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Message: <blockquote data-quote="Little DJ" data-source="post: 10888283" data-attributes="member: 320663">Norton’s TheoremNorton’s theorem is a circuit analysis technique which is similar to Thevenin’s theorem. By using this theorem the circuit is reduced to a single current source and one parallel resistor. As with the Thevenin equivalent circuit, the resulting two-terminal circuit is equivalent to the original circuit when connected to any external branch or component. In summary, Norton’s theorem may be simplified as follows: Any linear bilateral network may be reduced to a simplified two-terminal circuit consisting of a single current source and a single shunt resistor as shown in following figure.<img src="http://1.bp.blogspot.com/-K3f9DLmgv-0/TlsBRYm5-zI/AAAAAAAAAHM/LCSxZaLMEjs/s200/Norton%2527s+Theorem.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" />The following steps provide a technique which allows the conversion of any circuit into its Norton equivalent:1. Remove the load from the circuit.2. Label the resulting two terminals (ab). 3. Set all sources to zero. As before, voltage sources are set to zero by replacing them with short circuits and current sources are set to zero by replacing them with open circuits.4. Determine the Norton equivalent resistance, Rn, by calculating the resistance seen between terminals a and b. It may be necessary to redraw the circuit to simplify this step.5. Replace the sources removed in Step 3, and determine the current which would occur in a short if the short were connected between terminals a and b. If the original circuit has more than one source, it may be necessary to use the superposition theorem. In this case, it will be necessary to determine the short-circuit current due to each source separately and then determine the combined effect. The resulting short-circuit current will be the value of the Norton current In.6. Sketch the Norton equivalent circuit using the resistance determined in Step 4 and the current calculated in Step 5. As part of the resulting circuit, include that portion of the network removed in Step 1.The Norton equivalent circuit may also be determined directly from the Thevenin equivalent circuit by using the source conversion technique.<img src="http://2.bp.blogspot.com/-R_1VshRSuVo/TlsClO04mAI/AAAAAAAAAHQ/QssmKSCeZ8Q/s400/Norton%2527s+Theorem+01.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" />Relations between the circuits:<img src="http://3.bp.blogspot.com/-rBN0Pw9414c/TlsC-Cl7NsI/AAAAAAAAAHU/LnrgESfsAb8/s1600/Norton%2527s+Theorem+02.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" />Example:Determine the Norton equivalent circuit external to the resistor RL and find the current through RL.<img src="http://4.bp.blogspot.com/-Upgpt87444w/TlsDw78p4qI/AAAAAAAAAHY/VnK7abTjPe0/s1600/Norton%2527s+Theorem+03.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" />Steps 1 and 2: Remove load resistor RL from the circuit and label the remaining terminals as a and b.<img src="http://1.bp.blogspot.com/-fD4g4qqlI-Y/TlsEEGpHZfI/AAAAAAAAAHc/prFtzAqjFZM/s1600/Norton%2527s+Theorem+04.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" />Step 3: Zero the voltage and current sources.<img src="http://1.bp.blogspot.com/-r5DZf9bd7Q8/TlsEVxO7zQI/AAAAAAAAAHg/i0zANybCteU/s400/Norton%2527s+Theorem+05.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" />Step 4: The resulting Norton resistance between the terminals isRn= Rab = 24 OhmStep 5: The short-circuit current is determined by first calculating the current through the short due to each source.<img src="http://3.bp.blogspot.com/-nXD0RsFRzRs/TlsE_dXTpoI/AAAAAAAAAHk/y_Gox-LVbm4/s320/Norton%2527s+Theorem+06.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" />Iab = 20V / 24 Ohm = 0.833 A<img src="http://2.bp.blogspot.com/-Ktp7V-LOVlQ/TlsFcw7BXgI/AAAAAAAAAHo/z4g2vSl9EOY/s320/Norton%2527s+Theorem+07.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" />Iab(2) = -2.0 ANotice that the current Iab(2) is indicated as being a negative quantity. As seen before, this result merely indicates that the actual current is opposite to the assumed reference direction.Now, applying the superposition theorem, The Norton current is:IN = Iab(1) + Iab(2) = 0.833 A + -2.0 A = -1.167 AAs before, the negative sign indicates that the short-circuit current is actually from terminal b toward terminal a.<img src="http://4.bp.blogspot.com/-JMzV97aTlYo/TlsGm8wlWcI/AAAAAAAAAHs/txhlShIVJDw/s1600/Norton%2527s+Theorem+08.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" />Current through load resistor RL can be determined by current dividing rule.IL = 24 Ohm/(24 Ohm + 16 Ohm) x 1.167 A  =  0. 7002 A<img src="http://2.bp.blogspot.com/-nckIW0ofD-4/TlsHZUmKZEI/AAAAAAAAAHw/6YAeOKsE454/s400/Norton%2527s+Theorem+09.JPG" alt="" class="fr-fic fr-dii fr-draggable " style="" /></blockquote>

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