Thevenin and Norton Equivalents for Quick Analysis

Learning objectives

• Explain what a Thevenin equivalent and a Norton equivalent represent at a pair of terminals.
• Calculate $V_{Th}$, $R_{Th}$, and $I_N$ for a linear DC circuit with independent sources.
• Convert between Thevenin and Norton forms using source transformation.
• Use an equivalent circuit to find load voltage and load current quickly.
• Avoid the common exam mistakes that break the method.

Quick Review

From the last lesson, remember two habits. KCL sums currents at a node, and KVL sums voltages around a loop. For resistors, series values add, while parallel values combine with reciprocals: $1/R_{eq}=1/R_1+1/R_2+\cdots$. You will also reuse the voltage divider: for two series resistors, the voltage across $R_2$ is $V \cdot R_2/(R_1+R_2)$.

Theory

The black-box idea

A linear two-terminal circuit can be replaced at its terminals by a simple equivalent. Think of it as putting the messy circuit inside a black box. The load only cares about the voltage and current at the two terminals.

The Thevenin form is a voltage source $V_{Th}$ in series with a resistance $R_{Th}$. The Norton form is a current source $I_N$ in parallel with a resistance $R_N$.

Finding the Thevenin equivalent

1. Remove the load and label the open terminals $a$ and $b$.
2. Find $V_{Th}$ as the open-circuit voltage across $a$ and $b$. Keep the polarity consistent.
3. Turn off independent sources: replace voltage sources with shorts and current sources with opens. Then find the resistance looking into $a$ and $b$. That is $R_{Th}$.

A useful check is $R_{Th}=V_{oc}/I_{sc}$, where $I_{sc}$ is the short-circuit current through terminals $a$ and $b$.

[!IMPORTANT] Turn off only independent sources when finding $R_{Th}$. Dependent sources require a test-source method, which is an advanced topic. The basic exercises here use independent sources only.

The Norton equivalent

The Norton current is the short-circuit current: $I_N=I_{sc}$. The Norton resistance equals the Thevenin resistance: $R_N=R_{Th}$. To convert between forms, use:

$V_{Th}=I_N R_N$ and $I_N=V_{Th}/R_{Th}$.

Load analysis

Once you have the Thevenin equivalent, connect the load $R_L$ and use a voltage divider:

$I_L=\frac{V_{Th}}{R_{Th}+R_L}$

$V_L=V_{Th}\frac{R_L}{R_{Th}+R_L}$

Maximum power transfer occurs when $R_L=R_{Th}$. That is a useful exam fact, but it is not the same as maximum efficiency.

Worked examples

Example 1: A voltage divider with a load

Take a 12 V source in series with $R_1=6\ \Omega$, then node $a$, then $R_2=3\ \Omega$ to the return node $b$. The load terminals are $a$ and $b$ across $R_2$.

Step 1: Remove the load. The open-circuit voltage is the voltage across $R_2$:

$V_{Th}=12\frac{3}{6+3}=4\ \text{V}$

Step 2: Turn off the 12 V source by shorting it. Looking into $a$ and $b$, $R_1$ and $R_2$ are both connected between the terminals:

$R_{Th}=6\parallel 3=\frac{6\cdot 3}{6+3}=2\ \Omega$

Step 3: The Norton equivalent is $I_N=4/2=2\ \text{A}$ in parallel with $R_N=2\ \Omega$.

Step 4: Connect $R_L=6\ \Omega$:

$I_L=\frac{4}{2+6}=0.5\ \text{A}$

$V_L=4\frac{6}{2+6}=3\ \text{V}$

Check: with $a$ and $b$ shorted, $R_2$ is bypassed and $I_{sc}=12/6=2\ \text{A}$. This matches $I_N$.

Example 2: Convert a Norton source to Thevenin

A 6 mA current source is in parallel with $4\ \text{k}\Omega$. This is already a Norton equivalent: $I_N=6\ \text{mA}$ and $R_N=4\ \text{k}\Omega$.

Convert it:

$V_{Th}=I_N R_N=6\ \text{mA}\cdot 4\ \text{k}\Omega=24\ \text{V}$

So the Thevenin equivalent is 24 V in series with $4\ \text{k}\Omega$.

Guided practice

Use pen and paper. Draw each circuit, label the terminals, and remove the load before finding $V_{Th}$.

Exercise 1

A 9 V source is in series with $3\ \Omega$, then a node $a$, then $6\ \Omega$ to node $b$. Find $V_{Th}$ and $R_{Th}$ at $a$ and $b$.

Exercise 2

Using the equivalent from Exercise 1, connect $R_L=4\ \Omega$. Find $I_L$ and $V_L$.

Exercise 3

For the same Thevenin equivalent, write the Norton equivalent. Then state the load resistance for maximum power transfer.

