# Capacitors in Series and Parallel

*Why it’s important*: Building on what we’ve learned about capacitors in series and in parallel, we can start analyzing more complex arrangements.

#### Review of Capacitance in Series and Parallel

Capacitance is the ratio of the total charge stored in the capacitor to the voltage drop across it:

C = \frac{Q}{V}

We have previously learned about the equivalent capacitance of capacitors in parallel:

C_{Tparallel} = C_1 + C_2 + C_3 +...+C_N

As well as the relation for capacitors in series:

C_T = (\frac{1}{C_1}+\frac{1}{C_2}+...+\frac{1}{C_N})^{-1}

Let’s combine these to analyze more complex circuits.

#### Example 1:

What happens when we see capacitors in combinations like in this example? The easiest way to divide this circuit is by first analyzing each parallel branch independently.

Q: What is the equivalent capacitance of this circuit?

Let’s start with the first path, consisting of C_{1} and C_{2}. We’ll call the capacitance of this path C_{T1}. In this case, the two capacitors are in series so we’ll use the relation for capacitors in series:

C_{T1} = (\frac{1}{C_1}+\frac{1}{C_2})^{-1}=(\frac{1}{1\mu F}+\frac{1}{2\mu F})^{-1} = \frac{2}{3}\mu F

Let’s use the same procedure for the second branch:

C_{T2} = (\frac{1}{C_3}+\frac{1}{C_4})^{-1}=(\frac{1}{1\mu F}+\frac{1}{2\mu F})^{-1} = \frac{2}{3}\mu F

The two paths are identical, as they have the same capacitor arrangement.

The total capacitance for the circuit, then, treats each branch as a single capacitor with a value of 2/3 μF. We can now use the relation for capacitors in parallel to find the total equivalent capacitance of the circuit:

C_T = C_{T1} + C_{T2}= \frac{2}{3}\mu F+\frac{2}{3}\mu F=\frac{4}{3}\mu F

#### Example 2:

Q: What is the equivalent capacitance of this circuit?

In this case, we have two capacitors in series (C_{1} and C_{4}), with C_{2} and C_{3} in parallel. We can ‘collapse’ C_{2} and C_{3} into an equivalent capacitor and then use the series circuit rule.

Let’s start by finding the equivalent capacitance of C_{2} and C_{3}:which we will call C_{p1}

C_{p1} = C_2+C_3 = 4\mu F + 2\mu F = 6\mu F

Now we can use the formula for capacitors in series to find the equivalent capacitance of the whole circuit:

C_T = (\frac{1}{C_1}+\frac{1}{C_{p1}}+\frac{1}{C_4})^{-1}=(\frac{1}{1\mu F}+\frac{1}{6\mu F}+\frac{1}{6\mu F})^{-1}=.75\mu F

*Lesson 0*: Introduction to Module 3

*Lesson 1*: Introduction to DC Circuits

*Lesson 2*: Series and Parallel Circuits

*Lesson 3*: DC Power Sources and Batteries

*Lesson 4*: Resistors, Capacitors, and Inductors

*Lesson *5: Resistors in Series

*Lesson *6: Resistors in Parallel

*Lesson *7: Voltage Dividers

*Lesson *8: Kirchoff’s Current Law

*Lesson *9: Kirchoff’s Voltage Law

*Lesson *10: Capacitors

*Lesson *11: Dielectric Materials

*Lesson *12: Capacitors in Parallel

*Lesson *13: Capacitors in Series

*Lesson *14: Capacitors in Series and Parallel