Equivalent capacitance — and why combinations behave oppositely to resistors.
When capacitors are connected in parallel, both plates of each capacitor are connected to the same two nodes. Therefore, all capacitors have the same potential difference \( V \) across them. The total charge stored is:
\[ Q_{\text{total}} = Q_1 + Q_2 + Q_3 + \cdots = C_1 V + C_2 V + C_3 V + \cdots \]The equivalent capacitance is:
\[ \boxed{C_{\text{parallel}} = C_1 + C_2 + C_3 + \cdots} \]When capacitors are connected in series, the same charge \( Q \) is stored on each capacitor (the inner plates are isolated, so charge transfers by induction). The total potential difference is the sum:
\[ V_{\text{total}} = V_1 + V_2 + V_3 + \cdots = \frac{Q}{C_1} + \frac{Q}{C_2} + \frac{Q}{C_3} + \cdots \]The equivalent capacitance satisfies:
\[ \boxed{\frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \cdots} \]For two capacitors in series:
\[ C_{\text{series}} = \frac{C_1 C_2}{C_1 + C_2} \]Since charge \( Q \) is the same on all series capacitors, the voltage across each is:
\[ V_i = \frac{Q}{C_i} \]A smaller capacitor takes a larger share of the voltage. For two capacitors \( C_1 \) and \( C_2 \) in series across total voltage \( V \):
\[ V_1 = \frac{C_2}{C_1 + C_2} V, \qquad V_2 = \frac{C_1}{C_1 + C_2} V \]Note how the voltages are weighted by the other capacitor — the smaller capacitor gets the larger voltage.
Since voltage \( V \) is the same across all parallel capacitors, the charge on each is:
\[ Q_i = C_i V \]A larger capacitor stores more charge. For two capacitors in parallel, the charge splits in proportion to capacitance:
\[ \frac{Q_1}{Q_2} = \frac{C_1}{C_2} \]| Property | Series | Parallel |
|---|---|---|
| Same quantity | Charge \( Q \) | Voltage \( V \) |
| Equivalent \( C \) | \( \dfrac{1}{C} = \sum \dfrac{1}{C_i} \) | \( C = \sum C_i \) |
| \( C_{\text{eq}} \) vs individual | Less than smallest | Greater than largest |
| Analogy | Resistors in parallel | Resistors in series |