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Capacitors - Parallel and Serial Circuits

Parallel and serial connected capacitor circuits.

Capacitors in Parallel

Capacitors can be connected in parallel:

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

C = C1 + C2+ . . + 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 / C1 + 1 / C2+ . . + 1 / C n                       (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 ε o ε r l / ln(r2/ r1 )                  (3)

where

ε o = absolute permittivity, vacuum permittivity (8.85 10-12 F/m, farad/m)

ε r = relative permittivity

l = length of cylinders

r2= radius of inner cylinder

r1 = radius of outer cylinder

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