# Capacitors - Parallel and Serial 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 *

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