Page 55 - NETWORK ANALYSIS
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                                                                                                                       Thevenin Theorem

                                Vth = V4Ω = i x 4Ω


                          By using Current Divider Rule (CDR),


                                            (j5)
                               i      =             x  iT
                                         (j5 + 4− j8)

                                          V
                                  iT   =    =          20         =    20    = 1.22 − j0.976 A
                                         RT   6 +[j5 parallel to (4−j8)]  6 +[4+j8]

                                 i   =    (j5)    x iT =    (j5)    x (1.22 − j0.976) A = 0.049 + j1.562 A
                                    (j5 + 4− j8)     (j5 + 4− j8)




                                 Vth = V4Ω = i x 4 Ω   = (0.049 + j1.562) x (4) = (0.196 + j6.248) V



                       Step 3


                       Draw the Thevenin equivalent circuit with the impedance (5+j10) Ω connected to the

                       circuit.


















                                                 Figure 2.7(f): Thevenin equivalent circuit



                       Step 4


                       Calculate the current that flow through (5 + j10) Ω.


                           i(5+j10) Ω   =    (0.196 + j6.248 )    = (0.424 + j0.337) Amps
                                     (2.462 – j1.202) + (5+j10)


                       The current flowing through the impedance (5 + j10) Ω = (0.424 + j0.337) Amps
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