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                                                             Power electronic control in electrical systems 41










                      Fig. 2.8 Phasor diagram, resistive load.
















                      Fig. 2.9 Phasor diagram, inductive load.


                      Z ˆ 1 
 but the power factor can be unity, 0.8 lagging, or 0.8 leading. For each of
                      these three cases, the supply voltage E can be adjusted to keep the terminal voltage
                      V ˆ 100 V. For each case we will determine the value of E, the power-factor angle f,
                      the load angle d, the power P, the reactive power Q, and the volt±amperes S.
                        Unity power factor. In Figure 2.8, we have E cos d ˆ V ˆ 100 and E sin d ˆ X s I
                      ˆ 0:1   100/1 ˆ 10V. Therefore E ˆ 100 ‡ j10 ˆ 100:5e j5:71    V. The power-factor angle
                                                                               j0
                      is f ˆ cos  1  (1) ˆ 0, d ˆ 5:71 ,and S ˆ P ‡ jQ ˆ VI ˆ 100   100e ˆ 10kVA, with


                      P ˆ 10kW and Q ˆ 0.
                        Lagging power factor. In Figure 2.9, the current is rotated negatively (i.e. clockwise)
                      to a phase angle of f ˆ cos  1  (0:8) ˆ 36:87 . Although I ˆ 100 A and X s I is still

                      10 V, its new orientation `stretches' the phasor E to a larger magnitude: E ˆ V‡
                      jX s I ˆ (100 ‡ j0) ‡ j0:1   100e  j36:87    ˆ 106:3e j4:32    V. When the power-factor is lagging
                      a higher supply voltage E is needed for the same load voltage. The load angle is d ˆ 4:32

                      and S ˆ VI ˆ 100   100e ‡j36:87  ˆ 8000 ‡ j6000VA. Thus S ˆ 10kVA, P ˆ 8kW
                      and Q ˆ‡6kVAr (absorbed).
                        Leading power factor. The leading power factor angle causes a reduction in the
                      value of E required to keep V constant: E ˆ 100 ‡ j0:1   100e ‡j36:87    ˆ 94:3e  j4:86   V.
                      The load angle is d ˆ 4:86 , and S ˆ 10000e  j36:87    ˆ 8000   j6000; i.e. P ˆ 8kW

                      and Q ˆ 6 kVAr (generated).
                        We have seen that when the load power and current are kept the same, the
                      inductive load with its lagging power factor requires a higher source voltage E, and
                      the capacitive load with its leading power factor requires a lower source voltage.
                      Conversely, if the source voltage E were kept constant, then the inductive load would
                      have a lower terminal voltage V and the capacitive load would have a higher terminal
                      voltage. As an exercise, repeat the calculations for E ˆ 100 V and determine V in
                      each case, assuming that Z ˆ 1 
 with each of the three different power factors.
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