end{array}$

$\begin{array}{r}\frac{\mathrm{Q}}{\mathrm{c}}=\frac{\mathrm{Q}}{{\mathrm{c}}_1}+\frac{\mathrm{Q}}{{\mathrm{c}}_2}+\frac{\mathrm{Q}}{{\mathrm{c}}_3}\\ \frac{1}{\mathrm{c}}=\frac{1}{{\mathrm{c}}_1}+\frac{1}{{\mathrm{c}}_2}+\frac{1}{{\mathrm{c}}_3}\end{array}$