Capacitors  Parallel and Serial Connections
Parallel and serial connected capacitor circuits.
Capacitors in Parallel
Capacitors can be connected in parallel:
The equivalent capacitance for parallelconnected capacitors can be calculated as
C = C_{1} + C_{2} + . . + C_{n} (1)
where
C = equivalent capacitance for the parallel connected circuit (Farad, F, μF)
C_{1..n} = capacitance capacitors (Farad, F, μF)
It is common to use µF as the unit for capacitance.
Capacitors in Series
Capacitors can be connected in series:
The equivalent capacitance for seriesconnected capacitors can be calculated as
1 / C = 1 / C_{1} + 1 / C_{2} + . . + 1 / C_{n} (2)
For the special case with two capacitors in series  the capacitance can be expressed as
1 / C = (C_{1} + C_{2}) / (C_{1} C_{2}) (2b)
 or transformed to
C = C_{1} C_{2} / (C_{1} + C_{2}) (2c)
Example  Capacitors Connected in Parallel and in Series
The equivalent capacitance of two capacitors with capacitance 10 μF and 20 μF can be calculated as
in parallel
C = (10 μF) + (20 μF)
= 30 (μF)
in series
1 / C = 1 / (10 μF) + 1 / (20 μF)
= 0.15 (1/μF)
or
C = 1 / 0.15 (1/μF)
= 6.7 (μF)
Capacitors in Series
Three capacitors C_{1} = 3 μF , C_{2} = 6 μF and C_{3} = 12 μF are connected in series as indicated in the figure above. The voltage supply to the circuit is 230 V.
The equivalent circuit capacitance can be calculated with (2)
1 / C = 1 / (3 μF) + 1 / (6 μF) + 1 /_{ }(12 μF)
= (4 + 2 + 1) / 12
= 0.58 1/μF
 or transformed
C = 12 / (4 + 2 + 1)
= 1.7 μF
The total charge in the circuit can be calculated with
Q = U C
where
Q = charge (coulomb, C)
U = electric potential (V)
 or with values
Q = (230 V) (1.7 10^{6} F)
= 3.91 10^{4} C
= 391 μC
Since the capacitors are connected in series  the charge is 391 μC on each of them.
The voltage across capacitor 1 can be calculated
U_{1} = Q / C_{1 }
= (391 μC) / (3 μF)
= 130 V
The voltage across capacitor 2 can be calculated
U_{2} = Q / C_{2}
= (391 μC) / (6 μF)
= 65 V
The voltage across capacitor 3 can be calculated
U_{3} = Q / C_{3 }
= (391 μC) / (12 μF)
= 33 V
Capacitance of Two Coaxial Cylinders
The capacitance of two coaxial cylinders as indicated in the figure can be calculated as
C = 2 π ε_{o} ε_{r }l / ln(r_{2 }/ r_{1}) (3)
where
ε_{o} = absolute permittivity, vacuum permittivity (8.85 10^{12} F/m, Farad/m)
ε_{r }= relative permittivity
l = length of cylinders
r_{2} = radius of inner cylinder
r_{1} = radius of outer cylinder
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