Page 74 - NETWORK ANALYSIS
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                                                                                                         Norton Theorem














                                                             Figure 3.2(i)

                             By using Ohms Law;

                                     V        15
                                  IN =     =             = (0.6 – j1.2) A
                                     R     (5 +j10)




                       Step 3



                         Draw the Norton equivalent circuit with the 20 Ω resistor connected to the circuit.













                                                  Figure 3.2(j): Norton equivalent circuit




                       Step 4



                       Calculate the current flowing through the 20Ω resistor.

                          By using Current Divider Rule (CDR),

                                    R N              (22.5 +j7.5)
                               iL=        x IN      =              x (0.6 –  j1.2)  =  (0.423 –  j0.604) A
                                  R +R  L          ( (22.5 +j7.5)+(20)
                                   N


                          The current that flow through the 20 Ω resistor = (0.423 –    0.604) A
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