Result

Parallel / series

C_{par} = \sum C_{i}, \frac{1}{C _{ser}} = \sum \frac{1}{C _{i}}

Statement

Replace a group of capacitors with a single equivalent value.

Symbols and units

$C_{eq}$	single capacitor that replaces the group	F
$C_{i}$	each capacitor in the group	F

When it applies

Ideal capacitors, no leakage.

Don't use when

Assumes ideal capacitors with no leakage — a real capacitor's leakage current breaks the ideal series/parallel combination rules over time.

Derivation

Parallel — both are connected across the same two nodes, so they share the voltage and the charges add:

Q = Q_{1} + Q_{2} = C_{1} Δ V + C_{2} Δ V

Divide by ΔV:

C_{eq} = \frac{Q}{Δ V} = C_{1} + C_{2}

Series — the middle section is isolated, so induction forces the same Q on every capacitor while the voltages add:

Δ V = Δ V_{1} + Δ V_{2} = \frac{Q}{C _{1}} + \frac{Q}{C _{2}}

Divide by Q:

\frac{1}{C _{eq}} = \frac{1}{C _{1}} + \frac{1}{C _{2}}

Worked examples

2 µF with 3 µF

C_{par} = 5 μ F, C_{ser} = \frac{( 2 ) ( 3 )}{2 + 3} = 1.2 μ F

Watch out for

The opposite of resistors, which is easy to mix up. Sanity check: a series combination is always smaller than the smallest capacitor in it.

Connects to