Thursday, September 6, 2012


Norton’s theorem


Any two terminals of linear active bilateral network can be replaced by an equivalent current source in parallel to a resistance. The current source being the short circuit current through the load terminals and the resistance being internal resistance of the source network looking through the open circuit load terminals.

In thevenin’s theorem we simplified our circuit with a voltage source in series with a resistance. In Norton’s theorem we use a current source in parallel to a resistance instead of a voltage source in series to a resistance.

This theorem reduces a normally complicated network to a simple parallel circuit consisting of
   a)    An ideal current source IN Of infinite internal resistance and,
   b)   A resistance RN  in parallel with it.

IN is the current which would flow through a short circuit placed across terminals A&B. RN is equal to the thevenin’s resistance Rth . So RN is the resistance looking back from A-B terminals.

How to Nortonise a given circuit:

Consider we have to nortonise the network in figure 13.1 as seen from A-B terminals

First we short circuit the given two terminals (A&B) and find the current flowing through. 

See figure 13.2, this current is called the Norton’s current (IN).

Then again we open circuit the two terminals. And find the resistance as we do in the thevenin’s theorem. Note that for a given circuit, Norton’s resistance is equal to the thevenin’s resistance.
After that we can redraw the figure as shown in figure 13.3

Note that we connect the Norton’s resistance in parallel with Norton’s current source.

 We will discuss some solved problems on Norton's in my next post.

pabindu lakshitha
B.Sc(Engineering Undergraduate)

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