Thevenin and Norton: worked exercises

End of September 2026: in Year 3 of a technical institute, in the subject Electrical engineering and electronics (and in Telecommunications, in the computer science and telecommunications specialisation, where the circuit theory block is the same), after the first weeks on Ohm’s law and Kirchhoff’s principles (which I have already written about here, though pitched at the level of a vocational school) you come to a more powerful method for analysing circuits: the Thevenin and Norton theorems. The first written test on this block typically falls between the end of October and mid-November, so five to seven weeks from now.

The jump from Kirchhoff is conceptual even before it is technical: instead of writing a system of equations for the whole circuit, you isolate the part you are interested in (usually a load, a resistor or a specific branch) and replace everything else with a single equivalent source. It is a change of perspective that doesn’t click straight away for many students, because it seems “too convenient” to be true. The point of this post is to show you why it always works, and to give you a method that leaves no room for improvisation when the test arrives.

What the test really asks

Thevenin and Norton questions almost always ask for the same sequence of steps: given a circuit with one or more sources and resistors, calculate the equivalent resistance and source seen from a pair of terminals (usually those of a load), then use that equivalent to find the current and voltage on the load itself, and sometimes the power the load absorbs. A minority of questions also ask for the Thevenin↔Norton conversion or the condition for maximum power transfer. The rest (superposition, other methods of analysis) appears more often as a side exercise than as the central theme.

The method, step by step

1. The equivalent resistance seen from the terminals

The first step, and the one where most mistakes concentrate, is to calculate $R_{t h}$ : the resistance seen from the two terminals of interest with the load disconnected and with all sources turned off.

“Turning off” a source means:

an ideal voltage source becomes a short circuit (a wire, zero resistance);
an ideal current source becomes an open circuit (a break, no connection).

From here, $R_{t h}$ is simply the equivalent resistance of the resulting passive circuit, calculated with the normal series and parallel rules you already use with Kirchhoff.

2. The open-circuit voltage

$V_{t h}$ is the voltage you would measure across the terminals with the load disconnected, sources on. It is not the total voltage of the circuit: it is specifically the one that the rest of the circuit “offers” to whatever will be connected to those two terminals. It is calculated with Ohm and Kirchhoff on the circuit without the load.

3. The short-circuit current (for Norton)

$I_{N}$ is the current that would flow if the two terminals were connected directly to each other (short-circuited), again with the load disconnected. You don’t need to calculate it from scratch if you already have $V_{t h}$ and $R_{t h}$ : it is always

I_{N} = \frac{V _{t h}}{R _{t h}}

4. The Thevenin ↔ Norton equivalence

The two equivalent circuits (a voltage source $V_{t h}$ in series with $R_{t h}$ , or a current source $I_{N}$ in parallel with $R_{t h}$ ; the resistance is the same in both, $R_{N} = R_{t h}$ ) produce exactly the same behaviour as seen from the terminals. The relation that links them is the one just written, read backwards too:

R_{t h} = \frac{V _{t h}}{I _{N}}

You can switch from one to the other whenever the problem makes it more convenient: Norton is often more natural when the starting circuit already has current sources in parallel.

5. Reconnecting the load

Once you have the equivalent, reconnect the load resistor $R_{L}$ to the terminals. The circuit is now trivial: a single source, two resistances in series (Thevenin form) or in parallel (Norton form). Current and voltage on the load are found with Ohm in one line.

6. Maximum power transfer

A classic test question: for what value of $R_{L}$ does the load absorb the maximum possible power? The answer, which is proved by differentiating $P_{L} (R_{L})$ and setting the derivative to zero, is

R_{L} = R_{t h}

At that point the power on the load is

P_{ma x} = \frac{V _{t h}^{2}}{4 R _{t h}}

Note well: this condition maximises the power on the load, not the efficiency of the circuit; at $R_{L} = R_{t h}$ half of the power generated is dissipated inside $R_{t h}$ anyway.

Where the marks are lost

Forgetting to disconnect the load before calculating $R_{t h}$ and $V_{t h}$ : if you leave $R_{L}$ connected, both the equivalent resistance and the open-circuit voltage you get are wrong, because the load itself alters the circuit you are analysing.
Turning sources off wrongly: the most common slip is short-circuiting a current source (it should become an open circuit) or opening a voltage source (it should become a short circuit), exactly the opposite of what is needed.
Sign errors in the open-circuit voltage when there are several sources: always define a reference direction for $V_{t h}$ before starting the calculations, and keep it for the whole exercise.
Confusing $R_{t h}$ seen from the terminals with the total resistance of the circuit: they are the same thing only in very simple single-branch circuits; in a circuit with several branches the resistance seen from the terminals takes into account how they are connected to each other with respect to those two specific points, not the sum of all the resistances present.

Worked exercise 1: a divider with a load

An ideal source $V_{s} = 12 V$ feeds a divider: $R_{1} = 6 Ω$ in series between the source and a node A, and from A to ground $R_{2} = 3 Ω$ . A load $R_{L} = 2 Ω$ is connected between node A and ground. Find the current and power on the load using Thevenin.

