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+clc;
+clear;
+W1=500;
+R1=0.010;//Resistance
+XL1=0.05;//leakage reactance
+W2=750;
+disp('when both secondary voltages are 400V:')
+pf=0.8;//lag pf with 250KVA
+W3=250;
+R2=0.015;//Resistance value
+XL2=0.04;//Reactance value
+Z1=(R1+((XL1)*%i));
+Z2=(R2+((XL2)*%i));
+Z=Z1+Z2;
+disp(Z1,'The per unit impedance for common base value 500 KVA:')
+disp(Z2)
+disp(Z)
+theta=acos(0.8);
+S=W2*(pf-(sin(theta)*%i));
+S1=S*(Z2/Z);
+S2=S*(Z1/Z);
+SA=real(S1)+real(S2);//Real parts of the calculated power
+disp(SA,'The total active power is :')
+SR=W2*(sin(acos(0.8)));
+disp('When the open circuit secondary voltages are respectively 405 and 415')
+R3=0.0032;
+R4=0.0096;
+XL3=0.0160;
+XL4=0.0256;
+Z3=R3+((XL3)*%i);
+Z4=R4+((XL4)*%i);
+Z5=0.166+(0.125*%i);//Impedance value for the assured voltage 395V
+E1=405+(0*%i);
+E2=415+(0*%i);
+Ez=(E1/Z3)+(E2/Z4);
+Zo=(Z5*Z3*Z4)/((Z3*Z4)+(Z5*Z4)+(Z5*Z3));
+V=(Ez*Zo);
+disp(V,'The secondary terminal voltage is :')
+Vi1=E1-V;
+disp(Vi1,'The internal volt drop in the first transformer:')
+Vi2=E2-V;
+disp(Vi2,'The internal volt drop in the second transformer is :')
+I1=Vi1/Z3;
+I2=Vi2/Z4;
+S3=(340-(220*%i));
+S4=(270-(220*%i));
+S5=S1+S2;
+disp(S5,'The combined load is :')