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+// Calculating the maximum flux and no load current of the transformer
+clc;
+disp('Example 5.18, Page No. = 5.99')
+// Given Data
+Ep = 400;// Primary winding voltage (in volts)
+f = 50;// Frequency (in Hz)
+A = 2.5*10^(-3);// Area of cross section (in meter square)
+Sf = 0.9;// Stacking factor
+Tp = 800;// Primary winding turns
+li = 2.5;// Length of the flux path (in meter)
+u0 = 4*%pi*10^(-7);// Permeability of free space
+ur = 1000;// Relative ermeability
+D = 7.8*10^(3);// Density of iron (in kg per meter cube)
+FD_w = 2.6;// Working flux density (in W per kg)
+// Calculation of the maximum flux
+Ai = Sf*A;// Net iron area (in meter square)
+Bm = Ep/(4.44*f*Ai*Tp);// Maximum flux density of core (in Wb per meter square)
+Fm = Bm*Ai;// Maximum flux in the core (in Wb)
+disp(Fm,'Maximum flux in the core (Wb)=');
+// Calculation of the no load current
+AT0 = li/(ur*u0)*Bm;// Magnetic mmf (in A)
+Im = AT0/(2^(1/2)*Tp);// Magnetising current (in A)
+V = Ai*li;// Volume of the core (in meter cube)
+W = V*D;// Weight of core (in kg)
+Pi = W*FD_w;// Iron loss (in W)
+Il = Pi/Ep;// Loss component of no load current (in A)
+I0 =(Im*Im+Il*Il)^(1/2);// No load current (in A)
+disp(I0,'No load current (Ampere)=');
+//in book answers are 0.00225 (Wb) and 1.77 (Ampere) respectively.  The answers vary due to round off error