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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+//To find current in the 600 turn exciting coil
+
+clc;
+clear;
+
+N=600;
+mr=800;
+m0=4*%pi*(10^-7);
+
+phi=100*(10^-6);// Flux in air gap
+
+l1=10*(10^-2);
+l2=18*(10^-2);
+lg=2*(10^-3); // Air gap length
+Ac=(6.25)*(10^-4);// Central limb area
+As=3*(10^-4);// Side limb area
+
+Ra=lg/(m0*Ac);
+Ri=l1/(mr*m0*Ac);
+
+R=l2/(m0*mr*As);
+
+Rt=Ra+Ri; // Total reluctacne of the central limb
+
+AT1=Rt*phi; // MMF or Ampere turns for the central limb
+AT2=R*phi/2; // MMF, Two identical limbs hence flux becomes half and only one limb is considered
+
+AT=AT1+AT2;//Total MMF
+
+I=AT/N; // Current in the 600 turns
+
+printf('The current flowing in the 600 turns exciting coil = %g A\n',I)