**Series Connected Capacitors.**This changed the game play a bit, so lets look at what it really changes. The charging current (ic) flowing through the capacitors is

**THE SAME** for all capacitors as it only has one path to follow and iT = i1 = i2 = i3 etc. Then,

** Capacitors in Series** all have the same current so each capacitor stores the same amount of charge regardless of its capacitance. This is because the charge stored by an electrode of any one capacitor must have come from the electrode of its adjacent capacitor, therefor

**QT = Q1 = Q2 = Q3 .... ETC**

In the following circuit, capacitors, C1, C2 and C3 are all connected together in a series branch between point

**A** and

**B**Between point

**A** and

**B**, we have;

C1 = 0.1uF (VC1)

C2 = 0.2uF (VC2)

C3 = 0.3uF (VC3)

VAB = 12V (the over all voltage charge)

In a parallel circuit, what we thought Meyer had, the total capacitance, CT of the circuit was equal to the sum of all the individual capacitors added together. In a series connected circuit however, the total or equal capacitance CT is calculated differently. The voltage drop across each capacitor will be different depending upon the values of the individual capacitance - )we should see or measure the same if all the individual cells are made in the same manor) - Then by adding

**Kirchoff's Voltage Law (KVL)** to the above circuit, we get;

**VAB = VC1 + VC2 + VC3 = 12V**

**Kirchoff's Voltage Law (KVL)**This law is also called Kirchoff's second law, Kirchoff's loop (mesh) rule. The principle of conservation of energy implies that; The direct sum of the electrical potential differences (voltage) around any closed circuit is zero OR More simply, the sum of the emfs in any closed loop is equivalent to the sum of the potential drop in that loop OR The algebraic sum of the products of the resistance of the conductors and the currents in them in a closed loop is equal to the total emf available in that loop. http://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws#Kirchhoff.27s_voltage_law_.28KVL.29

**VAB** = VC1 + VC2 + VC3 = 12V

**VC1** = QT /C1 - VC2 = QT / C2 - VC3 = QT / C3

Since Q = CV or V = Q/C, substituting Q/C for each capacitors voltage VC in the above KVL equation gives us:

VAB = QT/CT = QT/C1 + QT/C2 + QT/C3

Dividing each term through by Q gives:

Series Capacitance Equation

CT = (1/C1) + (1/C2) + (1/C3) + ... etc

When adding together capacitors in series, the reciprocal (1/C) of the individual capacitors are all added together (just like resistors in parallel) instead of the capacitance's themselves. Then the total value for the capacitors in series equals the reciprocal of the sum of the reciprocal of the individual capacitance's.

**Example;**

Using Tony's number of 67nF over a single capacitor for a 10 capacitor cell:

1/CT =(1/67) = 0.014925 x 10 = 1/0,14925 = **6,7nF**

So we now all of a sudden only have 1/10'th of the initial capacitance. One important point to remember about capacitors connected in series is that the total capacitance (CT) of any number of capacitors connected together in series will **ALWAYS BE LESS** then the value of the smallest capacitor in the series and in our case **CT** = 6,7nF where's **one capacitor is 67nF**Adding cell's will of course lower the capacitance. If we had 20 capacitors, the number will only be **3,35nF**You can check the capacitance value for your cell's here: http://www.csgnetwork.com/seriescapacicalc.html