Thank you Two three-phase wye-connected loads are in parallel across a three-pha

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Thank you

Two three-phase wye-connected loads are in parallel across a three-phase supply. The first load draws a phase current of 20A at 0.9 power factor leading, and the second load draws a phase current of 30A at 0.8 power factor lagging. Calculate the following: The transmission line current The power factor of the load The real power supplied by the source if the line-to-line voltage of the transmission line is 400V. The line-to-line voltage of a three-phase transmission line is bc = 340 20degreeV Calculate the phase voltage an If a wye load impedance of = 10 60degreeOhmis connected to the transmission line, calculate the transmission line current b Calculate the current in the neutral line n The following are the voltage and current measured for a wye-connected load: ab = 200 50degreeV c =10 140degreeA Calculate the power factor angle Calculate the real power consumed by the load

Explanation / Answer

8.6))

a) |I| = sqrt( (20*0.9 + 30*0.8)^2 + (20*sqrt(1-0.9^2) - 30*sqrt(1-0.8^2))^2 )

= 43 A

b) 20*sqrt(1-0.9^2) < 30*sqrt(1-0.8^2)

=> lagging current dominates

pf = (20*0.9 + 30*0.8)/43 = 0.9767 lagging

c) P = sqrt(3)*|V|*|I|*pf

P = P1 + P2 = sqrt(3)*400*20*0.9 + sqrt(3)*400*30*0.8 = 29098.45 W

8.7))

a) Van = 340/sqrt(3) ANGLE (20 + 90) = 196.3 ANGLE (110)

b) Ia = Van/Z = 196.3/10 ANGLE (110-60) = 19.63 ANGLE(50)

Ib = 19.63 ANGLE(50 - 120) = 19.63 ANGLE(-70)

c) In = 0 since balanced 3-phase load

8.9))

a) Vcn = 200/sqrt(3) ANGLE (50 + 90) = 115.47 ANGLE(140)

pf angle = 140 - 140 = 0

b) pf = cos(0) = 1

power = sqrt(3)*200*10*1 = 3464.1 W

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