5. A star connected load characterized by a per-phase resistance of 12Ohm is con

Question

5. A star connected load characterized by a per-phase resistance of 12Ohm is connected in parallel with a second star connected load with a per-phase inductance of 76.4mH. The parallel load combination is energised via a 50Hz 415V line to line voltage. Calculate the line current in the supply lines, the real and reactive power absorbed by the oad, and the power factor of load. Ans: I = 22.32A, P=14.35kW, Q=7.18kVars, p.f.=0.89. 6. For question 5, calculate the capacitors required to be connected in star to return the load power factor to unity. What is the supply current now? Ans: C = 132.7MuF, I = 20A

Explanation / Answer

parallel impedance = 12*(2*3.14*50*76.4m)i/(12+24i) = 9.6+4.8i ohm

line current = (415/sqrt(3)) / (9.6+4.8i) = 22.32A with angle of -26.57o

real power = sqrt(3)*415*22.322*cos(26.57) = 14.35kW

reactive power = sqrt(3)*415*22.322*sin(26.57) = 7.18kVAR

power factor = cos(26.57) = 0.894

reactive power to be added for zero power = 7.18kVAR

V2/Xc = 7.18kVAR

4152/7.18x103 = Xc

Xc = 24 ohm

1/(2*3.14*50*C) = 24

C = 132.77uF

New current = 239.6V/12 ohm= 19.97 = 20A

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