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//Chapter 21, Problem 14
clc;
n1=600; //primary turns
n2=150; //secondary turns
r1=0.25; //primary resistance
r2=0.01; //secondary resistance
x1=1; //leakage reactance
x2=0.04;
re=r1+r2*(n1/n2)^2; //equivalent resistance
xe=x1+x2*(n1/n2)^2; //equivalent reactance
ze=sqrt(re^2+xe^2); //equivalent impedance
phie=acos(re/ze); //phase angle of the impedance
printf("(a) Equivalent resistance = %.2f ohms\n\n",re);
printf("(b) Equivalent reactance = %.2f ohms\n\n",xe);
printf("(c) Equivalent impedance = %.2f ohms\n\n",ze);
printf("(d) Phase angle of the impedance = %.2f deg",phie*180/%pi);
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