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                                                                                                                       Thevenin Theorem

                       Determine the new voltage source by using Ohm’s law and 3Ω connected to the new

                       source in series.





                           V = IR
                              = 2 x 3
                              = 6V




                                                             Figure 2.2(f)
















                                                          Figure 2.2(g)



                           Vth    =   i x (–j5Ω)


                                V        6
                           i   =     =         = (1.8 + j0.6) A
                                R    (3+j4−j5)

                            Vth = i x (–j5Ω) = (1.8 + j0.6)   x (–j5Ω) = (3 – j9) Volts






                       Method #3

                           By using Voltage Divider Rule (VDR),


                           Vth    =    −j5    x 6V  =  (3 – j9) Volts
                                    (3+j4−j5)
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