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//===========================================================================
//chapter 6 example 25
clc;
clear all;
//variable declaration
KT = 201; //turn ration
Ie = 3; //cross loss current in A
Im = 7; //magnetising component of exciting current in A
delta =0;
//calculations
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))/Is); //actual current in A
e = ((Kn-Kc)/Kc)*100; //ratio error
b =(180/%pi)*(((Im*z)-(Ie*z1))/(KT*Is));
//result
mprintf("ratio error = %3.3f percentage",e);
mprintf("\nphase angle error = %3.4f °",b);
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