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Verification of Norton Theorem
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 Objective: To Verify Norton Theorem

A linear active network consisting of independent and or dependent voltage and current sources and linear bilateral network elements   can be replaced by an equivalent circuit consisting of current sources in parallel with the resistance, the current source being the short circuited current across the load terminal and resistance being the internal resistance of the source network looking through the open circuited load terminals. In order to find the current through RL , the load resistance of the given figure by Norton's theorem, let, replace RL by short circuit.

 

Circuit Diagram:

 

 

 

                                                 

 

 

  

 

                                                 

 

 Obviously,

                                                       «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨left¨ rowspacing=¨0¨»«mtr»«mtd/»«/mtr»«mtr»«mtd»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mi»I«/mi»«mo»=«/mo»«mfrac»«msub»«mi»V«/mi»«mi»s«/mi»«/msub»«mrow»«msub»«mi»R«/mi»«mn»1«/mn»«/msub»«mo»+«/mo»«mfrac»«mrow»«msub»«mi»R«/mi»«mn»2«/mn»«/msub»«mo»*«/mo»«msub»«mi»R«/mi»«mn»3«/mn»«/msub»«/mrow»«mrow»«msub»«mi»R«/mi»«mn»2«/mn»«/msub»«mo»+«/mo»«msub»«mi»R«/mi»«mn»3«/mn»«/msub»«/mrow»«/mfrac»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«/mtr»«mtr»«mtd»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«msub»«mi»I«/mi»«mrow»«mi»s«/mi»«mo»/«/mo»«mi»c«/mi»«/mrow»«/msub»«mo»=«/mo»«mi»I«/mi»«mo»*«/mo»«mfrac»«msub»«mi»R«/mi»«mn»3«/mn»«/msub»«mrow»«msub»«mi»R«/mi»«mn»3«/mn»«/msub»«mo»+«/mo»«msub»«mi»R«/mi»«mn»2«/mn»«/msub»«/mrow»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»

 Next, the short circuit is removed and the independent source is deactivated .

 

Circuit Diagram:

 

 

                                       

 

 Here,                             «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨left¨ rowspacing=¨0¨»«mtr»«mtd/»«/mtr»«mtr»«mtd»«msub»«mi»R«/mi»«mrow»«mi»i«/mi»«mi»n«/mi»«mi»t«/mi»«/mrow»«/msub»«mo»=«/mo»«msub»«mi»R«/mi»«mn»2«/mn»«/msub»«mo»+«/mo»«mfrac»«mrow»«msub»«mi»R«/mi»«mn»1«/mn»«/msub»«mo»*«/mo»«msub»«mi»R«/mi»«mn»3«/mn»«/msub»«/mrow»«mrow»«msub»«mi»R«/mi»«mn»1«/mn»«/msub»«mo»+«/mo»«msub»«mi»R«/mi»«mn»3«/mn»«/msub»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«/mtr»«/mtable»«/math»

  As per Norton's theorem , the equivalent circuit would contain a current source in parallel to the internal resistance , the current source

  being the short circuitted current across the shorted terminals of the load resistor. 

 

Circuit Diagram:

 

 

                                        

 

 Obviously ,      «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨left¨ rowspacing=¨0¨»«mtr»«mtd/»«/mtr»«mtr»«mtd»«msub»«mi»I«/mi»«mi»L«/mi»«/msub»«mo»=«/mo»«msub»«mi»I«/mi»«mrow»«mi»s«/mi»«mo»/«/mo»«mi»c«/mi»«/mrow»«/msub»«mo»*«/mo»«mfrac»«msub»«mi»R«/mi»«mrow»«mi»i«/mi»«mi»n«/mi»«mi»t«/mi»«/mrow»«/msub»«mrow»«msub»«mi»R«/mi»«mrow»«mi»i«/mi»«mi»n«/mi»«mi»t«/mi»«/mrow»«/msub»«mo»+«/mo»«msub»«mi»R«/mi»«mi»L«/mi»«/msub»«/mrow»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»

 

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