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+//Calculate the most economical cross sectional area
+clear;
+clc;
+//soltion
+//given
+id=0.1;//interest & depreciation charges
+V=20000;//volts//supply voltage
+d=1.72*10^-6;//ohm cm//specific resistance
+cst=.6;//rs// cost of energy per unit
+p1=1500//kilowatts
+t1=8//hours
+pf1=.8//power factor
+p2=1000//kilowatts
+t2=10//hours
+pf2=.9//power factor
+p3=100//kilowatts
+t3=6//hours
+pf3=1//power factor
+t=365//no. of days
+i1=p1*1000/(sqrt(3)*V*pf1);//ampere//current at time t1
+i2=p2*1000/(sqrt(3)*V*pf2);//ampere//current at time t2
+i3=p3*1000/(sqrt(3)*V*pf3);//ampere//current at time t3
+R=d*100*1000;//ohm
+P2=8000*id;//Loss load factor
+printf("Annual cost of cable is Rs(80000a + 20000)per km\n");
+printf("Capital cost= P2*a= %d*a*l \n", P2);
+P3=(3*((i1^2*t1)+(i2^2*t2)+(i3^2*t3))*R*t*cst)/1000;//kWh//annual cost of energy loss
+printf("Energy loss per annum= (P3*l)/a= (%f*l)/a \n",P3);
+a=sqrt(P3/P2);
+printf("Economic cross section of conductor is= √(P3/P2)= %f square cm \n",a);