# Parallel and Serial Connected Capacitors

## Capacitance in parallel and series connected circuits

### Capacitors in Parallel

Capacitors can be connected in parallel:

The equivalent capacitance for parallel-connected 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 series-connected 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*