Page 94 - NETWORK ANALYSIS
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86
                                                                                                         Norton Theorem

                       Step 2


                       Calculate the Norton current source, IN by replacing 8Ω with a short circuit.


















                                                           Figure 3.10(d)


                               By using Ohms Law;


                                            V
                                 IT   = I1   =      =           (25)

                                            R     ([((−j2)//(j10))  +  (−j4)] // (10) ) + (j20)
                                   =          (25)         =       (25)      = (0.301 – j1.562) A
                                       (2.970−  4.569) + (  20)    (2.970 +   15.430)




                            By using Current Divider Rule (CDR);


                                 I2    =     (10)        x (0.301 – j1.562)
                                      10 + [(j10 //−j2) −j4]

                                   =     (10)    x (0.301 – j1.562) = (0.925 – j0.960) A
                                      10 + [−j6.5]

                                 IN    =    (j10)    x (0.925 – j0.960) A = (  .        – j1.2) A
                                      (j10 −j2)
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