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


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:

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.
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:



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:

c) Determine the current which would flow through a 1kΩ load
Let's start by drawing our Norton Equivalent Circuit with load Rl:
Questions:
1. a) Determine the Norton Current of the circuit below:

Answer
b) Determine the Norton Resistance.
Answer
c) Draw the Norton Equivalent Circuit.

d) What current would flow through a 200Ω load?
Answer
2. a) Determine the Norton Current of the circuit below:

Answer
b) Determine the Norton Resistance.
Answer
c) Draw the Norton Equivalent Circuit.

d) What would the voltage drop across a 2kΩ load be?
Answer
3. a) Determine the Norton Current of the circuit below:

Answer
b) Determine the Norton Resistance.
Answer
c) Draw the Norton Equivalent Circuit.

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