Resistor Networks: Series and Parallel Analysis

Recommended resources

The freely available resources this lesson will direct you to.

Hello! Welcome to the third and final lesson in our module on DC circuit foundations.

In the last two lessons, we established the fundamental laws governing DC circuits (Ohm's and Kirchhoff's Laws) and learned how to calculate the power and energy associated with them. While you can technically solve any DC circuit using a system of KVL/KCL equations, for complex circuits this can become computationally intensive—a familiar problem from your work with large systems in statistics and computing.

Today, we'll learn a more efficient method for analyzing many common circuits: simplification. We will focus on circuits composed of series and parallel resistor networks. By reducing these networks to a single equivalent resistance, we can turn a complex problem into a simple one that's easily solved with Ohm's Law. This skill is the capstone of basic DC circuit analysis.

By the end of this lesson, you will be able to identify series, parallel, and combination resistor networks, calculate their equivalent resistance, and use this to determine the voltage, current, and power for any component in the circuit.

1. Resistors in Series and the Voltage Divider Rule

A series circuit is defined by a single path for current to flow. Resistors are connected sequentially, like links in a chain.

A circuit diagram showing three resistors, R1, R2, and R3, connected in a single loop with a voltage source V.
A simple series circuit. The current I is the same through each resistor.

Because there is only one path, the key characteristic of a series circuit is that the current is the same through every component.

To find the equivalent resistance ( $R_{S}$ ), we can use KVL. The sum of the voltage drops across all resistors must equal the source voltage:
$V_{so u rce} = V_{1} + V_{2} + \dots + V_{N}$
Using Ohm's Law ( $V = I R$ ) for each resistor:
$V_{so u rce} = I R_{1} + I R_{2} + \dots + I R_{N} = I (R_{1} + R_{2} + \dots + R_{N})$
The entire circuit behaves as if there were a single resistor, $R_{S}$ , such that $V_{so u rce} = I R_{S}$ . Comparing these equations, we find the formula for equivalent resistance in series:

$R_{S} = R_{1} + R_{2} + \dots + R_{N} = \sum_{i = 1}^{N} R_{i}$

For a more detailed derivation and explanation, the following resource is a good, quick read.

Resistors in Series and Parallel

This section from the OpenStax textbook 'University Physics' provides a formal definition of series circuits and derives the formula for equivalent resistance.

Read the subsection titled 'Resistors in Series'. Focus on the derivation from Kirchhoff's loop law and the summary of the major features of series resistors at the end of the section.

The Voltage Divider Rule

A powerful shortcut emerges from the properties of series circuits. Since the voltage from the source is "divided" among the resistors, we can find the voltage across any specific resistor ( $R_{k}$ ) without first calculating the current.

The voltage across $R_{k}$ is $V_{k} = I R_{k}$ . The total current is $I = V_{t o t a l} / R_{S}$ . Substituting for $I$ :
$V_{k} = (\frac{V _{t o t a l}}{R _{S}}) R_{k} = V_{t o t a l} \frac{R _{k}}{R _{1} + R _{2} + \dots + R _{N}}$
This is the voltage divider rule. It states that the voltage across a resistor in a series circuit is proportional to its resistance relative to the total series resistance. This is a very handy tool for quick analysis.

2. Resistors in Parallel and the Current Divider Rule

In a parallel circuit, components are connected across the same two points, providing multiple paths for the current.

A circuit diagram showing three resistors, R1, R2, and R3, connected in parallel to a voltage source V.
A simple parallel circuit. The voltage V is the same across each resistor.

The key characteristic of a parallel circuit is that the voltage is the same across every component.

From Kirchhoff's Current Law, the total current leaving the source must equal the sum of the currents in each parallel branch:
$I_{t o t a l} = I_{1} + I_{2} + \dots + I_{N}$
Using Ohm's Law ( $I = V / R$ ) for each branch:
$I_{t o t a l} = \frac{V}{R _{1}} + \frac{V}{R _{2}} + \dots + \frac{V}{R _{N}} = V (\frac{1}{R _{1}} + \frac{1}{R _{2}} + \dots + \frac{1}{R _{N}})$
The equivalent circuit would have $I_{t o t a l} = V / R_{P}$ . Comparing these gives the formula for equivalent resistance in parallel:

$\frac{1}{R _{P}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \dots + \frac{1}{R _{N}} or R_{P} = (\sum_{i = 1}^{N} \frac{1}{R _{i}})^{- 1}$

A very common special case is for two resistors in parallel. The formula simplifies to the "product over sum" rule:
$R_{P} = \frac{R _{1} R _{2}}{R _{1} + R _{2}}$
Note that the equivalent resistance in a parallel network is always less than the smallest individual resistance.

This next section of the same textbook explains the parallel case.

Resistors in Series and Parallel

This section explains parallel circuits, deriving the formula for equivalent resistance and highlighting its key properties.

Read the subsection titled 'Resistors in Parallel'. Notice how the derivation stems from Kirchhoff's junction rule, in contrast to the series case.

The Current Divider Rule

Analogous to the voltage divider, we can derive a current divider rule for parallel circuits. It allows you to find the current in a specific branch ( $I_{k}$ ) if you know the total current ( $I_{t o t a l}$ ) entering the parallel combination. For a two-resistor case with $R_{1}$ and $R_{2}$ :
$I_{1} = I_{t o t a l} \frac{R _{2}}{R _{1} + R _{2}} and I_{2} = I_{t o t a l} \frac{R _{1}}{R _{1} + R _{2}}$
Notice that the current in one branch is proportional to the resistance of the other branch. This makes sense: more current will flow down the path of less resistance.

