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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+clear
+//
+//given
+vl=400 //line voltage
+va=vl/sqrt(3)
+vb=230.94 //angle(-120)
+vc=230.94 //angle(-240)
+//case a
+//the line currents are given by
+ia=12000/230.94 //with angle 0
+ib=10000/230.94 //with angle 120
+ic=8000/230.94 //with angle 240
+printf("\n ia= %0.3f A",ia)
+printf("\n ib= %0.5f A",ib)
+printf("\n ic= %0.5f A",ic)
+//case b
+//IN=ia+ib+ic
+//ia,ib and ic are phase currents hence contain with angles they are in the form sin(angle)+icos(angle)
+//IN=51.96*(sin(0)+i*cos(0))+43.3*(sin(120)+i*cos(120))+34.64*(sin(240)+i*cos(240))
+//IN=51.96+(-21.65+i*37.5)+34.64*(-0.5-i*0.866)
+//12.99+i*7.5 on which the sin+icos=sin**2+cos**2 operation is performed
+//therefore
+IN=15 //at angle 30
+printf("\n IN= %0.1f A",IN)