//Swinburne test on a dc shunt motor clc; clear; V=500; I=5; Rf=250; Ra=0.5; P=V*I; If=V/Rf; Ia=I-If; Pfc=(If^2)*Rf;// Field Copper Loss Pac=(Ia^2)*Ra; // Armature Copper Loss Pil=P-Pfc-Pac;// Iron loss // Generator Vg=500; Ig=100; Pog=Vg*Ig; // Power Output Iag=Ig+If; //Armature current Pgac=(Iag^2)*Ra; // Armature Copper loss slg=0.01*Pog;//stray loss Pgtl=Pgac+Pfc+slg+Pil; // Total losses effg=Pog*100/(Pog+Pgtl); // Motor Vm=500; Im=100; Pim=Vm*Im; // Power input to the motor Iam=Ig-If; // Armature current Pmac=(Iam^2)*Ra; // Armature Copper Loss Pom=Pim-Pmac-Pil-Pfc;// Ouput of the motor slm=0.01*Pom;// Stray loss Pmtl=Pmac+Pil+Pfc+slm; // Total loss of the motor effm=(Pom-slm)*100/(Pim); printf('i) The Efficiency of the machine as a generator delivering 100A at 500V = %g percent \n',effg) printf('ii) The Efficiency of the machine as a motor having a line current 100A at 500V = %g percent \n',effm)