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+//Book Name:Fundamentals of Electrical Engineering
+//Author:Rajendra Prasad
+//Publisher: PHI Learning Private Limited
+//Edition:Third ,2014
+
+//Ex10_5.sce
+
+clc;
+clear;
+f=50;
+N=285;
+Ns=300;  //which is near the value of N as slip lies b/w 0.03 to 0.05
+
+printf("\n (a)")
+p=(120*f)/Ns;
+printf("\n  Number of poles=%d \n",p)
+
+printf("\n (b)")
+s=(Ns-N)/Ns;
+s_percentage=s*100;
+printf("\n  Slip at full load=%d percentage \n",s_percentage)
+
+printf("\n (c)")
+//slip is proportional to rotor resistance
+s=2*s_percentage;
+printf("\n  Slip at full load if rotor resistance is doubled=%d percentage \n",s)
+
+printf("\n (d)")
+//copper loss=I^2*R; so copper loss doubles if rotor resistance doubles
+Pcu=280;
+Pcu_new=2*Pcu;
+printf("\n  The new value of rotor copper loss=%d watt \n",Pcu_new)