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Contoh Soal 7


                Dari rangkaian listrik di samping hitunglah:
                 a) Kuat arus pada masing-masing cabang
                 b) Beda potensial antara titik E juga antara E dan C



                  Jawab:

                 a)                                                 Menurut Hukum I Kirchoff:
                                                                     +  =  … … … (1)
                                                                               3
                                                                          2
                                                                     1
                                                                    Loop 1 (AEFDA):
                                                                    ∑ .  + ∑  = 0

                                                                      ( +  ) +  .  −  = 0
                                                                     1
                                                                         1
                                                                                       3
                                                                                    3
                                                                                            1
                                                                              1
                                                                      (2Ω + 1Ω) +  . 3Ω − 6  = 0
                     Loop II (BEFCB):                                1               3
                                                                    3Ω   + 3Ω  − 6  = 0
                     ∑ .  + ∑  = 0                                1        3
                                                                    3Ω   + 3Ω  = 6 
                       ( +  ) +  .  −  = 0                  1        3
                                     3
                               2
                                              1
                                        3
                      2
                          2
                                                                     1 Ω   + 1 Ω  = 2  … … … (2)
                       (5Ω + 1Ω) +  . 3Ω − 6  = 0                   1        3
                                      3
                      2
                     6Ω   + 3Ω  − 6  = 0                      2 x (2) :    2Ω   + 2 Ω  = 4 
                          2
                                  3
                                                                                   1
                                                                                            3
                     6Ω   + 3Ω  = 6                                 (3) : −2 Ω  + 1 Ω  = 2 V    +
                                  3
                          2
                                                                                   1
                                                                                            3
                     2 Ω   + 1 Ω  = 2  … … … (2)                                             3 Ω  = 6 V
                                    3
                          2
                                                                                            3
                     Berdasarkan Hukum I Kirchoff, maka:                                                   = ,  
                                                                                            
                      =  −                                    1 Ω  + 1 Ω  = 2 
                      2
                           3
                                1
                                                                                  3
                                                                         1
                     2 Ω ( −  ) + 1 Ω  = 2 V                  1 Ω  + 1 Ω .1,2 A  = 2 
                                          3
                                1
                           3
                                                                         1
                     2 Ω  − 2 Ω  + 1 Ω  = 2 V                  = 2  − (1 Ω .1,2 A) = 2 − 1,2 
                                   1
                                            3
                          3
                                                                     1
                     −2 Ω  + 3 Ω  = 2 V … … … . (3)              = ,  
                                     3
                            1
                                                                     
                                                                     +  = 
                                                  Jadi:              1    2    3
                                                   = ,      0,8  +  = 1,2 A
                                                                             2
                                                   
                                                   = ,       = 1,2 A − 0,8 A
                                                                     2
                                                   
                                                   = ,        = ,  
                                                   
                                                                     



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