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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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parentb1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff)
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+clear
+//
+//case a
+vl=11000
+il=50
+pf=0.85 //powerfactor
+p=vl*il*pf
+printf("\n Power supplied to the motor is= %0.5f kW",p)
+//case b
+vt=6350.85 //at angle 0
+zs=25.02 //at angle 0
+//subcase 1 powerfactor at 0.85 lag
+//e=vt-ia*zs
+//e=6350.85-50(at angle -31.79)*25.02(at angle 87.71)
+//substituting and solving as in x+iy form we get 5744.08 at angle -10.39 as the value of e
+printf("\n emf induced=5744.08 at angle -10.39")
+//subcase 2
+//for a 0.85 lead same process as above is followed except angles are considered positive due to lead
+printf("\n emf induced=7051.44 at angle -8.88")