Quick cloze check

The Thevenin voltage is the {{c1::open-circuit}} voltage across the load terminals.
To find $R_{Th}$, independent voltage sources are {{c2::shorted}} and independent current sources are {{c3::opened}}.
The Norton resistance $R_N$ equals {{c4::R_{Th}}}.
The load current from a Thevenin equivalent is $I_L =$ {{c5::V_{Th}/(R_{Th}+R_L)}}.
Maximum power transfer occurs when $R_L =$ {{c6::R_{Th}}}.

Expert Corner

• Remove the load first. Many wrong answers come from finding $V_{Th}$ with the load still connected.
• Short voltage sources and open current sources. Mixing these two up is the most common resistance calculation error.
• Keep units consistent. Convert mA to A or k$\Omega$ to $\Omega$ before using formulas.
• Check your answer with $I_{sc}=V_{Th}/R_{Th}$. If the short-circuit current does not match, recheck the source turn-off step.
• The equivalent preserves terminal behavior, not internal power. A resistor inside the original circuit may dissipate a very different power than $R_{Th}$.
• The method applies to linear circuits. Later, when you meet diodes, remember that a diode is not linear, so a plain Thevenin resistance is not enough.

Review deck

[
  { "front": "What is a Thevenin equivalent?", "back": "A linear two-terminal circuit replaced by V_Th in series with R_Th." },
  { "front": "How is V_Th defined?", "back": "The open-circuit voltage across the terminals with the load removed." },
  { "front": "How is R_Th found with independent sources?", "back": "Turn off the sources and find the resistance looking into the terminals." },
  { "front": "How do you turn off an independent voltage source?", "back": "Replace it with a short circuit." },
  { "front": "How do you turn off an independent current source?", "back": "Replace it with an open circuit." },
  { "front": "What is the alternative formula for R_Th?", "back": "R_Th equals the open-circuit voltage divided by the short-circuit current." },
  { "front": "What is a Norton equivalent?", "back": "A current source I_N in parallel with a resistance R_N." },
  { "front": "How is I_N defined?", "back": "The short-circuit current through the terminals." },
  { "front": "How are R_N and R_Th related?", "back": "They are equal." },
  { "front": "What is the source transformation formula from Norton to Thevenin?", "back": "V_Th equals I_N times R_N." },
  { "front": "What is the load current formula for a Thevenin equivalent?", "back": "I_L equals V_Th divided by the sum of R_Th and R_L." },
  { "front": "What is the load voltage formula for a Thevenin equivalent?", "back": "V_L equals V_Th times R_L divided by the sum of R_Th and R_L." },
  { "front": "When does maximum power transfer occur?", "back": "When the load resistance equals the Thevenin resistance." },
  { "front": "What does a Thevenin equivalent preserve?", "back": "The voltage and current behavior at its terminals, not internal power." }
]

Summary & Next Steps

You nowYou now hold the most useful shortcut in DC analysis. Any linear two-terminal network, no matter how tangled, behaves at its terminals like one voltage source in series with one resistor (Thevenin), or one current source in parallel with that same resistor (Norton).

The workflow to trust:

1. Label the terminals where the load connects, and pull the load out.
2. Compute V_Th as the open-circuit voltage across those terminals.
3. Turn off every independent source: voltage sources become shorts, current sources become opens.
4. Compute R_Th by combining the dead network into a single resistance, or check it with V_oc / I_sc.
5. Reattach the load to the equivalent and solve with a voltage divider.

Two habits worth keeping from this lesson:

• Always sanity check R_Th against a rough parallel or series estimate computed by eye. If your answer is wildly outside that range, you dropped a resistor or shorted the wrong branch.
• Remember what the equivalent does not preserve. Internal power dissipation is not modelled, so never use a Thevenin circuit to ask "how hot does R3 get". It answers terminal questions only.

Where this goes next

The next lesson is the Checkpoint: Circuit Theory Foundations. Expect a mixed problem set that folds Ohm's law, KCL and KVL, series and parallel reduction, and Thevenin or Norton reduction into single questions, typically "find the current in the load by any valid method". Practise choosing the method before you start writing equations. If a circuit asks only about one branch, reduction is almost always faster than full nodal analysis.

From there, Module 2 puts these equivalents to work. A diode plus resistor network, a transistor bias point, or an op-amp feedback loop all get easier once you can replace the messy part with a source and a resistance.

Quick self-check before you close the tab

1. Can you state, from memory, the two ways to find R_Th?
2. Can you convert between Thevenin and Norton in under ten seconds?
3. Given V_Th and R_Th, can you write I_L without pausing?

If any answer is a no, rerun one guided exercise above with the notes closed. That is the whole point of a quick-learn lesson: short, repeatable, and built for the exam room.