EE 303 Quiz 2, Spring 2008, Dr. McCalley Name:______SOLUTIONS______
Time for quiz: 20 minutes. Closed book, closed notes. You may only have a writing instrument. No calculators or laptops allowed. No other paper allowed. Show all work on the page below or on the back.
Three balanced three-phase loads are connected in parallel. Load 1 is Y-connected with an impedance of Z1=400 ; load 2 is D-connected with an impedance of Z2=-j1800 ; and load 3 is Y-connected and consumes a complex three-phase power denoted by S3. The loads are directly connected to a Y-connected generator (with positive, or a-b-c, phase sequence) having line-to-line voltage magnitude of volts. With the generator a-phase line-to-neutral voltage as reference (angle 0°), the current in the a-phase of the generator is amperes.
a) (40pts) Identify numerical values of the following phasors (magnitude and angle), voltages in volts and currents in amps:
Currents in b and c-phases of generatorà
b) (24pts) What is the power factor (give numerical value, and indicate leading, lagging, or unity)…
i) seen by the generator? _____1.0 (or unity)_____
ii) of load number 1? _____1.0 (or unity)_____
iii) of load number 2? _____0, leading______
iv) of load number 3? _____0, lagging______
c) (8pts) What is the total complex power supplied by the source? Be sure to indicate units.
; alternatively, we could compute
d) (8pts) What are the current in phase a of load 1, i.e., (magnitude and angle)?
amperes
e) (10pts) What is the current in phase a-b of load 2, i.e., (magnitude and angle)?
First, we convert load 2 to a Y through ZY=ZD/3=-j1800/3=-j600W. Then we obtain the current in phase a of this Y-connected load according to: amperes. Now the current in the a-phase of a Y-connected load is the same as the current in the line supplying this Y-connected load. So the current just computed is also the line current. Now, recalling that for delta-connected loads, the line current magnitudes are Ö3 times the phase current magnitudes, and the line currents LAG the phase currents by 30° (or phase currents LEAD the line currents by 30°), then we have that .
f) (10pts) What is the current in phase a of load 3, i.e., (magnitude and angle)?
Note that the source current is the same as the current to load number 1. The implication of this is that the source does not at all “see” loads numbers 2 and 3. Yet load 2 draws a current of amperes. The only way this can occur is if the load 3 current exactly cancels the load 2 current. This would be the case if .