What is Norton's Theorem?

A powerful shortcut for circuit analysis: any linear two-terminal network of sources and resistors can be replaced by one current source IN in parallel with one resistance RN. Learn how to find IN and RN, the step-by-step method, its link to Thévenin’s theorem, and a worked example.

Complete Learning Path — Norton's Theorem

From the statement and the Norton equivalent, to finding IN & RN, the steps, source transformation, load current and a worked example

What is Norton's Theorem?

Norton's theorem says that any linear two-terminal network — no matter how many sources and resistors it contains — behaves, as seen from its two output terminals, exactly like a single current source IN in parallel with a single resistance RN.

It is the current-source twin of Thévenin’s theorem (which uses a voltage source in series with a resistance). Both let you replace a messy circuit with a tiny, easy-to-analyse equivalent.

Norton's theorem: a linear network replaced by a current source I_N in parallel with resistance R_N feeding a load
Any linear two-terminal network = a current source IN in parallel with a resistance RN.
IN
Norton current
RN
Norton resistance
∥
In parallel
= RTh
Same as Thévenin R
Only for linear circuits

Norton's theorem applies to linear networks (resistors and independent/dependent linear sources). The load can be anything — the equivalent stays the same as you swap loads.

The Norton Equivalent Circuit

The Norton equivalent has just two parts: the source IN and the parallel resistance RN. Connect any load RL across it and it delivers exactly the same voltage and current as the original network.

Norton equivalent = IN ∥ RN

A current source IN in parallel with a resistance RN, presented at terminals a-b

Why it saves time

Once you have IN and RN, finding the current for dozens of different loads is a one-line current-divider calculation each time — no need to re-solve the whole network.

Finding the Norton Current IN

The Norton current is the short-circuit current: remove the load, place a wire (short) across the output terminals a-b, and find the current through that short.

Finding the Norton current by shorting the output terminals and measuring the short-circuit current I_N
Short the terminals a-b; the current through the short is the Norton current IN.

IN = Ishort-circuit (through a-b)

Use Ohm's law, mesh or nodal analysis to compute the current in the shorting wire

Finding the Norton Resistance RN

Deactivate every independent source, then look back into the terminals. Voltage sources become shorts; current sources become opens. The resistance you see is RN — identical to the Thévenin resistance.

Finding the Norton resistance by turning off all sources and measuring the resistance at terminals a-b
Turn all sources OFF (voltage → short, current → open) and find the resistance at a-b: that is RN = RTh.
Deactivating sources correctly

Replace an ideal voltage source with a short (0 V → a wire) and an ideal current source with an open (0 A → a gap). Leave resistors in place, then simplify series/parallel combinations.

Step-by-Step Method

Four steps turn any linear network into its Norton equivalent and give you the load current.

Remove the load

Take out RL and mark terminals a-b.

Short a-b → IN

Find the short-circuit current.

Sources off → RN

Resistance at a-b with sources deactivated.

Rebuild & solve

Draw IN∥RN, reconnect RL, use the current divider.

Norton vs Thévenin (Source Transformation)

Norton and Thévenin equivalents describe the same network and convert directly into each other by source transformation.

Source transformation between the Norton equivalent (I_N parallel R_N) and the Thevenin equivalent (V_Th series R_Th)
A current source IN ∥ RN (Norton) equals a voltage source VTh in series with RTh (Thévenin).

VTh = IN × RN  ·  RTh = RN  ·  IN = VTh / RTh

Convert between the two equivalents; the resistance is identical, and the sources relate by Ohm's law

Load Current from the Norton Model

With the Norton equivalent built, the current into any load RL follows straight from the current-divider rule.

Load current from the Norton equivalent using the current divider between R_N and R_L
The source current splits between RN and RL; the fraction into the load is the current-divider ratio.

IL = IN × RN / (RN + RL)

Current-divider rule: the load takes the share set by the parallel resistances

Worked Example

A 12 V source with R1 = 4 Ω in series and R2 = 4 Ω across the output. Find the Norton equivalent.

Worked Norton example: 12 V source with two 4 ohm resistors gives I_N = 3 A and R_N = 2 ohms
Short a-b for IN; deactivate the source for RN.
Norton current

Short a-b → R2 is bypassed, so IN = 12 V / R1 = 12 / 4 = 3 A.

Norton resistance

Short the source → R1 and R2 are in parallel: RN = 4 ∥ 4 = 2 Ω.

Check via Thévenin

VTh = IN × RN = 3 × 2 = 6 V — consistent with the Thévenin equivalent.

Where Norton's Theorem is Used

Any time you must analyse one part of a big circuit — especially for many load values — Norton (or Thévenin) is the tool.

Variable loads

Find load current quickly as RL changes — one divider per value.

Circuit design

Model a supply or amplifier output as IN ∥ RN.

Power transfer

Maximum power to the load when RL = RN.

AC networks

Works with impedances: IN ∥ ZN.

Key Terms at a Glance

The essential Norton vocabulary students search for.

Norton current (IN)

Short-circuit current at a-b.

Norton resistance (RN)

Resistance at a-b, sources off.

Norton equivalent

IN in parallel with RN.

Source transformation

Norton ↔ Thévenin.

Current divider

Splits IN between RN & RL.

Linear network

Resistors + linear sources only.

Frequently Asked Questions

Quick, expert answers to the questions people ask most about Norton's theorem.

What is Norton's theorem in simple words?

It says any linear circuit with two output terminals can be replaced by a single current source (IN) with a resistor (RN) in parallel — a simple model that behaves exactly like the original from the load's point of view.

How do you find the Norton current?

Short-circuit the two output terminals and find the current through the short. That is IN, the short-circuit current.

How do you find the Norton resistance?

Turn off all independent sources (voltage → short, current → open) and find the resistance looking into the terminals. That is RN, equal to the Thévenin resistance.

What is the relationship between Norton and Thévenin?

They are interchangeable: RN = RTh and VTh = IN × RN. Converting one to the other is called source transformation.

What are the steps of Norton's theorem?

Remove the load; short the terminals to get IN; deactivate sources to get RN; draw IN∥RN, reconnect the load and use the current divider.

How do you find the load current?

Apply the current divider: IL = IN × RN / (RN + RL).

Does Norton's theorem work for AC?

Yes — use impedances instead of resistances. The network reduces to a Norton current phasor IN in parallel with a Norton impedance ZN.

When should I use Norton instead of Thévenin?

Use whichever is more convenient — they give the same answer. Norton (a current source in parallel) is handy for parallel circuits and current-divider problems; Thévenin suits series/voltage problems.

Conclusion & Key Takeaways

Norton's theorem shrinks any linear two-terminal network to a current source and a parallel resistance — fast, reusable and exact.

IN ∥ RN

The Norton equivalent.

IN = short-circuit I

Short a-b to find it.

RN = RTh

Sources off, look in.

↔ Thévenin

VTh = INRN.

Current divider

Gives the load current.

Linear only

Great for variable loads.

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