Series and Parallel Resistors

Hello! Welcome to the sixth lesson in our module on DC Circuit Fundamentals.

In the past two lessons, we established the foundational laws of circuit analysis: Kirchhoff's Current Law (KCL), based on conservation of charge, and Kirchhoff's Voltage Law (KVL), based on conservation of energy. You learned how to write equations to describe the currents at nodes and the voltages around loops.

Today, we will use those fundamental laws to build our first set of powerful circuit analysis shortcuts. We will focus on how to simplify networks of resistors by finding their equivalent resistance. Your goal for this lesson is to be able to calculate the equivalent resistance for resistors connected in series, in parallel, and in combined networks. This is a crucial skill that dramatically simplifies the process of analyzing complex circuits.

1. What is Equivalent Resistance?

Before we dive in, what do we mean by "equivalent resistance"?

Imagine a complex network of resistors connected to a voltage source. From the perspective of the source, that entire network draws a certain amount of total current. The equivalent resistance, $R_{e q}$ , is the value of a single resistor that, if it replaced the entire network, would draw the exact same amount of total current from that same voltage source.

This is a powerful concept because it allows us to systematically reduce a complicated circuit into a very simple one, which we can then analyze easily using Ohm's Law.

Let's start with the two basic ways resistors can be connected: series and parallel.

Series and Parallel Circuits

This article from SparkFun provides excellent, clear definitions of series and parallel connections based on how their nodes are shared. Read the first two sections to understand the fundamental difference between the two configurations.

Read the sections 'Series Circuits Defined' and 'Parallel Circuits Defined'. Pay close attention to the key distinction: components in series share the same current, while components in parallel share the same voltage.

2. Resistors in Series

As you just read, resistors are in series when they are connected end-to-end in a single line, providing only one path for the current to flow.

Resistors in Series & Parallel — This diagram shows the basic configurations. On the left, resistors R1 and R2 are in series. The current has no choice but to flow through both sequentially.

The Series Formula

For resistors in series, the total equivalent resistance is simply the sum of their individual resistances:
$R_{S} = R_{1} + R_{2} + R_{3} + \dots + R_{N}$

Why is this true? It follows directly from KVL. If you apply a voltage $V$ across a series of resistors, KVL tells us that the total voltage must equal the sum of the individual voltage drops across each resistor: $V = V_{1} + V_{2} + V_{3}$ . Since the current $I$ is the same through all of them, Ohm's law gives us $V_{1} = I R_{1}$ , $V_{2} = I R_{2}$ , etc.

Substituting these in gives:
$V = I R_{1} + I R_{2} + I R_{3} = I (R_{1} + R_{2} + R_{3})$
For a single equivalent resistor, we'd have $V = I R_{S}$ . Comparing these two equations, it's clear that $R_{S} = R_{1} + R_{2} + R_{3}$ .

Resistors in Series and Parallel

This article provides a formal derivation of the series resistance formula using KVL, which should solidify the connection to our last lesson. It also includes a fully worked-out example.

Read the section 'Resistors in Series' up to (but not including) the parallel section. Focus on the derivation that starts with the KVL equation V = V₁ + V₂ + V₃.

Test your understanding!

You have three resistors connected in series: $R_{1} = 100 Ω$ , $R_{2} = 470 Ω$ , and $R_{3} = 2.2 k Ω$ . What is their total equivalent resistance? (Remember that $2.2 k Ω = 2200 Ω$ ).

Show answer

Since they are in series, we simply add them:
$R_{S} = 100 Ω + 470 Ω + 2200 Ω = 2770 Ω or 2.77 k Ω$

3. Resistors in Parallel

Resistors are in parallel when they are connected across the same two nodes. This means the voltage across each parallel resistor is identical, but the total current from the source splits to flow through the different branches.

The Parallel Formula

For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of the individual resistances:
$\frac{1}{R _{P}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}} + \dots + \frac{1}{R _{N}}$

This formula comes directly from KCL. At the node where the current splits, KCL states that the total current $I$ must equal the sum of the currents in each branch: $I = I_{1} + I_{2} + I_{3}$ . Since the voltage $V$ is the same across all of them, Ohm's law gives us $I_{1} = V / R_{1}$ , $I_{2} = V / R_{2}$ , etc.

Substituting gives:
$I = \frac{V}{R _{1}} + \frac{V}{R _{2}} + \frac{V}{R _{3}} = V (\frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}})$
For a single equivalent resistor, $I = V / R_{P}$ . Comparing these shows that $1/ R_{P} = 1/ R_{1} + 1/ R_{2} + 1/ R_{3}$ .

