d
                                                                      L M
                                         Flux density 
                                            B [T]
                                                                        N      A
                                    l
                                                                        
                                                      Current induces in the coil as the 
         Generator for single phase                   coil moves in the magnetic field
         Note
         Induction motor cannot start 
         by itself. This problem is 
         solved by introducing three 
         phase system                                    Current produced at terminal
                                                  
      Instead of using one coil only , three coils are used arranged in one 
                                   o
      axis with orientation of 120  each other. The coils are R-R  , Y-Y  
                                                                  1      1
      and B-B . The phases are measured in this sequence R-Y-B.  I.e Y 
               1
      lags R by 120o , B lags Y by 120o. 
                                         
                                                       Finish             R
                                                                                    e                               L 1
                                                                                       R
                                                         start          R 1
                                                       Finish            Y
                                                                                    e                               L 2                         Load
                                                                                       Y
                                                         start          Y 1
                                                       Finish            B
                                                                                    e B                             L 3
                                                       start            B
                                                                            1
              The three winding can be represented by the above circuit. In this 
              case we have six wires. The emf are represented by e  , e , e . 
                                                                                                                                                                                   R            Y         B
                           eR Em sint
                             e E sin(t-120)
                                 Y                   m
                           e E sin(t-240)
                               B                    m
                                                                                                                        
                The circuit can be simplified as follows, where R  can be 
                                                                                                                                                                                            1
                connected to Y  and Y1 can be connected to B. In this case 
                the circuit is reduced to 4 wires.
                                                                                                                                                                    Finish            R
           eRB1 eR eY eB                                                                                                                                                                     e R
                                                                                                                                                                     start           R 1
               E [sint sin(t  120)sin(t  240)]                                                                                                           Finish            Y
                          m                                                                                                                                                                     e                             e      +e        +e
                                                                                                                                                                                                    Y                             R         Y          B
                                                                                                                                                                     start          Y 1
               E [sintsintcos120  costsin120                                                                                                             Finish             B
                          m
                                                                                                                                                                                                e B
                                           sintcos240  costsin240]                                                                                          start            B
                                                                                                                                                                                        1
                 Em[sint 0.5sint 0.866cost 0.5sint0.866kost]
                  0
          Since the total emf is zero, R and B1 can be connected together, thus 
          we arrive with delta connection system.
                                                                                                                                       
                       R                                Fig. B
        Fig.A
                PL
                       R 1
                       Y
                                              e  s
                                              n  r
                                              i  o
                PM                            L  t
                                                 c
                                                 u
                                                 d
                                                 n
                                                 o
                      Y                          c
                        1
                      B
                PN
                       B 1                                                    Fig. C
      Since the total emf is zero, R and B  can be 
    •                                                  1
                                                                             R     Y          e Y
    connected together as in Fig.A , thus we arrive                                  1
    with delta connection system as in Fig. C.                          e                     Y
    •                                                                     R
      The direction of the emf can be referred to the                                             B
    emf waveform as in Fig. B where PL is +ve (R -                          R 1                     1
                                                                     1
    R), PM is –ve  (Y-Y ) and PN is –ve (B-B ).                                       B        e B
                                1                             1