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                                                                                                         Norton Theorem

                       Step 3



                       Draw the Norton’s equivalent circuit with ZL = (4 + j1) Ω connected to the circuit.















                                                Figure 3.5(f): Norton equivalent circuit




                       Step 4



                       Calculate the current that goes through (4 + j1) Ω.

                          By using Current Divider Rule (CDR),

                                         N            (3.657 – j1.886)
                                  iL  =      x IN =                    x (–  0.328  +  j1.393)A
                                       +   L      (3.657 – j1.886) + (4 + j1)
                                     N

                                 = (  .        +     .       ) A


                          The current that flow through the impedance (4 + j1) Ω = (  .        +     .       ) A





                       Example 6


                       Determine the current that goes through (30 – j30) Ω.















                                                      Figure 3.6(a)
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