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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
commit7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch)
treedbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /3835/CH9/EX9.8/Ex9_8.sce
parentb1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff)
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+clear
+//
+//given
+i=35
+v=220
+ra=0.15
+n1=1600
+//when motor is running at 1200rpm the back emf eb1 is given by eb1=v-(35*0.15)
+eb1=214.75
+//flux phy1 is proportional to armature current ia.Thus, at ia1=35 and ia2=15 n is proportional to eb/phy
+//2=(eb2*phy1)/(phy2*eb1)
+//therefore
+eb2=184.07
+//case a
+//resistance to be connected in series is rse ohm
+ia2=15
+rse=((v-eb2)/ia2)-ra
+printf("\n rse= %0.1f ohm",rse)
+//case b
+eb2=0.5*1.15*214.75
+ia2=50
+rse=((v-eb2)/ia2)-ra
+phy1=35
+eb2=220-50*0.15
+n2=(n1*eb2*phy1)/(1.15*phy1*eb1)
+printf("\n n2= %0.1f rpm",n2)