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//===========================================================================
//chapter 6 example 19
clc;
clear all;
//variable declraration
Ts = 200; //number of turns in secondary winding
Tp = 1; //number of turns in primary winding
Is = 5; //current in A
Zs = (1.2+0.2)+(%i*(0.5+0.3)); //secondary impedance Ω
MMF = 100;
Pi = 1.2; //iron loss in watts
Ie = 50; //energy component of eddy current in A
//calculations
KT =Ts/Tp //turn ratio
//Es = Is*Zs //secondary voltage in volts
Im =MMF/Tp //magnetising current in A
I0 = Im+%i*Ie //exciting current on primary side in A
I01 =sqrt(((real(I0))^2)+((imag(I0))^2))
alpha = atan(Ie/Im)
alpha1 = (alpha*180)/%pi
theta = atan(imag(Zs)/real(Zs))
theta1 = (theta*180)/%pi
Ip = (KT*Is)+(I01*sin(theta+alpha)) //primary current in A
e = ((-I01*sin(((theta1+alpha1)*%pi)/180))/Ip)*100 //ratio error
N = (I01*sin(((theta1+alpha1)*%pi)/180))/Is //number of secondary turns to be reduced
//result
mprintf("ratio error = %3.1f percentage",e);
mprintf("\nnumber of secondary turns to be reduced = %3.0f ",N);
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