Step 1: disconnect $R_{L}$ and calculate $R_{t h}$ . With the source turned off (replaced by a short circuit), $R_{1}$ and $R_{2}$ are in parallel as seen from the terminals A-ground:

R_{t h} = \frac{R _{1} \cdot R _{2}}{R _{1} + R _{2}} = \frac{6 \cdot 3}{6 + 3} = 2 Ω

Step 2: calculate $V_{t h}$ . With $R_{L}$ disconnected, the circuit is a simple voltage divider between $R_{1}$ and $R_{2}$ :

V_{t h} = V_{s} \cdot \frac{R _{2}}{R _{1} + R _{2}} = 12 \cdot \frac{3}{9} = 4 V

Step 3: reconnect $R_{L}$ to the equivalent. Now the circuit is $V_{t h} = 4 V$ in series with $R_{t h} = 2 Ω$ and $R_{L} = 2 Ω$ :

I_{L} = \frac{V _{t h}}{R _{t h} + R _{L}} = \frac{4}{2 + 2} = 1 A

P_{L} = I_{L}^{2} \cdot R_{L} = 1^{2} \cdot 2 = 2 W

Worked exercise 2: two sources on the same node

Two branches are connected in parallel between a node A and ground: the first has a source $V_{1} = 8 V$ in series with $R_{1} = 4 Ω$ , the second a source $V_{2} = 4 V$ in series with $R_{2} = 12 Ω$ . A load $R_{L} = 4 Ω$ is connected between A and ground. Find the current on the load.

Step 1: $R_{t h}$ . With both sources turned off (short-circuited), $R_{1}$ and $R_{2}$ are in parallel:

R_{t h} = \frac{R _{1} \cdot R _{2}}{R _{1} + R _{2}} = \frac{4 \cdot 12}{16} = 3 Ω

Step 2: $V_{t h}$ . With $R_{L}$ disconnected, the two branches share node A: I use Kirchhoff’s current law at the node, calling $V_{A}$ the (unknown) voltage of the node with respect to ground. The current leaving branch 1 towards A is $(V_{1} - V_{A}) / R_{1}$ , the one leaving branch 2 is $(V_{2} - V_{A}) / R_{2}$ , and their sum must be zero (no current goes to the load, which is disconnected):

\frac{V _{1} - V _{A}}{R _{1}} + \frac{V _{2} - V _{A}}{R _{2}} = 0

\frac{8 - V _{A}}{4} + \frac{4 - V _{A}}{12} = 0

I multiply everything by 12:

3 (8 - V_{A}) + (4 - V_{A}) = 0 \Rightarrow 24 - 3 V_{A} + 4 - V_{A} = 0 \Rightarrow 28 = 4 V_{A} \Rightarrow V_{A} = 7 V

So $V_{t h} = 7 V$ .

Step 3: reconnect $R_{L}$ .

I_{L} = \frac{V _{t h}}{R _{t h} + R _{L}} = \frac{7}{3 + 4} = 1 A

P_{L} = I_{L}^{2} \cdot R_{L} = 1^{2} \cdot 4 = 4 W

Worked exercise 3: from Thevenin to Norton and maximum transfer

Take the Thevenin equivalent of exercise 1 ( $V_{t h} = 4 V$ , $R_{t h} = 2 Ω$ , load disconnected). Find the Norton equivalent and check the condition for maximum power transfer.

Norton equivalent.

I_{N} = \frac{V _{t h}}{R _{t h}} = \frac{4}{2} = 2 A, R_{N} = R_{t h} = 2 Ω

The equivalent Norton circuit is a $2 A$ current source in parallel with $2 Ω$ .

Maximum power transfer. The load absorbs the maximum power when $R_{L} = R_{t h} = 2 Ω$ :

P_{ma x} = \frac{V _{t h}^{2}}{4 R _{t h}} = \frac{4 ^{2}}{4 \cdot 2} = \frac{16}{8} = 2 W

Direct check with $R_{L} = 2 Ω$ : $I_{L} = V_{t h} / (R_{t h} + R_{L}) = 4/4 = 1 A$ , so $P_{L} = 1^{2} \cdot 2 = 2 W$ , which matches, as it must.

Note that this is the same circuit as exercise 1, but there the load was $R_{L} = 2 Ω$ , which is already the value of maximum transfer: it is no accident, I chose it on purpose to show you that the $P_{L} = 2 W$ obtained there is already the maximum possible power for that circuit, whatever other value of $R_{L}$ you choose.

Where it leads next

Thevenin and Norton remain the reference tool even when, later in the year, you move from direct current to alternating current and phasors: the same method applies with complex impedances in place of pure resistances. Superposition too, which I have only mentioned here, comes in useful as an alternative method when a circuit has several sources and you want to calculate the contribution of each separately before adding them up.

If you first need a revision of series, parallel and Kirchhoff, you will find it in Ohm and Kirchhoff; the high-school version of circuits, with capacitors as well, is in electrostatics and circuits.

How I work on this topic in lessons

The problem I see most often is not understanding the theory in the abstract, but applying it to ever more complex diagrams without losing track of which sources are on or off at each step. In lessons we work on progressively more elaborate diagrams, and I always insist on the final check: if you have calculated correctly, the Thevenin equivalent and the Norton one must give exactly the same current and the same voltage on the load, whatever value of $R_{L}$ you pick to check.

There is no online practice test on circuits yet: the tests already available are in the high school practice tests category (questions in Italian). For the rest of the specialist physics you will find the overall picture on the physics page, and if you want a hand you can write to me via contacts: the first introductory call is free and lasts half an hour.

Thevenin and Norton theorems: the method with exercises