Thevenin and Norton Equivalent Circuits

One Black Box, Two Equivalents

Thévenin's and Norton's theorems are among the most useful results in all of circuit theory because they let you replace an arbitrarily complicated network of sources and resistors with a single source and a single resistor, as seen from one pair of terminals. Once a circuit has been reduced to its equivalent, questions about how it drives different loads become almost trivial. The theorems apply to any linear network, no matter how tangled the interior, and they come in two interchangeable flavors. The Thévenin equivalent is a voltage source $V_{th}$ in series with a resistance $R_{th}$. The Norton equivalent is a current source $I_N$ in parallel with the same resistance $R_N = R_{th}$. Both describe the identical black box; you simply pick whichever is more convenient for the problem at hand.

The power of the idea is that the outside world cannot tell the difference. If you measure voltage and current at the terminals of the original network for every possible load, you will get exactly the same readings from its Thévenin or Norton equivalent. That is what "equivalent" means here, and it is guaranteed by the linearity of the underlying elements.

The Three Quantities You Need

A complete equivalent is fixed by any two of three quantities, and the third follows from the other two. The open-circuit voltage at the terminals is the Thévenin voltage:

The short-circuit current that flows when you connect a wire across the terminals is the Norton current:

The equivalent resistance ties the two together. It is the resistance seen looking into the terminals with every independent source turned off, and it equals the ratio of the open-circuit voltage to the short-circuit current:

Rearranging that relation gives the source-transformation identity that converts freely between the two forms:

Finding R_th by Zeroing the Sources

The most reliable way to find $R_{th}$ when a circuit contains only independent sources is to deactivate them and compute the resistance looking back into the terminals. Deactivating a source means setting its value to zero:

• An independent voltage source becomes a short circuit. A zero-volt source imposes zero volts between its nodes, which is exactly what a wire does, so replace it with a wire.
• An independent current source becomes an open circuit. A zero-amp source forces zero current through its branch, which is exactly what a break does, so remove it.

With the sources zeroed, collapse the remaining resistor network using series and parallel combinations until a single equivalent resistance remains at the terminals. This is the same source-killing operation you use in the superposition theorem. One important caveat: if the network contains a dependent source, you cannot simply zero it, because its value tracks a circuit variable. In that case apply a 1 V test source (or a 1 A test source) at the terminals, solve for the resulting current (or voltage), and compute $R_{th} = V_{test}/I_{test}$.

A Step-by-Step Recipe

1. Identify the two terminals where you want the equivalent, and remove the load resistor that connects to them.
2. Find $V_{th}$ as the open-circuit voltage across those terminals, using nodal or mesh analysis on the remaining network.
3. Find $R_{th}$ by zeroing all independent sources and reducing the resistor network (or by the test-source method if dependent sources are present).
4. Obtain the Norton current from $I_N = V_{th}/R_{th}$, or independently as the short-circuit current, and use it as a cross-check.
5. Reattach the load to the simple equivalent and solve the now-trivial single-loop circuit.

Worked Example

Consider a 12 V source in series with a 4 kΩ resistor $R_1$, with a 12 kΩ resistor $R_2$ connected from the midpoint node down to ground. We want the Thévenin equivalent seen by a load attached between the midpoint node (call it terminal a) and ground (terminal b).

Step 1 — open-circuit voltage. With the load removed, no current is drawn from terminal a, so $R_1$ and $R_2$ form a simple voltage divider. The open-circuit voltage is:

Step 2 — equivalent resistance. Zero the 12 V source by replacing it with a short. Now $R_1$ and $R_2$ both connect from terminal a to ground, so they appear in parallel:

Step 3 — Norton current. Short the terminals. The shorted $R_2$ carries no voltage, so the current is set entirely by the source and $R_1$:

Cross-check. The source transformation must be consistent: $V_{th} = I_N R_{th} = (3\text{ mA})(3\text{ k}\Omega) = 9\text{ V}$, which matches Step 1 exactly. So the Thévenin equivalent is a 9 V source in series with 3 kΩ, and the Norton equivalent is a 3 mA source in parallel with 3 kΩ. As a final demonstration, attach a 6 kΩ load: the Thévenin model gives $V_L = 9 \cdot 6\text{k}/(3\text{k} + 6\text{k}) = 6\text{ V}$ and $I_L = 9/(9\text{k}) = 1\text{ mA}$.

Thévenin vs Norton at a Glance

Common Mistakes

• Leaving the load attached while finding $V_{th}$. The Thévenin voltage is the open-circuit voltage; you must remove the load first, or you will compute the loaded voltage instead.
• Zeroing the wrong way. A voltage source is replaced by a short and a current source by an open. Reversing these — opening a voltage source or shorting a current source — is the classic blunder that wrecks $R_{th}$.
• Deactivating dependent sources. Only independent sources are zeroed. Dependent sources stay active, so you must use a test source to find $R_{th}$ when they are present.
• Mismatched sign or reference for $I_{sc}$. The short-circuit current direction must be consistent with the polarity assumed for $V_{oc}$, or the ratio $V_{oc}/I_{sc}$ comes out with the wrong sign.
• Forgetting that the two equivalents share one resistance. $R_N$ always equals $R_{th}$; students sometimes recompute it differently for each form and get inconsistent answers.

Try It in CircuitMath

You can build the example network in the editor, place a probe across the terminals of interest, and let CircuitMath solve the node voltages so you can read off $V_{oc}$ directly. Reducing the source-killed network by hand and confirming it against the tool's results is a great way to build confidence with the method. Browse the full tutorial library for related analysis techniques.

Related Tutorials

• Superposition Theorem — the same source-killing idea, used to add per-source contributions.
• Nodal Analysis Step by Step — the workhorse for finding $V_{oc}$.
• Voltage Divider Equation — the shortcut used to get $V_{th}$ above.