neutral voltage with respect to earth. (v) Find the phase voltages and line voltages
across the load by symmetrical components. (vi) Find the total power delivered by
source and delivered to load using symmetrical components.
SOLUTION
(i) The subscript ‘s’ stands for source quantities. Substituting
a 1∠120º, V
sRN
230∠0º, V
sYN
190∠–120º and V
sBN
240∠140º,
(ii) Line voltages are obtained as below.
V
sRY
V
sRG
– V
sYG
230∠0º – 190∠–120º
325 j164.54 364.28∠26.85º V rms.
V
sYB
V
sYG
– V
sBG
190∠–120º – 240∠140º
88.85 – j318.81 330.96∠–74.43º V rms.
V
sBR
V
sBG
– V
sRG
240∠140º – 230∠0º
–413.85 j154.27 441.67
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