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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
commit7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch)
treedbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /3835/CH5/EX5.2/Ex5_2.sce
parentb1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff)
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+clear
+//
+//case a
+vab=400 //phase angle of 0
+vbc=400 //phase angle of 120
+vca=400 //phase angle of 240
+//the phase currents are given by iab,ibc,ica
+iab=400/150 //from the diagram
+printf("\n iab= %0.5f A",iab)
+//ibc=(400*314*50)/10**6 numerator with an angle of -120 and denominator angle of -90 which amounts to -30 in numerator
+//this leads to simplifying with the formula as the value obtained for ibc after simplification from above mutiplied by values of cos(-30)+jsin(-30)
+//therefore print as below
+printf("\n ibc=5.4414-j3.1416 A")
+//same method for ica
+printf("\n ica=3.1463+j4.2056 A")
+//case b
+//ia=iab-ica
+//ia=2.667-(3.1463+j4.2056)
+//leads to 4.2328 with an angle of -96.51
+//angle calculated using tan formula
+printf("\n ia=4.2328 with an angle of -96.51 A")
+//same for ib and ic
+printf("\n ib=4.1915 with angle of -48.55 A")
+printf("\n ic=7.6973 with an angle of 107.35 A")