Norton's Theorem

Norton's theorem says that any linear two-terminal network can be replaced, at its output terminals, by an equivalent current source in parallel with an equivalent resistance. For a student solving circuit problems, that means you can swap a messy network for a simpler one without changing the voltage or current seen by the load.

If the network is linear and you keep the same two terminals, the replacement gives the same terminal behavior for any load attached there. Norton form is especially handy when you want the load current.

Norton's Theorem in One Statement

For a linear two-terminal network, the external circuit can be replaced by

$$\text{Norton equivalent} = I_N \text{ in parallel with } R_N$$

Here, $I_N$ is the short-circuit current at the output terminals, and $R_N$ is the equivalent resistance seen looking back into the network.

If you already know the Thevenin equivalent of the same linear network, then

$$I_N = \frac{V_{TH}}{R_{TH}}, \qquad R_N = R_{TH}$$

That shortcut works only when both equivalents describe the same two-terminal linear circuit.

What the Equivalent Circuit Preserves

The inside of the circuit does not stay the same. What stays the same is the voltage-current behavior measured at the two terminals.

That distinction matters because two very different circuits can still be equivalent from the load's point of view. If they produce the same terminal voltage and current for every load connected at those terminals, they are equivalent there.

How To Find a Norton Equivalent

Use this sequence:

1. Remove the load and mark the two terminals.
2. Find $I_N$ by shorting those terminals and calculating the resulting current.
3. Find $R_N$ by looking back into the network with independent voltage sources replaced by shorts and independent current sources replaced by opens.
4. If the circuit contains dependent sources, do not simply turn everything off. Use a test source to find the effective resistance.
5. Draw the Norton equivalent and reconnect the load.

Worked Example With Numbers

Suppose the original network is already known in Thevenin form: an ideal $12 \, \text{V}$ source in series with a $6 \, \Omega$ resistor. Find the Norton equivalent, then connect a $3 \, \Omega$ load.

Start with the source current:

$$I_N = \frac{V_{TH}}{R_{TH}} = \frac{12}{6} = 2 \, \text{A}$$

So the Norton circuit is a $2 \, \text{A}$ current source in parallel with

$$R_N = 6 \, \Omega$$

Now attach the load $R_L = 3 \, \Omega$. The source, the $6 \, \Omega$ branch, and the $3 \, \Omega$ load are all in parallel, so first combine the two resistors:

$$R_{eq} = \frac{6 \cdot 3}{6 + 3} = 2 \, \Omega$$

The voltage across both branches is then

$$V = I_N R_{eq} = (2)(2) = 4 \, \text{V}$$

Since the load has the same $4 \, \text{V}$ across it, the load current is

$$I_L = \frac{V}{R_L} = \frac{4}{3} \, \text{A}$$

This matches the load behavior of the original circuit. That is exactly the point of Norton's theorem: different internal forms, same behavior at the terminals.

Common Mistakes Students Make

One common mistake is finding $R_N$ from the live circuit without saying what happens to the independent sources. For this step, independent voltage sources become shorts and independent current sources become opens.

Another mistake is assuming Norton form works for any circuit. The standard theorem applies to linear two-terminal networks. If the circuit is nonlinear, the same simple equivalent may not hold over every operating condition.

A third mistake is mixing up the source current with the load current. In a Norton circuit, the current source feeds the whole parallel network, and the load current is only one branch current.

When Norton's Theorem Is Useful

Norton's theorem is useful when the load changes and you want to recompute branch current quickly. It also helps when you simplify part of a larger circuit or when current division is more natural than voltage division.

It is also the natural partner to Thevenin's theorem. The two forms contain the same terminal information, but one form may make a specific calculation shorter.

Try a Similar Problem

Try your own version with a Norton source of $5 \, \text{A}$ in parallel with $10 \, \Omega$, then attach a $15 \, \Omega$ load. Find the terminal voltage first, then the load current. After that, convert the same circuit into Thevenin form and check that the load result stays the same.