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Capacitance and series capacitors

Short answer

Capacitance C = Q/V. In series, each capacitor stores the same charge Q, so total voltage V = Q/C1 + Q/C2 + Q/C3. Hence 1/Ceq = 1/C1 + 1/C2 + 1/C3; for two, Ceq = C1C2/(C1+C2). Equivalent capacitance is less than the smallest individual capacitance.

Define capacitance and derive relation for connection of capacitors in series.

Capacitance

Capacitance is the ability of a capacitor to store charge. It is the ratio of the charge stored to the potential difference across it:

C = Q/V (farad, F)

1 farad is the capacitance that stores 1 coulomb of charge at 1 volt. For a parallel plate capacitor, C = ε A/d.

Capacitors in series

 o--| |--| |--| |--o
    C1   C2   C3
  +V1-  +V2- +V3-
 <------- V ------->

In a series connection the same charge Q is stored on every capacitor (charge cannot go anywhere else), and the supply voltage divides:

V = V₁ + V₂ + V₃

Using V = Q/C for each capacitor:

V₁ = Q/C₁, V₂ = Q/C₂, V₃ = Q/C₃

For the equivalent capacitor C_eq, V = Q/C_eq. Substituting:

Q/C_eq = Q/C₁ + Q/C₂ + Q/C₃

Dividing by Q:

1/C_eq = 1/C₁ + 1/C₂ + 1/C₃

For two capacitors: C_eq = C₁C₂/(C₁ + C₂).

Notes

  • C_eq is less than the smallest capacitor.
  • The voltage divides in the inverse ratio of the capacitances: V₁ = V × C_eq/C₁ (the smaller capacitor takes the larger voltage).
  • Example: 4 μF and 12 μF in series give 3 μF; at 100 V, Q = 300 μC, V₁ = 75 V, V₂ = 25 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

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