Series & Parallel Capacitors

So we can say that:

\(Q_{Total} = Q_{C1} = Q_{C2} = {…} = Q_{Cn} \\\)

and from Kirchhoff’s Voltage Law we can also say that:

\(V_{Total} = V_{C1} + V_{C2} + {…} + V_{Cn} \\\)

and substituting for

\(V = \frac{Q}{C} \\\)

we get

\(\frac{Q_{Total}}{C_{Total}} = \frac{Q_{Total}}{C_1} + \frac{Q_{Total}}{C_2} + {…} + \frac{Q_{Total}}{C_n} \\\)

Then by dividing both sides by Q_Total we get:

\(\frac{1}{C_{Total}} = \frac{1}{C_1} + \frac{1}{C_2} + {…} + \frac{1}{C_n} \)

If that last equation looks eerily familiar, it should. It’s the same equation used for parallel resistors, but with Capacitance as the denominator!

Parallel Capacitors

Let’s consider what is happening here.

The same voltage must appear across each capacitor since their terminal are a common node. However, each capacitor does not necessarily have the same charge since Q = V.C

The total amount of charge stored in the network is distributed across the capacitors we can say that:

\(Q_{Total} = Q_{C1} + Q_{C2} + {…} + Q_{Cn} \\\)

Since V is common across all capacitors we can divide both sides by V to get:

\(\frac{Q_{Total}}{V} = \frac{Q_{C1}}{V} + \frac{Q_{C2}}{V} + {…} + \frac{Q_{Cn}}{V} \\\)

and since C = ^Q/_V we get:

\(C_{Total} = C_1 + C_2 + {…} + C_n \)

Again. If this looks eerily familiar, it should. It’s the same equation used for series resistors, but with capacitance substituted for resistance”

To Summarise

To calculate series/parallel capacitance. Just use the opposite formulae that you would have used for series/parallel resistors. So long as you have a working understanding of how capacitors work, this choice should feel fairly natural to you.

Examples

Example 1

Example 2