Norton's Theorem

Norton's Theorem, aka the Mayer–Norton theorem, states that any linear circuit can be represented by a single current source, In, in parallel with resistance Rn.

This is remarkably similar to Thévenin's theorem, the only difference is that we convert a complex network into a current source in parallel with a resistor instead of a voltage source in series with a resistor.

The circuit shown above on the left can be simplified to the circuit on the right "Norton Equivalent Circuit" using Nortons Theorem. If we attach a circuit to terminals X and Y, a load, then both circuits will drive the load identically.

Mayer and Norton

Hans Ferdinand Mayer
Edward Lawry Norton for whom the Norton equivalent circuit is named(Photograph taken October 13,1925 and reproduced courtesy of the AT&T Archives)

Norton's theorem was independently derived in the same year 1926 by Siemens and Halske researcher Hans Ferdinand Mayer and Bell Labs engineer Edward Lawry Norton.

Determining the Norton Equivalent Circuit

We can derive the Norton Equivalent Circuit using a process which can be broken down into two steps, calculating the Norton Current and the Norton Parallel Resistance.

Norton Current

The Norton Current is the short circuit current measured through terminals X and Y which can be connected to an external load.

Norton Resistance

The Norton Resistance, like the Thevenin Resistance, is the resistance measured across terminals X and Y when all of the sources are "zeroed".

We can think of this as being the resistance looking back into terminals X and Y.

Remember: To "zero" a voltage source we replace it with a short circuit and to "zero" a current source we replace it with an open circuit. Recall that the voltage across a short is zero and no current will flow through an open circuit hence effectively zeroing their effect.

Relationship between Thevenin Voltages and Norton Currents:

The Thevenin and Norton resistances are equivalent and can also be determined by using Vth and In with Ohm's law, i.e.

Examples:

1. a) Determine the Norton Current of the circuit below:

The calculate the Norton Current, we need to find the short circuit current which would flow between terminals X and Y. With a short through terminals X and Y we then have a simple current divider circuit with I1 being divided between branches containing R2 and R3 and the branch containing R3 is the current we need to calculate:

b) Determine the Norton Resistance.

To determine the Norton Resistance we need to "zero" sources and then redraw the circuit. In this case we simply replace the current source with an open circuit.
The Norton Resistance is then equal to the resistance seen across terminals X and Y which is simply R2 in parallel with R3 as no current will flow through R1 with the open:

c) Draw the Norton Equivalent Circuit

d) Determine the current which would flow through a 2kΩ load

With the Norton equivalent circuit at hand the current which would flow through a 2kΩ load would be a branch current in parallel with Rn which we can calulate with the current divider rule:

2. a) Determine the Norton Current of the circuit below:

Let's begin by shorting out X and Y which means that no current will flow through R5 so we can remove it and redraw the circuit as follows:
The simplification has helped but this is still by no means a simple circuit we can solve by inspection, and as always it is best to apply a systematic technique in this case. Let's use nodal analysis and redraw the circuit with nodes, currents and voltages shown:
At node Va:
At node Vb:
From here we can multiply [2] by 3 and then add [1] and [2] together to eliminate Va to solve for Vb which we need to determine In:
We can now use Vb and Ohm's Law to calculate In:

b) Determine the Norton Resistance.

To determine the Norton Resistance we need to "zero" sources and then redraw the circuit. In this case we simply replace the current sources with an open circuits:
From here we can simplify the circuit further by removing the open circuit paths:
The Norton resistance is then simply (R4+R3+R2)||R5:

c) Determine the current which would flow through a 1kΩ load

Let's start by drawing our Norton Equivalent Circuit with load Rl:
From here we can apply the current divider rule to determine the current through the load:

Questions:

1. a) Determine the Norton Current of the circuit below:

Answer

IN=95.238uA

b) Determine the Norton Resistance.

Answer

RN=2048.780Ω

c) Draw the Norton Equivalent Circuit.

d) What current would flow through a 200Ω load?

Answer

I=86.768uA

2. a) Determine the Norton Current of the circuit below:

Answer

IN=20mA

b) Determine the Norton Resistance.

Answer

RN=125Ω

c) Draw the Norton Equivalent Circuit.

d) What would the voltage drop across a 2kΩ load be?

Answer

VL=2.353V

3. a) Determine the Norton Current of the circuit below:

Answer

IN=2.4mA

b) Determine the Norton Resistance.

Answer

RN=3333.333Ω

c) Draw the Norton Equivalent Circuit.

d) What current would flow through a 5kΩ load?

Answer

IL=959.999uA

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