# Norton's theorem

 Missing imageThevenin_and_norton_step_1.png The original circuit Missing imageNorton_step_2.png Calculating the equivalent output current Missing imageThevenin_and_norton_step_3.png Calculating the equivalent resistance Missing imageNorton_step_4.png The equivalent circuit

Norton's theorem for electrical networks states that any collection of voltage sources and resistors with two terminals is electrically equivalent to an ideal current source I in parallel with a single resistor R. The theorem can also be applied to general impedances, not just resistors.

The theorem was published in 1926 by Bell Labs engineer Edward Lawry Norton (1898-1983).

To calculate the equivalent circuit:

1. Replace the load circuit with a short.
2. Calculate the current through that short, I, from the original sources.
3. Now replace voltage sources with shorts and current sources with open circuits.
4. Replace the load circuit with an imaginary ohm meter and measure the total resistance, R, with the sources removed.
5. The equivalent circuit is a current source with current I in parallel with a resistance R in parallel with the load.

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

[itex]

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} [itex]

The current through the load is then:

[itex]

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} [itex]

[itex]

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

And the equivalent resistance looking back into the circuit is:

[itex]

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

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

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