The ratio of full load current to short circuit current = 1/5 Xsc = j/(1/5) External reactance required = j*((1/5) - 0.05)) = j*0.15 pu Full load current = (30*1000)/(√3*11) = 1574.6 A Per unit reactance = j*0.15 = (I*Xr)/V j*0.15 = (1574.6*Xb)/((11/√3)*1000)) = 0.60 ohm
Voltage sag: It is usually associated with system faults but can also be caused by energization of heavy loads or starting of large motors. Voltage sag is the short reduction in the RMS voltage between 0.1 to 0.9 pu for a duration of 0.5 cycle to 1 minute
EXAMIANS