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+//CHAPTER 8- DIRECT CURRENT MACHINES
+//Example 37
+
+disp("CHAPTER 8");
+disp("EXAMPLE 37");
+
+//VARIABLE INITIALIZATION
+p_o=50*1000;                //in Watts
+v_t=250;                    //in Volts
+loss1=5000;                 //total core loss in Watts
+loss2=2000;                 //total core loss in Watts (when speed is reduced to half)
+speed=125/100;
+
+//SOLUTION
+
+//solution (a)
+
+//W_h=A*N, where W_h=hysteresis loss, A=constant and N=speed
+//W_e=B*(N^2), where W_e=eddy current loss, B=constant and N=speed
+//W_h+(W_e^2)=loss1 =>W_h+W_e=5000
+//(W_h/2)+(W_e/4)=loss2 =>(0.5*W_h)+(0.25*W_e)=2000 (when speed reduces to half)
+//So, we get two equations
+//W_h+W_e=5000.......................eq(i)
+//(0.5*W_h)+(0.25*W_e)=2000..........eq(ii)
+//solving the equations by matrix method
+A=[1 1;0.5 0.25];
+b=[5000;2000];
+x=inv(A)*b;
+W_h1=x(1,:);                //to access the 1st row of 2X1 matrix 
+W_e1=x(2,:);                //to access the 2nd row of 2X1 matrix 
+disp("Solution (a)");
+disp(sprintf("The hysteresis loss at full speed is %d W",W_h1));
+disp(sprintf("The eddy current loss at full speed is %d W",W_e1));
+
+//solution (b)
+W_h2=speed*W_h1;
+W_e2=(speed^2)*W_e1;
+disp("Solution (b)");
+disp(sprintf("The hysteresis loss at 125%% of the full speed is %d W",W_h2));
+disp(sprintf("The eddy current loss at 125%% of the full speed is %d W",W_e2));
+
+//END 
+