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Friday, 18 August 2017

NORTON ' S THEOREM

NORTON ' S THEOREM


Whereas thevenin's theorem was used to simplify a network to a constant - voltage source and a series resistance , Norton ' s theorem can be used to resolved a network into a constant current source and a parallel resistance . The interchange of voltage sources and current sources by use of Thevenin's and Norton's theorems is sometimes useful in circuit analysis .


            The theorem may be stated as follows ; -

              " Any two- terminal linear network containing independent voltage and current sources may be replaced by equivalent current I n in parallel with a resistance Rn where In is the short ckt current at network terminals and Rn is the equivalent resistance of network as seen from the terminals but with all voltage sources short ckted and all current sources open ckted . "


        The following procedure may be adopted to determine the Norton's equivalent ckt :

        

             1. calculate the short ckt current (In )at the network terminals . 

             2 . Redraw the network with each voltage source replaced by a short ckt in series with its                          internal resistance and each current source by an open ckt in parallel with its internal                              resistance .

              3. Calculate the resistance ( Rn ) of the redrawn network as seen from the network terminals                 (The resistance Rn is the same value as used in Thevenin 's equivalent ckt .)


      EXAMPLE 1. By using Norton's theorem find the current in the 12 ohms resistance of the ckt                                    shown in fig below :



   Solution . 

                    with 12 Ω resistance removed and terminals L-M short ckted , 

                  short ckt current , In = 20 / 8  = 2.5     

                 with 20V battery replaced by a short ckt < the resistance of the network   as seen from                                terminals L and M is

                            Rn = 8( 10 + 14 ) / 8 + 10 + 14 ) = 6Ω

                  

                        The Norton's equivalent ckt is shown in fig below 




The current through 12Ω resistance is 

I =2.5 x 6 / 6 + 12 = 0.833A    answer

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