Norton's Theorem

Reach for Norton's theorem when a linear two-terminal network is messy and you want the load current quickly, especially if the load changes and you need to recompute branch current several times. The theorem says any linear two-terminal network can be replaced, at its output terminals, by an equivalent current source in parallel with an equivalent resistance, without changing the voltage or current seen by the load. The Norton form is especially handy when current division is more natural than voltage division. As long as the network is linear and you keep the same two terminals, the replacement gives the same terminal behavior for any load:

where $I_N$ is the short-circuit current at the output terminals and $R_N$ is the equivalent resistance seen looking back into the network. What the equivalent preserves is not the inside of the circuit but the voltage-current behavior measured at the two terminals.

Step 1: Mark The Load Terminals

Start with the same two terminals where the external load connects, because the equivalent circuit is defined only from that viewpoint. Remove the load and mark the terminals.

Step 2: Find The Norton Current

Compute the short-circuit current across those terminals to get $I_N$: short the terminals and calculate the resulting current.

Step 3: Find The Norton Resistance

Look back into the network to find $R_N$ with independent sources turned off: independent voltage sources become shorts and independent current sources become opens. If the circuit contains dependent sources, do not simply turn everything off; use a test source to find the effective resistance.

If you already know the Thevenin equivalent of the same linear network, there is a shortcut:

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

Step 4: Reconnect The Load

Replace the original network with the Norton current source in parallel with $R_N$, then solve the simpler circuit for the load quantities you need.

Full Example, Step By Step

The original network is 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.

Step 3 (via the shortcut) — Norton current:

So the Norton circuit is a $2 \, \text{A}$ source in parallel with $R_N = 6 \, \Omega$.

Step 4 — reconnect the $3 \, \Omega$ load. The source, the $6 \, \Omega$ branch, and the $3 \, \Omega$ load are all in parallel, so first combine the two resistors:

The voltage across both branches is then

and the load current is

This matches the load behavior of the original circuit. Different internal forms, same behavior at the terminals.

Where Each Step Tends To Trip You Up

• Step 3 (Norton resistance): finding $R_N$ from the live circuit without stating what happens to the independent sources. Independent voltage sources become shorts, independent current sources become opens, and dependent sources require a test source instead.
• Step 1 (scope): assuming Norton form works for any circuit. The standard theorem applies to linear two-terminal networks; a nonlinear circuit may not hold the same simple equivalent over every operating condition.
• Step 4 (interpreting the result): mixing up the source current with the load current. The current source feeds the whole parallel network, and the load current is only one branch current.

A useful self-check: Norton is the natural partner to Thevenin's theorem. The two forms carry the same terminal information, so after solving, convert your Norton equivalent into Thevenin form and confirm the load result stays the same. A clean practice case is a $5 \, \text{A}$ source in parallel with $10 \, \Omega$ feeding a $15 \, \Omega$ load: find the terminal voltage first, then the load current, then convert and recheck.

Frequently Asked Questions