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Parallel and Serial Connected Capacitors

Capacitance in parallel and series connected circuits

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Capacitors in Parallel

Capacitors can be connected in parallel:

The equivalent capacitance for parallel-connected capacitors can be calculated as

C = C1 + C2 + . . + Cn                     (1)

where

C = equivalent capacitance for the parallel connected circuit  (Farad, F, μF)

C1..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 / C1 + 1 / C2 + . . + 1 / Cn                  (2)

For the special case with two capacitors in series - the capacitance can be expressed as

1 / C = (C1 + C2) / (C1 C2)                     (2b)

- or transformed to

C = C1 C2 / (C1 + C2)                     (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 C1 = 3 μF , C2 = 6 μF and C3 = 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

U1 = Q / C1

    = (391 μC) / (3 μF)

    = 130 V

The voltage across capacitor 2 can be calculated

U2 = Q / C2

    = (391 μC) / (6 μF)

    = 65 V

The voltage across capacitor 3 can be calculated

U3 = Q / C3

    = (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 εo εr l / ln(r2 / r1)                  (3)

where

εo = absolute permittivity, vacuum permittivity (8.85 10-12 F/m, Farad/m)

εr = relative permittivity

l = length of cylinders

r2 = radius of inner cylinder

r1 = radius of outer cylinder

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