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//============================================================================
//Chapter 6 Example 28
clc;
clear all;
//variable declaration
Vs = 100+0*%i; //secondary terminal voltage in V
Rp = 97.5; //primary resistance in Ω
Xp = 67.4; //primary reactance in Ω
X1 = 110; // total equivalent reactance in Ω
K =1000/100;
//calculations
//Es = Vs+(Is*(Rs+Xs*%i);
Es = Vs;
Ep = 10*(100+0*%i); //induced emf in primary winding in V
I0 = 0.02*(0.4-0.9165*%i); //no load current in A
Zp = Rp+Xp*%i;
Vd = I0*Zp;
Vp = Ep+Vd;
beta = (atan((imag(Vp))/real(Vp)))*180/%pi; //phase angle between primary and secondary voltage in °
Xs1 = X1-Xp; //reactance of secondary winding in Ω
//Es = Vs+(Is*Zs); //induced emf in secondary winding
//IP = (Is/10)+I0;
//V = Ip*Zp = (IS/10)+0.008-0.01833*i
//V = (9.75*Is)+2.015)-((1.2478-6.74*Is)*%i).....equation 1
//Vp = K*(ES+IP*ZP)
//VP =(1002.015+18.35*%i)-(1.2478-11*Is)*%i....equation 2
//comparing equation 1 and 2 we get
//1.2478-11*Is =0;
Is = 1.2478/11; //secondary current in A
v = Vs*Is;
//result
mprintf("phase angle between primary and secondary voltage = %3.2f ° lagging",beta);
mprintf("\nvolt ampere rating for zero phase angle = %3.2f",v);
mprintf("\nnote:Is values is taken as 0.114 wchich is approximate when answer is 0.1134");
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