Numford
Circuits

Capacitors in series

In series the reciprocals add: 1/C = 1/C₁ + 1/C₂ + 1/C₃. For 2, 3 and 6 μF, C = 1 μF; on 12 V each holds 12 μC and the voltages are 6 V, 4 V, 2 V.

Capacitors

In series the charge is the same on each and 1/C = 1/C₁ + 1/C₂ + 1/C₃. In parallel the voltage is the same and C = C₁ + C₂ + C₃. Energy U = ½CV². Leave a box empty to use fewer capacitors.

In series1 μF
Charge12 μC
Energy72 μJ
  1. C₁, C₂, C₃ in series: 1/C = 1/2 + 1/3 + 1/6 = 1, so C = 1 μF
  2. Charge from the cell: Q = CV = 1 × 12 = 12 μC
  3. In series every capacitor holds the same 12 μC, so V = Q/C splits the voltage: C₁: 6 V, C₂: 4 V, C₃: 2 V.
  4. Energy stored: U = ½CV² = ½ × 1 × 12² = 72 μJ
Derivation

Why it works

  1. In series the same charge Q sits on every capacitor: the inner plates are charged only by induction.
  2. The voltage across each is V₁ = Q/C₁, V₂ = Q/C₂ and V₃ = Q/C₃.
  3. They add to the supply: V = V₁ + V₂ + V₃ = Q(1/C₁ + 1/C₂ + 1/C₃).
  4. For the single equivalent capacitor V = Q/C, so 1/C = 1/C₁ + 1/C₂ + 1/C₃.
  5. C in series is smaller than the smallest capacitor, the same rule as resistors in parallel.
Questions

Capacitors in series: quick answers

What is the equivalent of 2, 3 and 6 μF in series?

1/C = 1/2 + 1/3 + 1/6 = 1, so C = 1 μF.

Which capacitor in series has the largest voltage?

The smallest one, since V = Q/C with the same Q on each. Here 2 μF takes 6 V and 6 μF only 2 V.

Why join capacitors in series?

To share a high voltage between several capacitors so none breaks down, at the cost of a smaller total capacitance.

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