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+// chapter 8
+// example 8.16
+// Determine duty cycle of chopper
+// page-480
+clear;
+clc;
+// given
+Edc=200; // in V (dc source)
+R=0.1; // in ohm
+L=10; // in mH
+Eb1=150, Eb2=-110; // in V (back emf)
+I0=10; // in A
+// calculate
+L=L*1E-3; // changing unit from mH to H
+// since I0=(E0-Eb)/R therefore we get,
+E01=I0*R+Eb1; // calculation of average load voltage when Eb=150 V
+E02=I0*R+Eb2; // calculation of average load voltage when Eb=-110 V
+// since E0=2*Edc(alpha-0.5), therefore we get
+alpha1=(E01/(2*Edc))+0.5; // calculation of duty cycle when Eb=150 V
+alpha2=(E02/(2*Edc))+0.5; // calculation of duty cycle when Eb=-110 V
+printf("\nThe duty cycle when Eb=%.f V is \t %.3f ",Eb1,alpha1);
+if alpha1>0.5 then
+    printf("\t\t It is forwarding mode");
+else
+    printf("\t\t It is reversing motoring mode");
+end
+printf("\nThe duty cycle when Eb=%.f V is \t %.3f ",Eb2,alpha2);
+if alpha2>0.5 then
+    printf("\t\t It is forwarding mode");
+else
+    printf("\t\t It is reversing motoring mode");
+end
+// Note : the answer vary slightly due to precise calculation
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