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authorprashantsinalkar2018-02-03 11:01:52 +0530
committerprashantsinalkar2018-02-03 11:01:52 +0530
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+//===========================================================================
+//chapter 6 example 26
+clc;
+clear all;
+
+//variable declaration
+KT = 199; //turn ration
+Is = 5; //secondary current in A
+Im = 7; //magnetising component of exciting current in A
+Ie = 4; //cross-loss component of exciting current in A
+delta =0;
+
+//calculations
+KN = 1000/5 //nominal ratio
+alpha =atan(Ie/Im) //angle in °
+alpha1 = (alpha*180)/%pi
+theta = delta+(((acos(0.8))*180)/%pi) //from figure taken the value of gamma
+z = cos((theta*%pi)/180)
+z1 = sin(((theta)*%pi)/180)
+Kc = KT+(((Ie*z)+(Im*z1))/Ie) //actual current in A
+e = ((Kn-Kc)/Kc)*100 //ratio error
+b =(180/%pi)*(((Im*z)-(Ie*z1))/(KT*Is))
+theta1 = delta-(((acos(0.8))*180)/%pi) //from figure taken the value of gamma
+z11 = cos((theta1*%pi)/180)
+z12 = sin(((theta1*%pi)/180))
+Kc1 = KT+(((Ie*z11)+(Im*z12))/Is) //actual current in A
+e1 = ((Kn-Kc1)/Kc1)*100 //ratio error
+b1 =(180/%pi)*(((Im*z11)-(Ie*z12))/(KT*Is))
+
+
+//result
+mprintf("ratio error = %3.2f percentage",e);
+mprintf("\nphase angle error = %3.1f °",b);
+mprintf("\nratio error = %3.2f percentage",e1);
+mprintf("\nphase angle error = %3.2f percentage°",b1);