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 = 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