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+
+//solving (i)
+Il=70//current drawn by the motor
+V=200//applied voltage
+Rsh=100//shunt field resistance
+Ish=V/Rsh//shunt field current
+Ia1=Il-Ish
+N1=500//initial speed
+Ra1=.2//armature resistance
+Eb1=200-Ia1*Ra1
+N2=350//reduced speed of motor
+Ia2=Ia1//armature current remains same
+//Eb2=200-68*(R+.2)
+//back emf is proportional to speed
+R=((V-Ia2*Ra1)*N1-Eb1*N2)/(Ia2*N1)
+mprintf("Additional resistance in the armature circuit=%f ohm\n",R)
+//solving (ii)
+Ra2=R+Ra1//armature resistance
+Ia3=35//armature current
+Eb3=V-Ia3*Ra2
+N3=N1*Eb3/Eb1
+mprintf("Speed=%d rpm",N3)
+//answer vary from the textbook due to round off error