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
Power factor for purely inductive circuit is 0 and power factor angle is 90 degree. Power factor for purely capacitive circuit is 0 and power factor angle is 90 degree Power factor for purely resistive circuit is 1 and power factor angle is 0 degree
EXAMIANS