pleas explane Consider a balanced 3 phi Wye-Wye system with line voltage VL = 20

Question

pleas explane

Consider a balanced 3 phi Wye-Wye system with line voltage VL = 208 60 degree V rms. If the load impedance is ZL = 10 + j6Ohm, determine the line current and total power absorbed by the load. Suppose that the same sized capacitor is placed in parallel with each load ZL of the previous problem. What value of capacitance is needed to increase the power factor of the load/capacitor combination to 0.95? Now determine the total power absorbed by the load. A Delta balanced 3 phi source, with source impedance Z Delta = 0.6 + j0.3Ohm is connected to a Wye load with impedance ZL = 12 + j7Ohm. If the line current in line I-aA = 9.6 - 20 degree Arms, determine the line voltage, phase voltages at the source, and total power absorbed by the load.

Explanation / Answer

For three phase star connected loads-

line current=Phase current

line current=V(phase)/load

and power absorbed by load=i(square)*r...

in this case r=10

when we need to increase power factor,we take voltage of capacitor such that net voltage makes an angl of cos inverse .95 with the current(along r)

FOR POWER-we always calculate real power,hence always do square of current in the part*resistance of the part..

plz ask if u have doubts

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