Illustrate balanced delta-connected load, Electrical Engineering

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Next, let us consider the case of a balanced delta-connected load with impedance of 5 45° supplied by a three-phase, three-wire 100-V system, as shown in Figure (a). We shall determine the line currents and draw the corresponding phasor diagram. With the assumed positive phase sequence (A-B-C) and with ¯VBC as the reference phasor, the line-to-line voltages ¯VAB, ¯VBC, and ¯VCA are shown in Figure. The rms value of the line-to-line voltages is 100 V for our example. Choosing the positive directions of the line and phase currents as in Figure(a), we have

2166_Illustrate Balanced Delta-Connected Load.png

By the application of Kirchhoff's current law at each of the vertices of the delta-connected load, we obtain

858_Illustrate Balanced Delta-Connected Load1.png

2349_Illustrate Balanced Delta-Connected Load2.png

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The phasor diagram showing the line-to-line voltages, phase currents, and line currents is drawn in Figure(b). The load power factor is lagging, and is given by cos 45°. For a balanced delta-connected load, the phase voltages and the line-to-line voltages are equal in magnitude, and the line voltages are in phase with the corresponding phase voltages.

1261_Illustrate Balanced Delta-Connected Load4.png

The line currents, in magnitude, are √3 times the phase currents, and the phase currents lead the corresponding line currents by 30°. The preceding example can also be solved by the one-line equivalent method for which the delta-connected load is replaced by its equivalent wye-connected load. The single-line equivalent circuit is shown in Figure(c). The details are left as an exercise for the student.


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