Applied Electrotechnology for Engineers by C.H. Laycock

By C.H. Laycock

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26b the sum of the three phase voltages VRN , VY N and VBN is clearly zero. 27a. Before connection to the load there will be no circulating currents in the generator coils because, being connected end-to-end, we have effectively summed the three voltages. In this delta (~) connection the line voltages and phase voltages are obviously equal VL = Vp For a balanced load R 1 = R 2 = R 3 and therefore the phasors representing the load currents 1RuIYL and IBL will be equal in magnitude and in phase with VR N , VY N and VBN respectively.

6. Because L and C dissipate no power, all the power must be dissipated in R. c . 11a, the current may be resolved into two components as shown. The component I cos <1> in phase with the voltage is called the active current whereas the quadrature component I sin <1> is called the reactive current. 11a is multiplied by V. 11a, the component VI cos <1> in phase with the voltage is clearly the power (in watts) whereas the quadrature component VI sin <1> is called the reactive power Q the unit for which is the reactive volt ampere (VAr) .

The vertical phasors being of equal length cancel exactly leaving V = fR. 15b that is, in phase and therefore the circuit is at resonance. J (R 2 ) =R at resonance. l5c. J (LC) while the circuit's impedance at this resonant frequency is ZD = R and is called the dynamic impedance. 2 showed that under certain circumstances the capacitor and inductor voltages for such a circuit could exceed the supply voltage. At resonance , in this example, the capacitor and inductor voltages Ve and VL are equal because since XL = Xe IXL = IXe The ratio of the reactance voltages to the supply voltage is thus VL/Vor VL/VR since V = VR .

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