Norton’s Theorem

Norton’s Theorem

We have already looked at what a current source is in a previous article on Voltage and Current Sources. Norton’s theorem builds on that knowledge stating that:

Any linear electrical network containing only voltage sources, current sources and resistances can be replaced at terminals A–B by an equivalent combination of a current source I_N in parallel with a resistance R_N.

As a reminder here is a schematic for a practical current source equivalent circuit:

What Norton’s theorem is saying is that any circuit comprised of only current sources and resistors can be represented by the above equivalent circuit containing only a single current source and a single parallel connected resistor. The power of this revelation is that it can be used to simplify the analysis of circuits.

Norton in Action

Consider the following, in which we will use R_L to represent some arbitrarily complex target that we wish to analyse that is being powered by the surrounding circuitry:

Deriving the Norton Equivalent

Step 1

Remove RL

Step 2

Use the voltage-current source equivalency principle to derive the equivalent current source to replace the existing 12V, 30Ω voltage source:

\(I = \frac{V}{R} = \frac{12}{30} = 0.4A = 400mA\\\) \(R_{is} = R_{vs} = 30\Omega\)

Step 3

Short out terminals A and B, then calculate the Norton current (I_N), by calculating the current that flows through the short:

\(I_N = IS_1 + IS_2 = 400mA + 1A = 1400mA\)

Step 4

Remove the short between A and B. Also remove all current sources leaving only their internal resistances:

Step 5

Calculate the Norton resistance (R_N) by calculating the total resistance of the circuit:

\(R_N = \frac{R_1 . R_2}{R_1 + R_2} = \frac{30 * 60}{30 + 60} = \frac{1800}{90} = 20\Omega\)

Step 6

Draw the Norton Equivalent Circuit: