Vab=25/15°V,
Ic = 10275⁰ A
Find the total real power consumed by the load. Note: read the subscripts of the voltage and
current carefully.
Ve
Vc:
5 points
P30: 5 points
25
√3
P36 = 3.
Vab
√3
P30 = 3|vc||1c| cos
25
√3
25
√3
00v. 01 105° - 75° = 30°
€490°
<15° +90°
10.cos(30°) =
-Z105°
= 375 W consumed by the load.
Fig: 1