Circuits often combine several capacitors. The rules look like the ones for resistors, but switched: the series rule for capacitors is the parallel rule for resistors, and the other way around.
Why it works
In parallel, each capacitor has the same voltage and their charges add: $Q = C_1V + C_2V$, so $C_{\text{eq}} = C_1 + C_2$. It is like making the plates bigger. In series, the inner plates are connected only to each other, so the charge they hold is pulled across from one to the other: every capacitor carries the same charge $Q$. Their voltages add: $V = \dfrac{Q}{C_1} + \dfrac{Q}{C_2}$, so $\dfrac{1}{C_{\text{eq}}} = \dfrac{1}{C_1} + \dfrac{1}{C_2}$. It is like making the gap wider.
Key idea
Parallel: $C_{\text{eq}} = C_1 + C_2 + \cdots$ (same $V$, charges add). Series: $\dfrac{1}{C_{\text{eq}}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \cdots$ (same $Q$, voltages add), smaller than any one of them.
| Resistors | Capacitors | |
|---|---|---|
| Series | $R_1 + R_2$ | $\dfrac{C_1C_2}{C_1 + C_2}$ |
| Parallel | $\dfrac{R_1R_2}{R_1 + R_2}$ | $C_1 + C_2$ |
| Shared in series | current | charge |
| Shared in parallel | voltage | voltage |
Worked examples
Worked example: Both ways
Find the equivalent capacitance of $3\,\mu\text{F}$ and $6\,\mu\text{F}$ in parallel, then in series.
- Parallel: $3 + 6 = 9\,\mu\text{F}$.
- Series: $\dfrac{3 \times 6}{3 + 6} = 2\,\mu\text{F}$, less than either.
Answer: $9\,\mu\text{F}$; $2\,\mu\text{F}$
Worked example: Charge and voltage in series
The $3\,\mu\text{F}$ and $6\,\mu\text{F}$ capacitors are in series across $12\,\text{V}$. Find the charge and each voltage.
- $Q = C_{\text{eq}}V = (2\,\mu\text{F})(12\,\text{V}) = 24\,\mu\text{C}$ on each.
- $V_3 = \dfrac{24}{3} = 8\,\text{V}$ and $V_6 = \dfrac{24}{6} = 4\,\text{V}$; they add to $12\,\text{V}$.
Answer: $24\,\mu\text{C}$; $8\,\text{V}$ and $4\,\text{V}$
Worked example: A combination
A $6\,\mu\text{F}$ capacitor is in series with a parallel pair of $2\,\mu\text{F}$ and $4\,\mu\text{F}$. Find $C_{\text{eq}}$.
- Parallel pair: $2 + 4 = 6\,\mu\text{F}$.
- In series with $6\,\mu\text{F}$: $\dfrac{6 \times 6}{6 + 6} = 3\,\mu\text{F}$.
Answer: $3\,\mu\text{F}$
Common mistake
Adding series capacitances directly, as you would series resistors. Putting capacitors in series is like widening the gap, which lowers capacitance. Check: a series combination is always smaller than its smallest capacitor, and a parallel one bigger than its biggest.
In series, the smaller capacitor ends up with the larger voltage, because both hold the same charge and $V = Q/C$. That is the opposite of series resistors, where the larger resistor takes the larger voltage.