Solving for Parallel Resistance — This image walks through a numerical example of calculating the equivalent resistance of three parallel resistors using the reciprocal formula.

A crucial consequence of this formula is that the equivalent resistance of a parallel combination is always less than the smallest individual resistance in the combination. This makes sense: by adding more paths for the current, you are making it easier for current to flow, thus lowering the total resistance.

Special Case: Two Resistors in Parallel

When you have only two resistors in parallel, a very common scenario, you can rearrange the formula into a more direct form known as "product over sum":
$R_{P} = \frac{R _{1} \times R _{2}}{R _{1} + R _{2}}$

This is often faster to use than the reciprocal method. Let's see a quick derivation.

Series & Parallel Resistors in Circuits

This segment of a video from 'Math and Science' shows how the 'product over sum' rule is derived from the general parallel formula. It's a handy algebraic trick to know.

Watch from 10:14 to 13:50. The instructor clearly demonstrates the algebraic steps to get from the reciprocal form to the product-over-sum form for two resistors.

Test your understanding!

You have a $10 k Ω$ resistor and a $2.5 k Ω$ resistor.

What is their equivalent resistance if connected in parallel? Use the "product over sum" formula.
Is the result less than the smallest individual resistor ( $2.5 k Ω$ )?

Show answer

Using product over sum:
$R_{P} = \frac{10 k Ω \times 2.5 k Ω}{10 k Ω + 2.5 k Ω} = \frac{25 ( k Ω ) ^{2}}{12.5 k Ω} = 2 k Ω$
Yes, $2 k Ω$ is less than $2.5 k Ω$ , which confirms our calculation is reasonable.

4. Combining Series and Parallel Resistors

Most real-world circuits aren't purely series or purely parallel. They are combinations of both. To solve these, we use a step-by-step reduction strategy:

Identify a simple group of resistors that are purely in series or purely in parallel.
Calculate the equivalent resistance of just that group.
Redraw the circuit, replacing the group with its single equivalent resistor.
Repeat this process, looking for the next simple series or parallel group in the newly simplified circuit.
Continue until only one resistor remains. This is the total equivalent resistance of the entire network.

This process requires careful identification of which resistors are truly in series or parallel. Redrawing the circuit at each step is the key to avoiding mistakes.

Let's watch a comprehensive example where this strategy is applied from start to finish.

Series & Parallel Resistors in Circuits

This is the most important part of the lesson. In this video, the instructor takes a complex resistor network and methodically reduces it step by step, redrawing the circuit each time. Follow his process closely.

Watch the example from 14:49 to 25:38. Notice how he starts from the part of the circuit furthest from the source, identifies a series pair, combines them, and then redrawing reveals a new parallel pair. He repeats this until the entire circuit is simplified.

5. From Theory to Practice

Given your interest in hardware, it's valuable to see how these theoretical calculations connect to the real world. Using a breadboard and a multimeter, you can physically build these resistor networks and measure their equivalent resistance to verify that the formulas work exactly as predicted.

Series and Parallel Circuits

The following article contains a short, hands-on experiment you could perform with basic components. It walks you through measuring series and parallel combinations on a breadboard.

Read the sections 'Experiment Time - Part 1' and 'Experiment Time - Part 2'. Even if you don't have the components right now, reading through the procedure will help you visualize how these abstract concepts are measured and verified in practice.

Conclusion

Today you've learned a fundamental skill for simplifying DC circuits. By systematically identifying and combining series and parallel resistor groups, you can reduce even a dauntingly complex network to a single equivalent resistance.

Key Takeaways:

Series Resistors: Share the same current. Their resistances add directly: $R_{S} = R_{1} + R_{2} + \dots$ .
Parallel Resistors: Share the same voltage. Their reciprocals add: $1/ R_{P} = 1/ R_{1} + 1/ R_{2} + \dots$ .
The equivalent resistance of parallel resistors is always less than the smallest resistor in the group.
For two resistors in parallel, the product-over-sum shortcut is often easiest: $R_{P} = (R_{1} \cdot R_{2}) / (R_{1} + R_{2})$ .
Combined networks are solved by iteratively identifying, calculating, and redrawing until a single equivalent resistance remains.

In our next lesson, we will build directly on the concept of series resistors to derive the Voltage Divider Rule, another essential shortcut for quickly finding the voltage across a specific component in a series chain without first having to calculate the current.

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