Capacitors in Series

When capacitors are connected in series, the total capacitance is less than any one of the series capacitors’ individual capacitances. If two or more are connected, the overall effect is that of a single (equivalent) capacitor having the sum total of the plate spacings of the individual capacitors. This example will determine the equivalent capacitance for series connections of capacitors. As you will see, equivalent capacitance is in direct analogy with equivalent conductance.

Lets start with the series connection of N capacitors

It can be written as this:

v(t) = \frac{1}{C_{1}}\int_{t0}^{t} i dt + v_{1}(t_0) + \frac{1}{C}_{2} \int_{t0}^{t} i dt + v_{2}(t_0) + … + \frac{1}{C}_{N}\int_{t0}^{t} i dt + v_{N}(t_0) = (\frac{1}{C}_{1}+\frac{1}{C}_{2}+ … +\frac{1}{C}_{N})\int_{t0}^{t} i dt + v_{1}(t_0) + v_{2}(t_0) + … + v_{N}(t_0)

or,

v(t) = (\sum_{n=1}^{N}\frac{1}{C}_{n}) \int_{t0}^{t} i dt + v(t_0)

In the equation above we see

v(t) = \frac{1}{C}_{s} \int_{t0}^{t} i dt + v(t_0)

where v(t_0) is the voltage on {C}_{s} at t = t_0.
The equivalent capacitance is the product over the sum of the two individual capacitances.