Norton's theorem
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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:
- Replace the load circuit with a short.
- Calculate the current through that short, I, from the original sources.
- Now replace voltage sources with shorts and current sources with open circuits.
- Replace the load circuit with an imaginary ohm meter and measure the total resistance, R, with the sources removed.
- 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:
- <math>
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} <math>
The current through the load is then:
- <math>
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} <math>
- <math>
= 2/3 \cdot 5.625 \mathrm{mA} = 3.75 \mathrm{mA} <math>
And the equivalent resistance looking back into the circuit is:
- <math>
R = 1\,\mathrm{k}\Omega + 2\,\mathrm{k}\Omega \| (1\,\mathrm{k}\Omega + 1\,\mathrm{k}\Omega) = 2\,\mathrm{k}\Omega <math>
So the equivalent circuit is a 3.75 mA current source in parallel with a 2 kΩ resistor.
See also
External links
- Origins of the equivalent circuit concept (http://tcts.fpms.ac.be/cours/1005-01/equiv.pdf)
- Norton's theorem at allaboutcircuits.com (http://www.allaboutcircuits.com/vol_1/chpt_10/8.html)fr:Théorème de Norton