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+clc,clear
+printf('Example 5.19\n\n')
+
+VA=2000*1000
+V_L=11000,V_ph=V_L/sqrt(3)
+R_a=0.3,X_s=5 //armature resistance and synchronous reactance
+
+//case (i)
+phi=acos(0.8) //lagging
+I_L=VA/(sqrt(3)*V_L) , I_a=I_L
+E_ph = sqrt((V_ph*cos(phi)+I_a*R_a)^2 +(V_ph*sin(phi)+ I_a*X_s)^2)
+
+//Using E_ph = sqrt((V_ph*cos(phi)+I_aph*R_a)^2 +(V_ph*sin(phi)+ I_aph*X_s)^2)
+//we get V_ph^2 -579.4455 V_ph -44653301.91=0
+p=[1 -579.4455 -44653301.91]
+roots(p)
+V_ph=ans(1) //second root is ignored as its -ve
+printf('Terminal voltage is %.4f V',V_ph)