Page 47 - NETWORK ANALYSIS
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                                                                                                                       Thevenin Theorem

                       Figure 2.4(f) shows the new circuit;
















                                                          Figure 2.4(f)



                              By using Voltage Divider Rule (VDR),


                               Vth    =   V4Ω =     4   x (-j150) = (60 – j120) Volts
                                              (4 + j3 – j5)



                       Step 3


                       Draw the Thevenin equivalent circuit with the impedance (2 + j4) Ω connected to the

                       circuit.


















                                             Figure 2.4(g): Thevenin equivalent circuit


                       Step 4


                       Calculate the current that flow through (2 + j4) Ω.


                                        (60−j120 )
                          i(2 + j4)Ω   =   (0.8 − j1.6) + (2+j4)   = (-8.823 – j35.294) Amps

                       The current flowing through the impedance (2 + j4) Ω = (- 8.823 – j35.294) Amps
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