Norton's theorem

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Norton's theorem for electrical networks states that any collection of voltage sources, current sources, and resistors with two terminals is electrically equivalent to an ideal current source, I, in parallel with a single resistor, R. For single-frequency AC systems the theorem can also be applied to general impedances, not just resistors. The Norton equivalent is used to represent any network of linear sources and impedances, at a given frequency. The circuit consists of an ideal current source in parallel with an ideal impedance (or resistor for non-reactive circuits).

Norton's theorem is an extension of Thévenin's theorem and was introduced in 1926 separately by two people: Hause-Siemens researcher Hans Ferdinand Mayer (1895-1980) and Bell Labs engineer Edward Lawry Norton (1898-1983). Mayer was the only one of the two who actually published on this topic, but Norton made known his finding through an internal technical report at Bell Labs.

Any black box containing only voltage sources, current sources, and resistors can be converted to a Norton equivalent circuit.
Any black box containing only voltage sources, current sources, and resistors can be converted to a Norton equivalent circuit.

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[edit] Calculation of a Norton equivalent circuit

To calculate the equivalent circuit:

  1. Calculate the output current, IAB, when a short circuit is the load (meaning 0 resistance between A and B). This is INo.
  2. Calculate the output voltage, VAB, when in open circuit condition (no load resistor - meaning infinite resistance). RNo equals this VAB divided by INo.
  • The equivalent circuit is a current source with current INo, in parallel with a resistance RNo.

Step 2 can also be thought of like this:

  • 2a. Now replace independent voltage sources with short circuits and independent current sources with open circuits.
  • 2b. For circuits without dependent sources RNo is the total resistance with the independent sources removed.*

* Note: A more general method for determining the Norton Impedance is to connect a current source at the output terminals of the circuit with a value of 1 Ampere and calculate the voltage at its terminals; this voltage is equal to the impedance of the circuit. This method must be used if the circuit contains dependent sources. This method is not shown below in the diagrams.

[edit] Conversion to a Thévenin equivalent

To convert to a Thévenin equivalent circuit, one can use the following equations:

R_{Th} = R_{No} \!
V_{Th} = I_{No} R_{No} \!

[edit] Example of a Norton equivalent circuit

Step 0: The original circuit
Step 0: The original circuit
Step 1: Calculating the equivalent output current
Step 1: Calculating the equivalent output current
Step 2: Calculating the equivalent resistance
Step 2: Calculating the equivalent resistance
Step 3: The equivalent circuit
Step 3: The equivalent circuit

In the example, the total current Itotal is given by:


I_\mathrm{total} = {15 \mathrm{V} \over 2\,\mathrm{k}\Omega + 1\,\mathrm{k}\Omega \| (1\,\mathrm{k}\Omega + 1\,\mathrm{k}\Omega)} = 5.625 \mathrm{mA}

The current through the load is then, using the current divider rule:


I = {1\,\mathrm{k}\Omega + 1\,\mathrm{k}\Omega \over (1\,\mathrm{k}\Omega + 1\,\mathrm{k}\Omega + 1\,\mathrm{k}\Omega)} \cdot I_\mathrm{total}

= 2/3 \cdot 5.625 \mathrm{mA} = 3.75 \mathrm{mA}

And the equivalent resistance looking back into the circuit is:


R = 1\,\mathrm{k}\Omega + 2\,\mathrm{k}\Omega \| (1\,\mathrm{k}\Omega + 1\,\mathrm{k}\Omega) = 2\,\mathrm{k}\Omega

So the equivalent circuit is a 3.75 mA current source in parallel with a 2 kΩ resistor.

[edit] In popular culture

Both Norton's theorem and Thévenin's theorem were featured in the May 4, 2006 and May 10, 2006 Doonesbury comic strip panels [1], [2]. Registration required to view 1 comic. Paid registration required to view both.

[edit] See also

[edit] External links