Introduction to Electrical Engineering  Basic vocational knowledge (Institut für Berufliche Entwicklung, 213 p.) 
6. Electrical Field 
6.2. Capacity 

Figs. 6.2. to 6.4. show certain phenomena between two charged plates. Two plates provided with connections and separated by a dielectric are called capacitor (Fig. 6.6.). This component is capable of storing a certain charge when a certain voltage is present. This storage capability is called capacity of a capacitor.
Fig. 6.6. Design of a capacitor
1  Conducting metal plates (electrodes)
2  dielectric
3  Connections
C = Q/U
[C] = (A·· s)/V = F
where:
C 
capacity 
Q 
charge 
U 
voltage 
Since the unit 1 F (farad) is very great, the capacity of the capacitors manufactured only reaches fractions of 1 F. These fractions are designated by the prefixes specified by legal regulation:
1 pF = 1 picofarad = 10^{12} F
1 nF = 1 nanofarad = 10^{9} F
1 µF = 1 microfarad = 10^{6} F
The storage capacity of a capacitor is dependent on the area of the electrodes, the distance between them and the type of dielectric.
C = eA/d
where:
e ^{1)} 
dielectric constant 
A 
area of the electrodes 
d 
distance between the electrodes 
^{1)} e Greek letter epsilon
The material constant is usually stated for the dielectric inquestion in the form of the product of the absolute dielectric constant times the relative dielectric constant.
e = e_{0} · e_{r}
where:
e_{0} 
absolute dielectric constant 
e_{r} 
relative dielectric constant 
The absolute dielectric constant applies to vacuum and is
e_{0} = 8.86 · 10^{12} (A·· s)/(V·· m)
Table 6.2. e_{r} of Some Insulating Materials
Insulating material 
e_{r} 
air 
1 
paper 
2 
transformer oil 
2.5 
rubber 
2.7 
porcelain 
5 
Epsilon (special ceramic compound for the production of capacitors) 
up to 10,000 
Like resistors, capacitors can be connected in series or in parallel. The total capacity obtained in this way is to be determined. Fig. 6.7. shows the series connection of two capacitors. The two capacitors have the same charge Q. The following holds:
Q_{AB} = Q_{1} = Q_{2}
U_{AB} = U_{1} + U_{2}
Fig. 6.7. Capacitors connected in
series
When dividing the voltage equation by the charge, we have
U_{AB}/Q = U_{1}/Q + U_{2}/Q
After inversion, we obtain from equation 6.2.
1/C = U/Q
and, for the total capacity of a series connection of capacitors we have
1/C_{equ} = 1/C_{1} +1/C_{2}
This equation has the same structure as the equation for the determination of R_{equ} of a parallel connection of resistors.
The parallel connection of two capacitors is shown in Fig. 6.8. The same voltage is applied to the two capacitors, and each capacitor has tored a charge in accordance with its capacity.
Fig. 6.8. Capacitors connected in
parallel
Thus, we have
U = U_{C1} = U_{C2}
Q_{AB} = Q_{1} + Q_{2}
After division, we obtain
Q_{AB}/U = Q_{1}/U +Q_{2}/U
and with equation 6.2. we have
C_{equ} = C_{1} + C_{2}
This equation has the same structure as the equation for the determination of R_{equ} of a series connection of resistors.
From the equations 6.5. and 6.6., the following general statement can be derived: In a series connection of capacitors, the total capacity is always smaller than the smallest individual capacity, and in a parallel connection of capacitors, the total capacity is always greater than the greatest individual capacity.
Example 6.1.
Two capacitors with a capacity of 470 nF and of 680 nF have to be connected in series and then in parallel. Determine the total capacity of each of the two types of connections!
Given:
C_{1} = 470 nF
To be found:
C in series connection and in parallel connection
Solution:
Series connection of C_{1} and C_{2}
1/C_{equ} = 1/C_{1} +1/C_{2} = (C_{2} + C_{1})/(C_{1} · C_{2})
C_{equ} = (C_{1} · C_{2})/(C_{1} + C_{2})
C_{equ} = (470 nF · 680 nF)/(470 nF + 680 nF)
C_{equ} = 277.9 nF
Parallel connection of C1 and C2
C_{equ} = C_{1} + C_{2}
C_{equ} = 470 nF + 680 nF
C_{equ} = 1150 nF
C_{equ} = 1/15 µF
In series connection, a total capacity of 277.9 nF is obtained while in parallel connection the total capacity is 1.15/µF.