Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Suppose that the load in the problem below is split into two--one Delta connecte

ID: 2079282 • Letter: S

Question

Suppose that the load in the problem below is split into two--one Delta connected load and another Y connected load--whose impedances consume equal lone currents.

not b and c, require separate phasor diagrams for the two loads.

The voltage and current of a balanced three-phase source is Vbc 480 L180 V and I load is A-connected, answer the following: 5460 A. If the a) (10 points) Draw the circuit schematics and mark the given parameters with labels and values b) (10 points) All the line and phase currents of the load; draw the phasor diagram for these currents. c) (10 points) All the line and phase voltages of the load; draw the phasor diagram for these voltages. d) (5 points) State whether the load is inductive or capacitive e) (10 points) Total real and reactive power consumed by the load f) (10 points) Total complex power and apparent power consumed by the load

Explanation / Answer

a. Vbc=480<1800 so Vab=480<3000, Vca=480<600

Ia= (Vab+Vac)/Z = 5<600

so Z=(Vab+Vac) / Ia =(480<3000-480<600) / 5<600 =(-144-j83.13)

b. Ia=5<60 , Ic=5<180 , Ib=5<300

Ibc=Vbc/Z=2.5-j1.44, =2.8<-30

Iab=-0.00012-j2.88=2.88<-90

Ica=-2.49-1.44j=2.88<-150

c. phase voltage

Vbc=480<1800 so Vab=480<3000, Vca=480<600

line voltage are

Vbc=480<1800 so Vab=480<3000, Vca=480<600

Vcb=480<-1800 so Vba=480<-3000, Vac=480<-600

d. since Z as postive imaginary part its inductive load

e.total real power=Iab*Vab cos(angle between them)=1200W

total reactive power=Iab*Vab sin(angle between them)=691VAR

f. total complex power=1200W-j691VAR

total apperent power=1384VA<-30

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote