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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 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);