In this post we will be looking at Capacitors in series calculations. Capacitors in series are connected end to end so that the same charge (*Q*) appears across each of them. Meanwhile, in this connection, different potential difference (Pd) appears across each of the capacitors. The sum of voltage drops across each of the capacitor is equal to the potential difference (Pd) applied to the whole circuit.

The figure below shows three capacitors in series. Each has charge of the same magnitude on its plates, in agreement with principles of conservation of charge.

Hence the potential differences across the capacitors are respectively

And so, if *C* is the equivalent capacitance of the set, we have;

For any number of capacitors in series,

The reciprocal of the equivalent capacitance of a series arrangement of capacitors is equal to the sum of the reciprocals of the individual capacitors in general. Evidently *C* is smaller than the capacitance of any of the individual capacitors.

###### Capacitors in Series Calculations

###### Example

Two capacitors, one of 10µF and the other 30µF are connected in series across the terminals of a 12*V* battery.

- Find the equivalent capacitance of the combination
- The potential difference across each capacitor,
- Charge on each of its plats, and
- The energy stored in it.

###### Solutions

The equivalent capacitance of the two capacitors is

The charge on each of its plates is Q1=Q2 because the same charge (Q) will be present on all the plates of the capacitors. This is true for all capacitors in series connection.

The potential difference across each capacitor

*Note that the sum of V _{1} and V_{2 }must be equal to the potential difference (V) applied to the circuit (V = V_{1} +V_{2}).*

Energy stored in a capacitor is given by

W=1/2QV

*Where “* *” is the energy stored in Joules (J), “Q” is the charge in Coulomb (C), and “V” is the potential difference in Volts (V)*

∴ Energy stored in the capacitors are respectively

See also Parallel Capacitors Calculations