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+clc
+P=5e+5//rated power output in Watts
+P1=25e+4// power at half rated torque
+f=50//frequency in Hz
+If=10//rated firld current in amp
+Xs=10//reactance in ohm
+p=4//no.of poles
+Vl=33e+2//line voltage in volts
+N1=1500
+//Solution
+Vph=Vl/sqrt(3)//phase voltage in volts
+Is=P/(sqrt(3)*Vl*0.8)//Current in amp
+theta1=acosd(0.8)
+E=Vph-(-%i*Xs*(Is*(%i*sind(theta1)+cosd(theta1))))
+y=imag(E)
+x=real(E)
+Er=sqrt((y^2)+(x^2))
+theta2=atand(y/x)
+Ia=Is
+E2=Vph+(%i*Ia*Xs)
+y2=imag(E2)
+x2=real(E2)
+Er2=sqrt((y2^2)+(x2^2))
+theta3=atand(y2/x2)
+P=3*Vph*Er2*sind(-theta3)/Xs
+Wms=2*%pi*N1/f
+T=P/Wms
+If1=Er2*If/Er
+If2=12
+Er3=Er*If2/If
+P2=-500e+3
+d1=asind(P2*Xs/(3*Vph*Er3))
+Is=(Vph-Er3*(cosd(d1)+(%i*sind(d1))))/(%i*Xs)
+Isr=abs(Is)
+u=imag(Is)
+v=real(Is)
+pf=cosd(atand(u/v))
+printf('\n\n The Breaking Torque for machine operation at rated current and upf=%0.1f N-m\n\n',T)
+printf('\n\n The Field Current for machine operation at rated current and upf=%0.1f Amp\n\n',If1)
+printf('\n\n The Armature Current at 12 A Field Current=%0.1f Amp\n\n',Isr)
+printf('\n\n The power factor at 12 A Field Current=%0.1f lead\n\n',pf)
+//The answers vary due to round off error