# Capacitors - Parallel and Serial Connections

### 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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