Equivalent capacitance of series capacitors.

$\overline{)\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{e}\mathbf{q}}}{\mathbf{=}}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{1}}}{\mathbf{+}}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{2}}}{\mathbf{+}}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{3}}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{+}}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{n}}}}$

**The equivalent capacitance of the combination is less than the capacitance of any of the capacitors.**

When two or more capacitors are connected in series across a potential difference:

a) the potential difference across the combination is the algebraic sum of the potential differences across the individual capacitors.

b) the equivalent capacitance of the combination is less than the capacitance of any of the capacitors.

c) each capacitor carries the same amount of charge.

d) All of the above choices are correct.

e) None of the above choices are correct.

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