For formal definitions of both voltage and current divider rules, you can refer to the document Electrical Engineering Fundamentals. The rule for Voltage Division is on page 66 (Eq. 2.5) and Current Division is on page 69 (Eq. 2.7).

3. Combination Circuits: The "Break it Down, Build it Up" Method

Most real-world circuits are not purely series or parallel but a combination of both. To analyze these, we use a systematic reduction technique. The strategy is to:

Break it Down: Identify distinct series or parallel sub-networks within the larger circuit.
Calculate the equivalent resistance for each sub-network.
Redraw the circuit with the sub-networks replaced by their single equivalent resistors.
Repeat this process until the entire circuit is reduced to a single equivalent resistance.

Once the circuit is broken down, you can find the total current from the source using Ohm's Law. Then, you:
5. Build it Up: Work backward through your simplification steps. At each step, use the known total current or voltage for an equivalent resistor to find the currents and voltages of the individual resistors that formed it. The voltage and current divider rules are especially useful here.

The following video provides an excellent, detailed walkthrough of this entire process on a non-trivial circuit. The methodology is robust and will serve you well.

How to Solve Any Series and Parallel Circuit Problem

This video from Jesse Mason demonstrates a powerful, systematic method for solving any combination series-parallel circuit.

Watch the entire video (00:00 - 13:37). Pay close attention to the two main phases: 'Break it down' (00:00 - 07:03): Focus on how the complex circuit is redrawn and systematically simplified to find the single equivalent resistance. 'Build it up' (07:03 - 13:37): Observe how he works backward from the simplified circuit, using Ohm's law and the properties of series/parallel connections at each step to find all individual voltages and currents.

Test your understanding!

Let's apply the first half of that method. Consider the circuit below. What is the equivalent resistance, $R_{e q}$ , between points A and B?

A combination series-parallel circuit.

Show answer

Step 1: Identify resistors in series. $R_{1}$ and $R_{2}$ are in series.
$R_{12} = R_{1} + R_{2} = 30 Ω + 50 Ω = 80 Ω$
Step 2: Redraw the circuit. Now we have $R_{12}$ in parallel with $R_{3}$ . Calculate their equivalent resistance.
$R_{123} = \frac{R _{12} \times R _{3}}{R _{12} + R _{3}} = \frac{80 Ω \times 40 Ω}{80 Ω + 40 Ω} = \frac{3200}{120} Ω \approx 26.67 Ω$
Step 3: Redraw the circuit again. Now we have $R_{123}$ in series with $R_{4}$ .
$R_{e q} = R_{123} + R_{4} = 26.67 Ω + 5 Ω = 31.67 Ω$

The total equivalent resistance between points A and B is approximately 31.67 Ω.

4. Practice and Further Examples

The key to mastering this topic is practice. Seeing the simplification process applied to various configurations helps build intuition. The next video offers several more examples, starting from simple cases and moving to more complex ones.

Equivalent Resistance of Complex Circuits - Resistors In Series and Parallel Combinations

This video from The Organic Chemistry Tutor provides a series of rapid-fire examples of calculating equivalent resistance for various circuit configurations.

Watch the segment from 02:55 to 15:24. You don't need to follow every calculation in detail. Instead, focus on how each circuit is visually assessed and broken down into smaller, solvable parts. Notice the frequent use of redrawing to clarify the relationships between resistors.

5. Connection to Formal Circuit Analysis

The simplification techniques learned today are your first line of attack when analyzing a circuit. For any circuit that can be reduced to a single equivalent resistance, this method is far more efficient than the full KVL/KCL analysis you learned in the first lesson.

However, some circuits, like a Wheatstone bridge, cannot be simplified this way. For those, you must fall back on mesh or nodal analysis. Often, a hybrid approach is best: you simplify parts of a multi-loop circuit first to reduce the number of unknown currents and thus the number of simultaneous equations you need to solve.

Given your background in linear algebra, you might find it interesting to see how mesh analysis leads to a system of linear equations that can be solved using matrix methods like Cramer's rule. The PDF resource Electrical Engineering Fundamentals demonstrates this in the "Multi-loop circuit analysis" section starting on page 70 (Example 2.7). You are not required to study this now, but it's a good example of where the formal methods are indispensable.

Conclusion

You have now completed the foundational toolkit for DC circuit analysis. You can simplify complex resistor networks, solve for the voltage and current anywhere in the circuit, and calculate the power being dissipated or supplied.

Key Takeaways:

Series Resistors: Share the same current. Their resistances add up: $R_{S} = \sum R_{i}$ . The voltage across them is determined by the Voltage Divider Rule.
Parallel Resistors: Share the same voltage. The reciprocals of their resistances add up: $1/ R_{P} = \sum 1/ R_{i}$ . The current through them is determined by the Current Divider Rule.
Combination Circuits: Solved by systematically reducing series and parallel sub-networks until a single equivalent resistance is found. Then, work backward to find individual currents and voltages.

Preview of the Next Lesson:
We will now pivot from circuit analysis to the fundamental physics that makes it all work. In the next module, "Electric and Magnetic Fields," we will start from the very beginning: the electric field created by a single point charge. This will build the foundation for understanding Coulomb's Law, electric potential, and eventually, the deeper principles of electromagnetism that you were interested in exploring.